Tag: RD Sharma Class 10 Maths Solutions

RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.2
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.3
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7
• RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise
• RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS
• RD Sharma Class 10 Solutions Chapter 7 Triangles MCQS

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :
Question 1.
State basic proportionality theorem and its converse.
Solution:
Basic Proportionality Theorem : If a line is drawn parallel to one side of a triangle intersects the other two sides in distinct points, then the other two sides are divided in the same ratio.
Conversely : In a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side of the triangle.

Question 2.
In the adjoining figure, find AC.

Solution:

Question 3.
In the adjoining figure, if AD is the bisector of ∠A, what is AC ?

Solution:
In the figure, AD is the angle bisector of ∠A of ∆ABC
AB = 6 cm, BC = 3 cm, DC = 2 cm

Question 4.
State AAA similarity criterion.
Solution:
If in two triangles, corresponding angles are respectively equal then the triangles are similar.

Question 5.
State SSS similarity criterion.
Solution:
If in two triangles, corresponding sides are in the same ratio, then the two triangles are similar.

Question 6.
State SAS similarity criterion.
Solution:
If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio, then the triangles are similar.

Question 7.
In the adjoining figure, DE is parallel to BC and AD = 1 cm, BD = 2 cm. What is the ratio of the area of ∆ABC to the area of ∆ADE?
Solution:
In ∆ABC, DE || BC

Question 8.
In the figure given below, DE || BC. If AD = 2.4 cm, DB = 3.6 cm and AC = 5 cm. Find AE.
Solution:
In ∆ABC, DE || BC
AD = 2.4 cm, DB = 3.6 cm, AC = 5 cm

x = 2
AE = 2 cm

Question 9.
If the areas of two similar triangles ABC and PQR are in the ratio 9 : 16 and BC = 4.5 cm, what is the length of QR ?
Solution:
∆ABC ~ ∆PQR
Area of ∆ABC : area of ∆PQR = 9 : 16
BC = 4.5 cm
Let QR = A
The area of two similar triangles are in the ratio of the squares of their corresponding sides

Question 10.
The areas of two similar triangles are 169 cm² and 121 cm² respectively. If the longest side of the larger triangle is 26 cm, what is the length of the longest side of the smaller triangle ?
Solution:
Let ∆ABC be the larger triangle and ∆PQR
be the smaller triangle and their longest sides be BC and QR respectively
Area of ∆ABC = 169 cm²
area of ∆PQR = 121 cm²
BC = 26 cm
Let QR = x cm
∆ABC ~ ∆PQR

Question 11.
If ABC and DEF are similar triangles such that ∠A = 57° and ∠E = 73°, what is the measure of ∠C ?
Solution:
∆ABC ~ ∆DEF
Their corresponding angles are equal ∠A = ∠D, ∠B = ∠E and ∠C = ∠F
But ∠A = 57°

Question 12.
If the altitude of two similar triangles are in the ratio 2 : 3, what is the ratio of their areas ?
Solution:
Let ∆ABC ~ ∆PQR
and let AL ⊥ BC and PM ⊥ QR

Question 13.

Solution:

Question 14.
If ∆ABC and ∆DEF are similar triangles such that AB = 3 cm, BC = 2 cm, CA = 2.5 cm and EF = 4 cm, write the perimeter of ∆DEF.
Solution:

Question 15.
State Pythagoras Theorem and its converse.
Solution:
In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Conversely : If in a triangle, the square of one side is equal to the sum of squares of the remaining two sides, then the angle opposite to the first side is a right angle.

Question 16.
The lengths of the diagonals of a rhombus are 30 cm and 40 cm. Find the side of the rhombus. (C.B.S.E. 2008)
Solution:
In rhombus ABCD, BD = 30 cm, AC = 40 cm

Question 17.
In figure, PQ || BC and AP : PB = 1 : 2.

Solution:
In ∆ABC, PQ || BC
∆APQ ~ ∆ABC
But AP : PB = 1 : 2
The ratio of the areas of two similar triangles are proportional to the square of their corresponding sides

Question 18.
In the figure, S and T are points on the sides PQ and PR respectively of ∆PQR such that PT = 2 cm, TR = 4 cm and ST is parallel to QR. Find the ratio of the areas of ∆PST and ∆PQR. [CBSE 2010]

Solution:
In ∆PQR, ST || QR and PT = 2 cm, TR = 4 cm
PR = PT + TR = 2 + 4 = 6 cm

Question 19.
In the figure, ∆AHK is similar to ∆ABC. If AK = 10 cm, BC = 3.5 cm and HK = 7 cm, find AC. [CBSE 2010]

Solution:
In the given figure, AK = 10 cm, BC = 3.5 cm, HK = 7 cm

Question 20.
In the figure, DE || BC in ∆ABC such that BC = 8 cm, AB = 6 cm and DA = 1.5 cm. Find DE.

Solution:
In the given figure,
DE || BC
BC = 8 cm, AB = 6 cm and DA = 1.5 cm

Question 21.
In the figure, DE || BC and AD = \(\frac { 1 }{ 2 }\) BD. If BC = 4.5 cm, find DE. [CBSE 2010]

Solution:
In the given figure,
DE || BC

Question 22.
In the figure, ∠M = ∠N = 46°. Express x in terms of a, b and c where a, b, c are lengths of LM, MN and NK respectively.

Solution:
In the figure, ∠M = ∠N = 46°
∠M = a, PN = x, MN = b, NK = c
∠M = ∠N = 46°
But there are corresponding angle
PN || ML
∆PKN ~ ∆LKM

Hope given RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise

Other Exercises

• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.2
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.3
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7
• RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise
• RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS
• RD Sharma Class 10 Solutions Chapter 7 Triangles MCQS

Question 1.
In each of the figures. [(i) – (iv)] given below, a line segment is drawn parallel to one side of the triangle and the lengths of certain line-segments are marked. Find the value of x in each of the following:


Solution:

Question 2.
What values of x will make DE || AB In the figure

Solution:

Question 3.
In ∆ABC, points P and Q are on CA and CB, respectively such that CA = 16 cm, CP = 10 cm, CB = 30 cm and CQ = 25 cm. Is PQ || AB ?
Solution:
In ∆ABC, P and Q are the points on two sides CA and CB respectively
CA = 16 cm, CP = 10 cm, CB = 30 cm and CQ = 25 cm

Question 4.
In the figure, DE || CB. Determine AC and AE.

Solution:

Question 5.
In the figure, given that ∆ABC ~ ∆PQR and quad ABCD ~ quad PQRS. Determine the values of x, y, z in each case.



Solution:

Question 6.
In ∆ABC, P and Q are points on sides AB and AC respectively such that PQ || BC. If AP = 4 cm, PB = 6 cm and PQ = 3 cm, determine BC.
Solution:
In ∆ABC, P and Q are points on AB and AC respectively such that PQ || BC AP = 4 cm, PB = 6 cm, PQ = 3 cm

Question 7.
In each of the following figures, you find two triangles. Indicate whether the triangles are similar. Give reasons in support of your answer.



Solution:
(i) In figure (i)
Let in ∆ABC, AB = 4.6, BC = 10, CA = 8
and in ∆DEF, DE = 2.3, EF = 5 and FD = 4

Question 8.
In ∆PQR, M and N are points on sides PQ and PR respectively such that PM = 15 cm and NR = 8 cm. If PQ = 25 cm and PR = 20 cm. state whether MN || QR.
Solution:
In ∆PQR, P and Q are points on PQ and PR such that
PM = 15 cm, NR = 8 cm PQ = 25 cm
and PR = 20 cm PN = PR – NR = 20 – 8 = 12 cm

Question 9.
In ∆ABC, P and Q are points on sides AB and AC respectively such that PQ || BC. If AP = 3 cm, PB = 5 cm and AC = 8 cm, find AQ.
Solution:
In ∆ABC, P and Q are points on the sides AB and AC such that PQ || BC and AP = 3 cm, PQ = 5 cm, AC = 8 cm

8x = 24
=> x = 3
AQ = 3 cm

Question 10.
In the figure, ∆AMB ~ ∆CMD; determine MD in terms of x, y and z.

Solution:

Question 11.
In ∆ABC, the bisector of ∠A intersects BC in D. If AB = 18 cm, AC = 15 cm and BC = 22 cm, find BD.
Solution:
In ∆ABC, AD is the bisector of ∠A meeting BC in D
AB = 18 cm, AC = 15 cm and BC = 22 cm

Question 12.
In the figure, l || m
(i) Name three pairs of similar triangles with proper correspondence; write similarities.

Solution:

Question 13.
In the figure, AB || DC
(i) ∆DMU ~ ∆BMV
(ii) DM x BV = BM x DU

Solution:
Given : In the figure,
ABCD is a trapezium in which

Question 14.
ABCD is a trapezium in which AB || DC. P and Q are points on sides AD and BC such that PQ || AB. If PD = 18, BQ = 35 and QC = 15, find AD.
Solution:
In trapezium ABCD,
AB || DC
P and Q are points on AD and BC respectively such that
PQ || BC PD = 18, BQ = 35, QC = 15

Question 15.
In ∆ABC, D and E are points on sides AB and AC respectively such that AD x EC = AE x DB. Prove that DE || BC.
Solution:
Given : In ∆ABC,
D and E are points on sides AB and AC respectively and AD x EC = AE x DB

Question 16.
ABCD is a trapezium having AB || DC. Prove that O, the point of intersection of diagonals, divides the two diagonals in the same ratio. Also prove that \(\frac { area(\triangle OCD) }{ area(\triangle OAB) } =\frac { 1 }{ 9 }\) , if AB = 3CD
Solution:
Given: ABCD is a trapezium in which AB || DC and diagonals AC and BD intersect each other at O.


Hence proved.

Question 17.
Corresponding sides of two triangles are in the ratio 2 : 3. If the area of the smaller triangle is 48 cm², determine the area of the larger triangle.
Solution:
Let the corresponding sides of two triangles are 2x : 3x
The ratio of the areas of two similar triangles is the ratio of the squares of their corresponding sides

Question 18.
The areas of two similar triangles are 36 cm² and 100 cm². If the length of a side of the smaller triangle in 3 cm, find the length of the corresponding side of the larger triangle.
Solution:
Area of smaller triangle = 36 cm²
and area of larger triangle = 100 cm²
One side of smaller triangle = 3 cm
Let the corresponding side of larger triangle = x
∆s are similar

Question 19.
Corresponding sides of two similar triangles are in the ratio 1 : 3. If the area of the smaller triangle in 40 cm², find the area of the larger triangle.
Solution:
The corresponding sides of two similar triangles are in the ratio 1 : 3
Let their sides be x, 3x
Area of the smaller triangle is 40 cm²
Triangles are similar

Question 20.
In the figure, each of PA, QB, RC and SD is perpendicular to l. If AB = 6 cm, BC = 9 cm, CD = 12 cm and PS = 36 cm, then determine PQ, QR and RS.

Solution:
Given : In the figure,
PA, QB, RC and SD are perpendiculars on l
AB = 6 cm, BC = 9 cm, CD = 12 cm and PS = 36 cm

Hence PQ = 8 cm, QR = 12 cm and RS = 16 cm

Question 21.
In each of the figures given below, an altitude is drawn to the hypotenuse by a right-angled triangle. The length of different line-segments are marked in each figure. Determine x, y, z in each case.

Solution:
(i) In figure (i)
In ∆ABC, ∠B = 90°
BD ⊥ AC
∆ABD ~ ∆CBD

Question 22.
Prove that in an equilateral triangle, three times the square of a side is equal to four times the square of its altitudes.
Solution:
Given : In an equilateral ∆ABC,
AD ⊥ BC
To prove : 3AB² = 4AD²
Proof : The altitude of an equilateral triangle bisects the opposite side

Question 23.
In ∆ABC, AD and BE are altitudes. Prove that :

Solution:
Given : In ∆ABC,
AD ⊥ BC and BE ⊥ AC

Question 24.
The diagonals of quadrilateral ABCD intersect at O. Prove that

Solution:
Given : ABCD is quadrilateral in which diagonals AC and BD intersect each other atO


= \(\frac { BO }{ DO }\) {From (i)}
Hence proved.

Question 25.
In ∆ABC, ray AD bisects ∠A and intersects BC in D. If BC = a, AC = b and AB = c, prove that:
(i) BD = \(\frac { ac }{ b + c }\)
(ii) DC = \(\frac { ab }{ b + c }\)
Solution:
Given: In ∆ABC
AD is the bisector of ∠A
AB = c, BC = a, CA = b

Question 26.
There is a staircase as shown in the figure, connecting points A and B. Measurements of steps are marked in the figure. Find the straight line distance between A and B.

Solution:
There are 4 steps in staircase AB
Taking first step,
In ∆ALP

Question 27.
In ∆ABC, ∠A = 60°. Prove that BC² = AB² + AC² – AB.AC.
Solution:


= AB² + AC² – AB.AC.
Hence proved.

Question 28.
In ∆ABC, ∠C is an obtuse angle, AD ⊥ BC and AB² = AC² + 3BC². Prove that BC = CD.
Solution:
Given : In ∆ABC, ∠C is an obtuse angle AD ⊥ BC and AB² = AC² + 3BC²
To prove : BC = CD

Question 29.
A point D is on the side BC of an equilateral triangle ABC such that DC = \(\frac { 1 }{ 4 }\) BC. Prove that AD² = 13 CD²
Solution:
Given : In the equilateral ∆ABC,
D is a point on BC such that DC = \(\frac { 1 }{ 4 }\) BC

Question 30.
In ∆ABC, if BD ⊥ AC and BC² = 2 AC.CD, then prove that AB = AC.
Solution:
Given : In ∆ABC,
BD ⊥ AC
BC² = 2 AC.CD

Question 31.
In a quadrilateral ABCD, given that ∠A + ∠D = 90°. Prove that AC² + BD² = AD² + BC².
Solution:
Given : In quadrilateral ABCD,
∠A + ∠D = 90°
AC and BD are joined
To prove: AC² + BD² = AD² + BC²

Question 32.
In ∆ABC, given that AB = AC and BD ⊥ AC. Prove that BC² = 2 AC.CD.
Solution:
Given: In ∆ABC,
AB = AC
BD ⊥ AC

Question 33.
ABCD is a rectangle. Points M and N are on BD such that AM ⊥ BD and CN ⊥ BD. Prove that BM² + BN² = DM² + DN².
Solution:
Given : In rectangle ABCD,
BD is the diagonal
AM ⊥ BD and CN ⊥ BD

Question 34.
In ∆ABC, AD is median. Prove that AB² + AC² = 2AD² + 2DC².
Solution:
Given : In ∆ABC, AD is the median of BC
To prove : AB² + AC² = 2AD² + 2DC²

Question 35.
In ∆ABC, ∠ABC = 135°. Prove that: AC² = AB² + BC² + 4 ar (∆ABC).
Solution:
Given : In ∆ABC, ∠ABC = 135°,


Hence proved.

Question 36.
In a quadrilateral ABCD, ∠B = 90°. If AD² = AB² + BC² + CD², then prove that ∠ACD = 90°.
Solution:
Given : In quadrilateral ABCD, ∠B = 90° and AD² = AB² + BC² + CD²

Question 37.
In a triangle ABC, N is a point on AC such that BN ⊥ AC. If BN² = AN.NC, prove that ∠B = 90°.
Solution:
Given : In ∆ABC, BN ⊥ AC and BN² = AN.NC
To prove : ∠B = 90°

Question 38.
Nazima is fly Ashing in a stream. The tip of her Ashing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out ? If she pulls the string at the rate of 5 cm per second, what will the horizontal distance of the fly from her after 12 seconds.

Solution:
Height of the rod from stream level = 1.8 m
and of string from the point under the tip of rod = 2.4 m
Let the length of string = x
x² = (1.8)² + (2.4)² = 3.24 + 5.76 = 9.00 = (3.0)²
x = 3.0
Length of string = 3 m
Rate of pulling the string = 5 cm per second
Distance covered in 12 seconds = 5 x 12 = 60 cm.
At this stage, length of string = 3.0 – 0.6 = 2.4 m
Height = 1.8 m
Let base = y then
(2.4)² = y² + (1.8)²
=> 5.76 = y² + 3.24
=> y² = 5.76 – 3.24 = 2.52
y = 1.59
and distance from her = 1.59 + 1.2 = 2.79 m

Hope given RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 12 Heights and Distances MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 12 Heights and Distances MCQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 12 Heights and Distances Ex 12.1
• RD Sharma Class 10 Solutions Chapter 12 Heights and Distances VSAQS
• RD Sharma Class 10 Solutions Chapter 12 Heights and Distances MCQS

Mark the correct alternative in each of the following :
Question 1.
The ratio of the length of a rod and its shadow is 1 : \(\sqrt { 3 } \) The angle of elevation of the sum is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
Let AB be rod and BC be its shadow
So that AB : BC = 1 : \(\sqrt { 3 } \)
Let θ be the angle of elevation

Question 2.
If the angle of elevation of a tower from a distance of 100 metres from its foot is 60?, the height of the tower is

Solution:
Let AB be the tower and a point P at a distance of 100 m from its foot, angle of elevation of the top of the tower is 60°

Let height of the tower = h

Question 3.
If the altitude of the sun is at 60?, then the height of the vertical tower that will cast a shadow of length 30 m is

Solution:
Let AB be tower and a point P distance of 30 m from its foot of the tower which form an angle of elevation pf the sun of 60°

Question 4.
If the angles of elevation of a tower from two points distance a and b (a > b) from its foot and in the same straight line from it are 30? and 60?, then the height of the tower is

Solution:
Let AB be the tower and P and Q are such points that PB = a, QB = b and angles of elevation at P and Q are 30° and 60° respectively

Question 5.
If the anglr of elevation of the top of a tower from two points distant a and b from the base and in the same straight line with It are complementary, then the height of the tower is

Solution:
Let AB be the tower and P and Q are two points such that PB = a and QB = b and angles of elevation are θ and (90° – θ)
Let height of tower = h

Question 6.
From a light house the angles of depression of two ships on opposite sides of the light house are observed to be 30° and 45°. If the height of the light house is h metres, the distance between the ships is

Solution:
Let AB be light house and P and Q are two ships on its opposite sides which form angle of elevation of A as 45° and 30° respectively AB = h
Let PB = x and QB = y

Question 7.
The angle of elevation of the top of a tower stahding on a horizontal plane from a point A is a. After walking a distance d towards the foot of the tower the angle of elevation is found to be p. The height of the tower is

Solution:
Let AB be the tower and C is a point such that the angle of elevation of A is a. After walking towards the foot B of the tower, at D the angle of elevation is p Let h be the height of the tower and DB = x Now in right AACB,

Question 8.
The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30° with horizontal, then the length of the wire is
(a) 12 m
(b) 10 in
(c) 8 m
(d) 6 m
Solution:
Let AB and CD be two poles
AB = 20 m, CD = 14 m
A and C are joined by a wire
CE || DB and angle of elevation of A is 30° Class 10 Solutions Chapter 12 Heights and Distances MCQS – 9.png
Let CE = DB = x and AC = l

Question 9.
From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower is
(a) 25 m
(b) 50 in
(c) 75 m
(d) 100 m
Solution:
Let AB be the tower and CD be cliff Angle of elevation of A is equal to the angle of depression of B at C
Let angle be Q and CD = 25 m

Question 10.
The angles of depression of two ships from the top of a light house are 45° and 30° towards east. If the ships are 100 m apart, the height of the light house is

Solution:
Let AB be the light house C and D are two ships whose angles of depression on A are 30° and 45° respectively

Question 11.
if the angle of elevation of a cloud from a point 200 m above a lake is 30° and the angle of depression of its reflection in the lake is 60°, then the height of the cloud above the lake, is
(a) 200 m
(b) 500 m
(c) 30 m
(d) 400 m
Solution:
Let C be the cloud and R is its reflection in the lake
L is a point 200 m above the lake. Such that
LM = 200 m
Angle of elevation of C with L is 30° and angle of depression of R is 60°

Let height of cloud CB = h
∴ BR = h and NB = LM = 200 m
∴ CN = (h – 200) m and NR = (h + 200) m

Question 12.
The height of a tower is 100 m. When the angle of elevation of the sun changes from 30° to 45°, the shadow of the tower becomes x metres less. The value of x is

Solution:
Let AB be tower and AB = 100 m and angles of elevation of A at C and D are 30° and 45° respectively and CD = x
Let BD =y

Question 13.
Two persons are a metres apart and the height of one is double that of the other. If from the middle point of the line joining their feet, an observer finds the angular elevation of their tops to be complementary, then the height of the shorter post is

Solution:
Let AB and CD are two persons standing ‘a’ metres apart
M is the mid-point of BD and from M, the angles of elevation of A and C are complementary

Question 14.
The angle of elevation of a cloud from a point h metre above a lake is θ. The angle of depression of its reflection in the lake is 45°. The height of the cloud is
(a) It tan (45° + θ)
(b) h cot (45° – θ)
(c) h tan (45° – θ)
(d) h cot (45° + θ)
Solution:
Let C is the cloud and R is its reflection in the lake
From the lake, ‘7’ m aboves it, E is point
where angle of elevation of C is θ
and angle of depression of reflection is 45°

Question 15.
A lower subtends an angle of 30° at a point on the same level as its foot. At a second point h metres above the first, the depression of the foot of the tower is 60°. The height of the tower is

Solution:
Let CD is the tower and A is a point such that the angle of elevation of C is 30°
B is and their point h m high of A and angle of depressio of D is 60°

Question 16.
It is found that on walking x metres towards a chimney in a horizontal line through its base, the elevation of its top changes from 30° to 60° . The height of the chimney is

Solution:
In the figure, AB is chimney and CB and DB are its shadow

Question 17.
The length of the shadow of a tower standing on level ground is found to be 2.v metres longer when the sun’s elevation is 30° than when it was 45° . The height of the tower in metres is

Solution:
AB is a tower
BD and BC are its shadows and CD = 2x

Question 18.
Two poles are ‘a’ metres apart and the height of one is double of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the smaller is


Solution:
Let height of pole CD = h
and AB = 2h, BD = a
M is mid-point of BD

Question 19.
The tops of two poles of height 16 m and 10 m are connected by a wire of length l metres. If the wire makes an angle of 30° with the horizontal, then l =
(a) 26
(b) 16
(c) 12
(d) 10
Solution:
Let AB and CD are two poles AB = 10 m and CD = 16 m

Question 20.
If a 1.5 m tall girl stands at a distance of 3 m from a lamp-post and casts a shadow of length 4.5 m on the ground, then the height of the lamp-post is
(a) 1.5 m
(b) 2 m
(c) 2.5 m
(d) 2.8 m
Solution:
Let AB is girls and CD is lamp-post AB = 1.5
which casts her shadow EB

Question 21.
The length of shadow of a tower on the plane ground is \(\sqrt { 3 } \) times the height of the tower. The angle of elevation of sun is
(a) 45°
(b) 30°
(c) 60°
(d) 90° [CBSE 2012]
Solution:
Let AB be tower and BC be its shadow
∴Let AB = x

Question 22.
The angle of depression of a car, standing on the ground, from the top of a 75 m tower, is 30°. The distance of the car from the base of the tower (in metres) is
(a) 25 \(\sqrt { 3 } \)
(b) 50 \(\sqrt { 3 } \)
(c) 75 \(\sqrt { 3 } \)
(d) 150 [CBSE 2013]
Solution:
AB is a tower and AB = 75 m
From A, the angle of depression of a car C
on the ground is 30°

Question 23.
A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall is

Solution:
Let AB is a wall and AC is the ladder 15 m long which makes an angle of 60° with the ground
∴In ∆ABC, ∠B = 90°
Let height of wall AB = h

Question 24.
The angle of depression of a car parked on the road from the top of a 150 m high tower is 30°. The distance of the car from the tower (in metres) is
(a) 50\(\sqrt { 3 } \)
(b) 150 \(\sqrt { 3 } \)
(c) 150 \(\sqrt { 3 } \)
(d) 75
Solution:
Let AB be the tower of height 150 m
C is car and angle of depression is 30°

Question 25.
If the hei8ht of a vertical pole is \(\sqrt { 3 } \) times the length of its shadow on the ground, then the angle of elevation of the sun at that time is
(a) 30°
(b) 60°
(c) 45°
(d) 75° [CBSE 2014]
Solution:
Let AB be a vertical pole and let its shadow be BC

Question 26.
The angle of elevation of the top of a tower at a point on the ground 50 m away from the foot of the tower is 45°. Then the height of the tower (in metres) is

Solution:
Let AB be tower and C is a point on the ground 50 m away

Question 27.
A ladder makes an angle of 60° with the ground when placed against a wall. If the foot of the ladder is 2 m away from the wall, then the length of the ladder (in metres) is

Solution:
Suppose AB is the ladder of length x m
∴ OA = 2m, ∠OAB = 60°

Hope given RD Sharma Class 10 Solutions Chapter 12 Heights and Distances MCQS are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7

Other Exercises

• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.2
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.3
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7
• RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise
• RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS
• RD Sharma Class 10 Solutions Chapter 7 Triangles MCQS

Question 1.
If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is a right-angled triangle. (C.B.S.E. 1992)
Solution:
We know that if the square of the hypotenuse (longest side) is equal to the sum of squares of other two sides then it is right triangle
Now the sides of a triangle are 3 cm, 4 cm and 6 cm
(Longest side)² = (6)² = 36
and sum of two smaller sides = (3)² + (4)² = 9 + 16 = 25
36 ≠ 25
It is not a right-angled triangle

Question 2.
The sides of certain triangles are given below. Determine which of them are right triangles :
(i) a = 1 cm, b = 24 cm and c = 25 cm
(ii) a = 9 cm, b = 16 cm and c = 18 cm
(iii) a = 1.6 cm, b = 3.8 cm and c = 4 cm
(iv) a = 8 cm, b = 10 cm and c = 6 cm (C.B.S.E. 1992)
Solution:
We know that if the square of hypotenuse is equal to the sum of squares of other two sides, then it is a right triangle
(i) Sides of a triangle are a = 7 cm, b = 5.24 cm and c = 25 cm
(Longest side)² = (25)² = 625
Sum of square of shorter sides = (7)² + (24)² = 49 + 576 = 625
625 = 625
This is right triangle
(ii) Sides of the triangle are a = 9 cm, b = 16 cm, c = 18 cm
(Longest side)² = (18)² = 324
and sum of squares of shorter sides = (9)² + (16)² = 81 + 256 = 337
324 ≠ 337
It is not a right-angled triangle
(iii) Sides of the triangle are a = 1.6 cm, 6 = 3.8 cm, c = 4 cm
(Longest side)² = (4)² =16
Sum of squares of shorter two sides + (1.6)² + (3.8)² = 2.56 + 14.44 = 17.00
16 ≠ 17
It is not a right triangle
(iv) Sides of the triangle are a = 8 cm, b = 10 cm, c = 6 cm
(Longest side)² = (10)² = 100
Sum of squares of shorter sides = (8)² + (6)² = 64 + 36 = 100
100 = 100
It is a right triangle

Question 3.
A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point ?
Solution:
Let a man starts from O, the starting point to west 15 m at A and then from A, 8 m due north at B
Join OB

Now in right ∆OAB
OB² = OA² + AB² (Pythagoras Theorem)
OB² = (15)² + (8)² = 225 + 64 = 289 = (17)²
OB = 17
The man is 17 m away from the starting point

Question 4.
A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.
Solution:
Length of ladder = 17 m
Height of window = 15 m

Let the distance of the foot of ladder from the building = x
Using Pythagoras Theorem
AC² = AB² + BC²
=> (17)² = (15)² + x²
=> 289 = 225 + x²
=> x² = 289 – 225
=> x² = 64 = (8)²
x = 8
Distance of the foot of the ladder from the building = 8m

Question 5.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops. (C.B.S.E. 1996C, 2002C)
Solution:
Two poles AB and CD which are 6 m and 11 m long respectively are standing oh the ground 12 m apart

Draw AE || BD so that AE = BD = 12 m and ED = AB = 6 m
Then CE = CD – ED = 11 – 6 = 5 m
Now in right ∆ACE
Using Pythagoras Theorem,
AC² = AE² + EC² = (12)² + (5)² = 144 + 25 = 169 = (13)²
AC = 13
Distance between their tops = 13 m

Question 6.
In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC. (C.B.S.E. 1994)
Solution:
∆ABC is an isosceles triangle in which AB = AC = 25 cm .
AD ⊥ BC BC = 14 cm

Perpendicular AD bisects the base i.e . BD = DC = 7 cm
Let perpendicular AD = x
In right ∆ABD,
AB² = AD² + BD² (Pythagoras Theorem)
=> (25)² = AD² + (7)²
=> 625 = AD² + 49
=> AD² = 625 – 49
=> AD² = 576 = (24)²
=> AD = 24
Perpendicular AD = 24 cm

Question 7.
The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach ?
Solution:
In first case,
The foot of the ladder are 6 m away from the wall and its top reaches window 8 m high
Let AC be ladder and BC = 6 m, AB = 8 m

Now in right ∆ABC,
Using Pythagoras Theorem
AC² = BC² + AB² = (6)² + (8)² = 36 + 64 = 100 = (10)²
AC = 10 m
In second case,
ED = AC = 10 m
BD = 8 m, let ED = x
ED² = BD² + EB²
=> (10)² = (8)² + x²
=> 100 = 64 + x²
=> x² = 100 – 64 = 36 = (6)²
x = 6
Height of the ladder on the wall = 6 m

Question 8.
Two poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
Solution:
Let CD and AB be two poles which are 12 m apart
AB = 14 m, CD = 9 m and BD = 12 m
From C, draw CE || DB
CB = DB = 12 m
EB = CD = 9 m
and AE = 14 – 9 = 5 m

Now in right ∆ACE,
AC² = AE² + CE² (Pythagoras Theorem)
= (5)² + (12)²
= 25 + 144 = 169 = (13)²
AC = 13
Distance between their tops = 13 m

Question 9.
Using Pythagoras theorem, determine the length of AD in terms of b and c shown in the figure. (C.B.S.E. 1997C)

Solution:
In right ∆ABC, ∠A = 90°
AB = c, AC = b
AD ⊥ BC

Question 10.
A triangle has sides 5 cm, 12 cm and 13 cm. Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is 13 cm. (C.B.S.E. 1992C)
Solution:
A triangle has sides 5 cm, 12 cm and 13 cm
(Longest side)² = (13)² = 169
Sum of squares of shorter sides = (5)² + (12)² = 25 + 144= 169
169 = 169
It is a right triangle whose hypotenuse is 13 cm

BD = 4.6 cm

Question 11.
ABCD is a square, F is the mid point of AB. BE is one third of BC. If the area of ∆FBE = 108 cm² find the length of AC. (C.B.S.E. 1995)
Solution:
In square ABCD, F is mid point of AB i.e.,

Question 12.
In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC. (C.B.S.E. 2000)
Solution:
In ∆ABC, AB = AC
AD ⊥ BC
AB = AC = 13 cm, AD = 5 cm
AD ⊥ BC
AD bisects BC at D
BD = \(\frac { 1 }{ 2 }\) BC
=> BC = 2BD

Question 13.
In a ∆ABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that
(i) AD = a √3
(ii) area (∆ABC) = √3 a² (C.B.S.E. 1991)
Solution:
In ∆ABC, AB = BC = AC = 2a
AD ⊥ BC
AD bisects BC at D
BD = DC = \(\frac { 1 }{ 2 }\) BC = a

Question 14.
The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus. (C.B.S.E. 1993C)
Solution:
ABCD is a rhombus whose diagonals AC = 24 cm and BD = 10 cm
The diagonals of a rhombus bisect eachother at right angles

AO = OC = \(\frac { 24 }{ 2 }\) = 12 cm
and BO = OD = \(\frac { 10 }{ 2 }\) = 5 cm
Now in right ∆AOB,
AB² = AO² + BO² (Pythagoras Theorem)
(12)² + (5)² = 144 + 25 = 169 = (13)²
AB = 13
Each side of rhombus = 13 cm

Question 15.
Each side of a rhombus is 10 cm. If one of its diagonals is 16 cm, find the length of the other diagonal.
Solution:
In rhombus ABCD, diagonals AC and BD bisect eachother at O at right angles
Each side = 10 cm and one diagonal AC = 16 cm

AO = OC = \(\frac { 16 }{ 2 }\) = 8 cm
Now in right angled triangle AOB,
AB² = AO² + OB² (Pythagoras Theorem)
(10)² = (8)² + (BO)²
=> 100 = 64 + BO²
=> BO² = 100 – 64 = 36 = (6)²
BO = 6
BD = 2BO = 2 x 6 = 12 cm

Question 16.
Calculate the height of an equilateral triangle each of whose sides measures 12 cm.
Solution:
Each side of the equilateral ∆ABC = 12 cm
AD ⊥ BC which bisects BC at D


BD = DC = \(\frac { 12 }{ 2 }\) = 6 cm

Question 17.
In the figure, ∠B < 90° and segment AD ⊥ BC. Show that:
(i) b² = h² + a² + x² – 2ax
(ii) b² = a² + c² – 2ax

Solution:
Given : In ∆ABC, ∠B < 90°
AD ⊥ BC
AD = c, BC = a, CA = b AD = h, BD = x, DC = a – x
To prove: (i) b² = h² + a² + x² – 2ax
(ii) b² = a² + c² – 2ax
Proof: (i) In right ∆ADC, AC² = AD² + DC² (Pythagoras Theorem)
=> b² = h² + (a – x)² = h² + a² + x² – 2ax
(ii) Similarly in right ∆ADB
AB² = AD² + BD²
c² = h² + x² ….(i)
b² = h² + a² + x² – 2ax = h² + x² + a² – 2ax
= c² + a² – 2ax {From (i)}
= a² + c² – 2ax
Hence proved.

Question 18.
In an equilateral ∆ABC, AD ⊥ BC, prove that AD² = 3 BD². (C.B.S.E. 2002C)
Solution:
Given : ∆ABC is an equilateral in which AB = BC = CA .
AD ⊥ BC

Question 19.
∆ABD is a right triangle right-angled at A and AC ⊥ BD. Show that
(i) AB² = BC.BD
(ii) AC² = BC.DC
(iii) AD² = BD.CD
(iv) \(\frac { { AB }^{ 2 } }{ { AC }^{ 2 } }\) = \(\frac { BD }{ DC }\) [NCERT]
Solution:
In ∆ABD, ∠A = 90°
AC ⊥ BD

Question 20.
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?
Solution:
Let AB be the vertical pole whose height = 18 m

Question 21.
Determine whether the triangle having sides (a – 1) cm, 2 √a cm and (a + 1) cm is a right angled triangle. [CBSE 2010]
Solution:
Sides of a triangle are (a – 1) cm, 2 √a cm and (a + 1) cm
Let AB = (a – 1) cm BC = (a + 1) cm
and AC = 2 √a

Question 22.
In an acute-angled triangle, express a median in terms of its sides.
Solution:
In acute angled ∆ABC,
AD is median and AL ⊥ BC
Using result of theorem, sum of the squares of any two sides is equal to the twice the square of half of the third side together with twice the square of the median which bisects the third side

Question 23.
In right-angled triangle ABC in which ∠C = 90°, if D is the mid point of BC, prove that AB² = 4AD² – 3AC².
Solution:
Given : In right ∆ABC, ∠C = 90°
D is mid point of BC

Question 24.
In the figure, D is the mid-point of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that :


Solution:

Question 25.
In ∆ABC, ∠A is obtuse, PB x AC and QC x AB. Prove that:
(i) AB x AQ = AC x AP
(ii) BC² = (AC x CP + AB x BQ)
Solution:
Given : In ∆ABC, ∠A is an obtuse angle PB x AC and QC x AB on producing them

Question 26.
In a right ∆ABC right-angled at C, if D is the mid-point of BC, prove that BC² = 4 (AD² – AC²).
Solution:
Given : In ∆ABC, ∠C = 90°
D is mid-point of BC
AD is joined

=> 4AD² = 4AC² + BC²
=> BC² = 4AD² – 4AC²
=> BC² = 4 (AD² – AC²)
Hence proved.

Question 27.
In a quadrilateral ABCD, ∠B = 90°, AD² = AB² + BC² + CD², prove that ∠ACD = 90°.
Solution:
Given : ABCD is a quadrilateral in which ∠B = 90° and AD² = AB² + BC² + CD²
To prove : ∠ACD = 90°
Construction : Join AC

proof : In right ∆ABC, ∠B = 90°
AC² = AB² + BC² ….(i) (Pythagoras Theorem)
But AD² = AB² + BC² + CD² (given)
=> AD² = AC² + CD² {From (i)}
∆ACD is a right angle with right angle ACD
Hence proved.

Question 28.
An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/ hr. How far apart will be the two planes after 1\(\frac { 1 }{ 2 }\) hours?
Solution:

Speed of the first plane = 1000 km/hr
Distance travelled in 1\(\frac { 1 }{ 2 }\) hour due north = 1000 x \(\frac { 3 }{ 2 }\) = 1500 km
Speed of the second plane = 1200 km/hr
Distance travelled in 1\(\frac { 1 }{ 2 }\) hours due west

Hope given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 12 Heights and Distances VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 12 Heights and Distances VSAQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 12 Heights and Distances Ex 12.1
• RD Sharma Class 10 Solutions Chapter 12 Heights and Distances VSAQS
• RD Sharma Class 10 Solutions Chapter 12 Heights and Distances MCQS

Answer each of the following questions either in one word or one sentence or per requirement of the questions :
Question 1.
The height of a tower is 10 m. What is the length of its shadow when Sun’s altitude is 45° ?
Solution:
Let AB be the tower and BC is its shadow
Height of AB = 10 m
Let length of BC = x
Angle of elevation of A at C is 45°

Question 2.
If the ratio of the height of a angle tower and the length of 7 its shadow is \(\sqrt { 3 } \) : 1, what is the of elevation of the Sun?
Solution:
Ratio in the height of a tower and its shadow = \(\sqrt { 3 } \) : 1

Question 3.
What is the angle of elevation of the Sun when the length of the shadow of a vertical pole is equal to its height ?
Solution:
Let height of a vertical pole = h
Then its shadow = x

Question 4.
From a point on the ground, 20 m away from the foot of a vertical tower, the angle of elevation of the top of the tower is 60°, what is the height of the tower ?
Solution:
Let AB be the tower and the angles of elevation of its top from a point C on the ground is 60° and BC = 20 m
Let height of the tower AB = h

Question 5.
If the angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line, with it are complementary, find the height of the tower.
Solution:
Let CD be the tower and let the angles of elevation of the top C of the tower from two points A and B are θ and (90° – θ) as the angles are complementary

Let height of the tower CD = h
and AD = 9 m, BD = 4 m

Question 6.
In the figure, what are the angles of depression from the observing positions O₁ and O₂ of the object at A ?
Solution:
In the given figure, draw XY || ABC from O₁,O₂ (by joinging them)
∵∠AO₁C = 60°
∠AO₁O₂ = 90° – 60° = 30°
Similarly,
∵∠O₂ AB = 45°
∠XO₂A = ∠O₂AB = 45° (alternate angles)
Hence angles of depression are 30° and 45°

Question 7.
The tops of two towers of height x and y, standing on level ground, subtend angles of 30° and 60° respectively at the centre of the line joining their feet, then find x : y. [CBSE2015]
Solution:
Let AB and CD be two towers which are apart from each other at a distance of BD and M is mid point of BD.
Angles of elevation are 30° and 60°

Question 8.
The angle of elevation of the top of a tower at a point on the ground is 30°. What will be the angle of elevation, if the height of the tower is tripled?     [CBSE 2015]
Solution:
Let AB be the tower and apgle of elevation of A at C is 30°

Question 9.
AB is a pole of height 6 m standing at a point B and CD is a ladder inclined at angle of 60° to the horizontal and reaches upto a point D of pole. If AD = 2.54 m, find the length of the ladder. (Use \(\sqrt { 3 } \)  = 1.73)       [CBSE 2016]
Solution:
BD = AB – AD = 6 – 2.54 = 3.46 m

Question 10.
An observer, 1.7 m tall, is 20\(\sqrt { 3 } \) m away from a tower. The angle of elevation from the eye of an observer to the top of tower is 30°. Find the height of the tower. [CBSE 2016]
Solution:

Question 11.
An observer, 1.5 m tall, is 28.5 m away from a 30 m high tower. Determine the angle of elevation of the top-of the tower from the eye of the observer. [CBSE 2017]
Solution:

Here, AE = 28.5 m
ED = BC = 28.5 m
In ΔADE,
tan ∠ADE = \(\frac { 28.5 }{ 28.5 }\)
= ∠ADE = tan^-1(1) – 45°
∴ The angle of elevation at the top of the tower from the eye of the observer is 45°.

Hope given RD Sharma Class 10 Solutions Chapter 12 Heights and Distances VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6

Other Exercises

• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.2
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.3
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7
• RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise
• RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS
• RD Sharma Class 10 Solutions Chapter 7 Triangles MCQS

Question 1.
Triangles ABC and DEF are similar.
(i) If area (∆ABC) = 16 cm², area (∆DEF) = 25 cm² and BC = 2.3 cm, find EF. (C.B.S.E. 1992)
(ii) If area (∆ABC) = 9 cm², area (∆DEF) = 64 cm² and DE = 5.1 cm, find AB.
(iii) If AC = 19 cm and DF = 8 cm, find the ratio of the area of two triangles. (C.B.S.E. 1992C)
(iv) If area (∆ABC) = 36 cm², area (∆DEF) = 64 cm² and DE = 6.2 cm, find AB. (C.B.S.E. 1992)
(v) If AB = 1.2 cm and DE = 1.4 cm, find the ratio of the areas of ∆ABC and ∆DEF. (C.B.S.E. 1991C)
Solution:

Question 2.
In the figure, ∆ACB ~ ∆APQ. If BC = 10 cm, PQ = 5 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ. Also, find the area (∆ACB) : area (∆APQ).

Solution:

Question 3.
The areas of two similar triangles are 81 cm² and 49 cm² respectively. Find the ratio of their corresponding heights, what is the ratio of their corresponding medians ?
Solution:
Areas of two similar triangles are 81 cm² and 49 cm²
The ratio of the areas of two similar triangles are proportion to the square of their corresponding altitudes and also squares of their corresponding medians
Ratio in their altitudes = √81 : √49 = 9 : 7
Similarly, the ratio in their medians = √81 : √49 = 9 : 7

Question 4.
The areas of two similar triangles are 169 cm² and 121 cm² respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.
Solution:
Triangles are similar Area of larger triangle = 169 cm²
and area of the smaller triangle =121 cm²
Length of longest sides of the larger triangles = 26 cm
Let the length of longest side of the smaller triangle = x

Question 5.
The areas of two similar triangles are 25 cm² and 36 cm² respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.
Solution:
Area of first triangle = 25 cm²
Area of second = 36 cm²
Altitude of the first triangle = 2.4 cm
Let altitude of the second triangle = x
The triangles are similar

Question 6.
The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.
Solution:
Length of the corresponding altitude of two triangles are 6 cm and 9 cm
triangles are similar

Question 7.
ABC is a triangle in which ∠A = 90°, AN ⊥ BC, BC = 12 cm and AC = 5 cm. Find the ratio of the areas of the ∆ANC and ∆ABC.
Solution:
In ∆ABC, ∠A = 90°
AN ⊥ BC
BC = 12 cm, AC = 5 cm

Question 8.
In the figure, DE || BC
(i) If DE = 4 cm, BC = 6 cm and area (∆ADE) = 16 cm², find the area of ∆ABC.
(ii) If DE = 4 cm, BC = 8 cm and area of (∆ADE) = 25 cm², find the area of ∆ABC. (C.B.S.E. 1991)
(iii) If DE : BC = 3 : 5, calculate the ratio of the areas of ∆ADE and the trapezium BCED

Solution:

Question 9.
In ∆ABC, D and E are the mid-points of AB and AC respectively. Find the ratio of the areas of ∆ADE and ∆ABC.
Solution:
In ∆ABC, D and E are the mid points of AB and AC respectively

Question 10.
The areas of two similar triangles are 100 cm² and 49 cm² respectively. If the altitude of the bigger triangle is 5 cm, find the corresponding altitude of the other. (C.B.S.E. 2002)
Solution:
∆ABC ~ ∆DEF
area ∆ABC = 100 cm²
and area ∆DEF = 49 cm²

Question 11.
The areas of two similar triangles are 121 cm² and 64 cm² respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other. (C.B.S.E. 2001)
Solution:
∆ABC ~ ∆DEF
area of ∆ABC = 121 cm² area of ∆DEF = 64 cm²
AL and DM are the medians of ∆ABC and ∆DEF respectively
AL = 12.1 cm

Question 12.
In ∆ABC ~ ∆DEF such that AB = 5 cm and (∆ABC) = 20 cm² and area (∆DEF) = 45 cm², determine DE.
Solution:
∆ABC ~ ∆DEF
area (∆ABC) = 20 cm²
area (∆DEF) = 45 cm²

Question 13.
In ∆ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divides ∆ABC into two parts equal in area. Find \(\frac { BP }{ AB }\).
Solution:
In ∆ABC, PQ || BC and PQ divides ∆ABC in two parts ∆APQ and trap. BPQC of equal in area
i.e., area ∆APQ = area BPQC

Question 14.
The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR. (C.B.S.E. 2004)
Solution:
∆ABC ~ ∆PQR
area (∆ABC) : area (∆PQR) = 9 : 16
and BC = 4.5 cm

Question 15.
ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of ∆APQ is one sixteenth of the area of ∆ABC. (C.B.S.E. 2005)
Solution:
In ∆ABC, P and Q are two points on AB and AC respectively such that
AP = 1 cm, PB = 3 cm, AQ = 1.5 cm and QC = 4.5 cm

Question 16.
If D is a point on the side AB of ∆ABC such that AD : DB = 3 : 2 and E is a point on BC such that DE || AC. Find the ratio of areas of ∆ABC and ∆BDE. (C.B.S.E. 2006C)
Solution:
In ∆ABC, D is a point on AB such that AD : DB = 3 : 2

Question 17.
If ∆ABC and ∆BDE are equilateral triangles, where D is the mid point of BC, find the ratio of areas of ∆ABC and ∆BDE. [CBSE 2010]
Solution:
∆ABC and ∆DBE are equilateral triangles Where D is mid point of BC

Question 18.
Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights.
Solution:
Two isosceles triangles have equal vertical angles
So their base angles will also be the equal to each other
Triangles will be similar Now, ratio in their areas = 36 : 25

Question 19.
In the figure, ∆ABC and ∆DBC are on the same base BC. If AD and BC intersect

Solution:
Given : Two ∆ABC and ∆DBC are on the same base BC as shown in the figure
AC and BD intersect eachother at O

Question 20.
ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. Prove that
(i) ∆AOB ~ ∆COD
(ii) If OA = 6 cm, OC = 8 cm, find

Solution:
Given : ABCD is a trapezium in which AB || CD
Diagonals AC and BD intersect each other at O

Question 21.
In ∆ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ∆APQ and trapezium BPQC.
Solution:
In ∆ABC, P is a point on AB such that AP : PQ = 1 : 2
PQ || BC
Now we have to find the ratio between area ∆APQ and area trap BPQC

Question 22.
AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed. Prove that Area (∆ADE) : Area (∆ABC) = 3 : 4. [CBSE 2010]
Solution:
Given: In equilateral ∆ABC, AD ⊥ BC and with base AD, another equilateral ∆ADE is constructed

Hope given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2

Other Exercises

• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

Evaluate each of the following (1-19) : 1.
Question 1.
sin45° sin30° + cos45° cos30°.
Solution:
sin 45° sin 30° + cos 45° cos 30°

Question 2.
sin60° cos30° + cos60° sin30°.
Solution:

Question 3.
cos60° cos45° – sin60° sin45°.
Solution:

Question 4.
sin²30° + sin²45° + sin²60° + sin²290°.
Solution:

Question 5.
cos²30° + cos²45° + cos²60° + cos²90°.
Solution:

Question 6.
tan²30° + tan²60° + tan²45°.
Solution:

Question 7.
2sin²30° – 3cos²45° + tan²60°.
Solution:

Question 8.
sin²30°cos²4S° + 4tan²30° + \(\frac { 1 }{ 2 }\) sin²90° -2cos²90° + \(\frac { 1 }{ 24 }\) cos20°.
Solution:

Question 9.
4 (sin⁴ 60° + cos⁴ 30°) – 3 (tan² 60° – tan² 45°) + 5cos² 45°
Solution:

Question 10.
(cosecc² 45° sec² 30°) (sin² 30° + 4cot² 45° – sec² 60°).
Solution:

Question 11.
cosec³ 30° cos 60° tan³ 45° sin2 90° sec² 45° cot 30°.
Solution:

Question 12.
cot² 30° – 2cocs² 60° – \(\frac { 3 }{ 4 }\)sec2 45° – 4sec² 30°.
Solution:

Question 13.
(cos0° + sin45° + sin30°) (sin90° – cos45° + cos60°)
Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.
4 (sin⁴ 30° + cos² 60°) – 3 (cos² 45° – sin² 90°) – sin² 60°
Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Find the value of x in each of the following : (20-25)

Question 20.
2sin 3x = √3
Solution:

Question 21.
2sin \(\frac { x }{ 2 }\) = 1
Solution:

Question 22.
√3 sin x=cos x
Solution:

Question 23.
tan x = sin 45° cos 45° + sin 30°
Solution:

Question 24.
√3 tan 2x = cos 60° +sin 45° cos 45°
Solution:

Question 25.
cos 2x = cos 60° cos 30° + sin 60° sin 30°
Solution:

Question 26.
If θ = 30°, verify that :

Solution:

Question 27.
If A = B = 60°, verify that:
(i) cos (A – B) = cos A cos B + sin A sin B
(ii) sin (A – B) = sin A cos B – cos A sin B tan A – tan B
(iii) tan (A – B) = \(\frac { tanA-tanB }{ 1+tanA-tanB }\)
Solution:

Question 28.
If A = 30° and B = 60°, verify that :
(i) sin (A + B) = sin A cos B + cos A sin B
(ii) cos (A + B) = cos A cos B – sin A sin B
Solution:

Question 29.
If sin (A + B) = 1 and cos (A,-B) = 1,0° < A + B < 90°, A > B find A and B.
Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.
In a ∆ABC right angle at B, ∠A = ∠C. Find the values of
(i) sin A cos C + cos A sin C
(ii) sin A sin B + cos A cos B
Solution:

Question 33.
Find acute angles A and B, if sin (A + 2B)=\(\frac { \sqrt { 3 } }{ 2 }\) and cos (A + 4B) = 0, A > B.
Solution:

Question 34.
In ΔPQR, right-angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠P and ∠R.
Solution:

Question 35.
If sin (A – B) = sin A cos B – cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values of sin 15° and cos 15°.
Solution:

Question 36.
In a right triangle ABC, right angled at ∠C if ∠B = 60° and AB – 15 units. Find the remaining angles and sides.
Solution:

Question 37.
If ΔABC is a right triangle such that ∠C = 90°, ∠A = 45° and BC = 7 units. Find ∠B, AB and AC.
Solution:

Question 38.
In a rectangle ABCD, AB = 20 cm, ∠BAC = 60°, calculate side BC and diagonals AC and BD.
Solution:

Question 39.
If A and B are acute angles such that

Solution:

Question 40.
Prove that (√3 + 1) (3 – cot 30°) = tan³ 60° – 2sin 60°. [NCERT Exemplar]
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 12 Heights and Distances Ex 12.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 12 Heights and Distances Ex 12.1

Other Exercises

• RD Sharma Class 10 Solutions Chapter 12 Heights and Distances Ex 12.1
• RD Sharma Class 10 Solutions Chapter 12 Heights and Distances VSAQS
• RD Sharma Class 10 Solutions Chapter 12 Heights and Distances MCQS

Question 1.
A tower stands vertically on the ground. From a point on the ground, 20 m away from the foot of the tower, the angle of elevation of the top of the tower is 60°. What is the height of the tower ?
Solution:
Let TS is the tower and P is a point which is 20 m away from the foot of the tower and angle of elevation of T is 60°

Question 2.
The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 9.5 m away from the wall. Find the length of the ladder.
Solution:
Let LM is the ladder which makes an angle of 60° with the wall LM

Question 3.
A ladder is placed along a wall of a house such that its upper end is touching the top of the wall. The foot of the ladder is 2 m away from the wail and the ladder is making an angle of 60° with the level of the ground. Determine the height of the wall.
Solution:
Let LM be the ladder which makes an angle of 60° with the wall LN and at a distance of 2 m from the foot of the wall

Question 4.
An electric pole is 10 m high. A steel wire tied to top of the pole is affixed at a point on the ground to keep the pole up right. If the wire makes an angle of 45° with the horizontal through the foot of the pole, find the length of the wire.
Solution:
Let AB be the pole and a wire AC is tied to the top of the pole with a point C on the ground which makes an angle of 45° with the ground

Question 5.
A kite is flying at a height of 75 metres from the ground level, attached to a string inclined at 60° to the horizontal. Find the length of the string to the nearest metre.
Solution:
Let K be the kite flying in the sky at a height of 75 m from the ground LM. The string KL makes an angle of 60° to the ground
Let length of string KL = x m
KM = 75 m

Question 6.
A ladder 15 m long just reaches the top of a vertical wall. If the ladders makes an angle of 60° with the wall, then find the height of the wall.
Solution:
Given that, the height of the ladder = 15 m
Let the height of the vertical wall = h
and the ladder makes an angle of elevation 60° with the wall i.e., θ = 60°.

Question 7.
A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff. At a point on the plane 70 metres away from the tower, an observer notices that the angles of elevation of the top and the bottom of the flag-staff are respectively 60° and 45°. Find the height of the flag-staff and that of the tower.
Solution:
Let TR be the tower and FT is the flag on it. P is an point on the ground 70 m away from the foot of the tower. From P, the angle of elevation of the top and bottom of the flag are 60° and 45° respectively Let h be the height of flag staff and x be the height of the tower

Question 8.
A vertically straight tree, 15 m high, is broken by the wind in such a way that its top just touches the ground and makes an angle of 60° with the ground. At what height from the ground did the tree break ? [CBSE 1995]
Solution:
Let TR be the tree whose height is 15 m
Let it is broken from A and its top T touches the ground at B making an angle of 60° with the ground

Question 9.
A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff of height 5 metres. At a point on the plane, the angles of elevation of the bottom and the top of the flag-staff are respectively 30° and 60°. Find the height of the tower. (C.B.S.E. 1995)
Solution:
Let TR be the tower and FT is a flag staff on it. A is a point on the ground such that it makes angles of elevation of the bottom and top of the flag staff of 30° and 60° respectively

Question 10.
A person observed the angle of elevation of the top of a tower as 30°. He walked 50 m towards the root of the tower along level ground and found the angle of elevation of the top of the tower as 60°. Find the height of the tower.
Solution:
Let TR be the tower. A person at A on the ground observes the angle of elevation of the top T of the tower as 30° and then moves towards the foot of the tower. At a distance of 50 m at B, the angle of elevation becomes 60°

Let h be the height of the tower and BR = x, then AR = (50 + x)

Question 11.
The shadow of a tower, when the angle of elevation of the sun is 45°, is found to be 10 m longer than when it was 60°. Find the height of the tower.
Solution:
Let TR be we know
Let the shadow of TR at the elevation of the sun 45° be x m and at 60°, will be (x – 10) m Now in right ΔTAR,

Question 12.
A parachutist is descending vertically and makes angles of elevation of 45° and 60° at two observing points 100 m apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of the point where he falls on the ground from the just observation point.
Solution:
Let P be the parachutist landing to Q on the ground
A and B are two observations such that AB = 10 m

Let h be the height of the parachutist from the ground and x be the distance of B from Q and (100 + x) is the distance from A to Q Now in right ΔAPQ,

Question 13.
On the same side of a tower, two objects are located. When observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 150 m, find the distance between the objects. (C.B.S.E. 1992)
Solution:
Let TR be the tower and A, B are two objects which makes angles of elevation with top of the tower as 45° and 60° respectively

Question 14.
The angle of elevation of a tower from a point on the same level as the foot of the tower is 30°. On advancing 150 metres towards the foot of the tower, the angle of elevation of the tower becomes 60°. Show that the height of the tower is 129.9 metres (Use\(\sqrt { 3 } \) = 1-732).          (C.B.S.E. 2006)
Solution:
Let TR be the tower. A is a point on the same level which makes an angle of elevation of 30° with the top of the tower TR, and 150 m from A towards the foot of the tower the angle of elevaton is 60°, Let TR = h m

Question 15.
The angle of elevation of the top of a tower as observed from a point in a horizontal plane through the foot of the tower is 32°. When the observer moves  towards the tower a distance of 100 m, he finds the angle of elevation of the top to be 63°. Find the height of the tower and the distance of the first position from the tower. [Take tan 32° = 0.6248 and tan 63° = 1.9626]              (C.B.S.E. 2001C)
Solution:
Let PQ be the tower and from a points A, and B the angles of elevations of top P of the tower are 32° and 63° respectively and AB= 100 m

Question 16.
The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metres towards the foot of the tower to a point B the angle of elevation increases to 60°. Find the height of the tower and the distance of the tower from the point A. (C.B.S.E. 2002)
Solution:
Let CD be the tower and from a point A on the same ground, the angle of elevation of the top of the tower is 30°. B is another point such that AB = 20 m and from B, the angle of elevation is 60°

Let h be the height of the tower CD and x is the length of AD
∴  BD = (x – 20) m
Now in right ΔCAD

Question 17.
From the top of a building 15 m high the angle of elevation of the top of a tower is found to be 30°. From the bottom of the same building, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower and the distance between the tower and building. (C.B.S.E. 2002)
Solution:
Let TR be the tower and AB be the building The angles of elevation of the top of the tower T, from A is 30° and from B is 60°
Height of AB = 15 m

Let the height of tower TR = h and the distance between the tower and building = x
In right ΔTBR,

Question 18.
On a horizontral plane there is a vertical tower with the flag pole on the top of the tower. At a point 9 metres away from the foot of the tower the angle of elevation of the top and bottom of the flag pole are 60° and 30° respectively. Find the height of the tower and the flag pole mounted on it.      (C.B.S.E. 2005)
Solution:
Let TR be the tower and F is the flag pole on it. A is a point 9 m away from the foot of the tower and angles of elevation of the top and bottom of the flag pole from A are 60° and 30° respectively

Question 19.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with the ground. The distance between the foot of the tree to the point where the top touches the ground is 8 in. Find the height of the tree.
Solution:
Let TR be the tree and it is broken from A and broken part of the tree makes an angle of 30° on the ground at B

Question 20.
From a point P on the ground the angle of elevation of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flag-staff from P is 45°. Find the length of the flag-staff and the distance of the building from the point P. (Take \(\sqrt { 3 } \)  = 1.732
Solution:
Let BA is the building such that BA = 10 in CB is a flag-staff on the building P is a point on the ground such that the angles of the elevation of the top of the building is 30° and that of the top of the flag-staff is 45°

Question 21.
A 1.6 m tall girl stands at a distance of 3.2 m from a lamp-post and casts a shadow of 4.8 m on the ground. Find the height of the lamp-post by using
(i) trigonometric ratios
(ii) property of similar triangles.
Solution:
Let LP is the lamp-post and GR is the girl who is at a distance of 3.2 m
From the lamp-post and its shadow is AR which is 4.8 m long. Let ∠A = 0
Height of girl GR = 1.6 m and let height of lamp post = h

Question 22.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increasees from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution:
Let BA is the boy whose height is 1.5 m and CD is building whose height is 30 m Angle of elevation of C from B (eyes of the boy) is 30°

and on moving some distance towards the building at L, the angle of elevation of C is 60°
Let AM = BL = x and AD = y
Then LE = MD = y – x
and CE = CD – ED = 30 – 1.5 = 28.5 m
Now in right ΔCBE

Question 23.
The shadow of a tower standing on a level ground is found to be 40 m longer when Sun’s altitude is 30° than when it was “60°. Find the height of the tower.
Solution:
Let TR be the tower and RB and RA are its shadows at the elevation of 60° and 30° respectively. Such that BA = 40 m

Let h be the height of the tower and let shadow RA = x m
Then shadow RB = (x – 40) m
Now in right ΔTAR,

Question 24.
From a point on the ground the angles of elevation of the bottom and top of a transmission tower fixed at the top of 20 m high building are 45° and 60° respectively. Find the height of the transmission tower.
Solution:
BC is the building and AB is the transmission on it the height of the building is 20 m From a point P, the angles of elevation of B and A are 45° and 60° respectively

Let height of the transmission AB = h and let PC = x
Now in right ΔBPC,

Question 25.
The angles of depression of the top and bottom of 8 m tali building from the top of a multistoried building are 30° and 45° respectively. Find the height of the multistoried building and the distance Between the two buildings.
Solution:
Let AB be the multistoried building and CD is another building. From A, the angles of depression of the top C and bottom D of the other building are 30° and 45° respectively Height of building is 8 m i.e. CD = 8 m

Question 26.
A statue 1.6 m tall stands on the top of pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angles of elevation of the top of thd\pedestal is 45°. Find the height of the pedestal. (C.B.S.E. 2008)
Solution:
Let AB be the statue standing on the top of a pedestal BC
From a point P on the ground the angles of elevation of the top of the statue is 60° and top of the pedestal BC is 45°
Let height of BC = h and PC = x

Question 27.
A T.V. Tower stands vertically on a bank of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From a point 20 m away this point on the same bank, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the river.
Solution:
Let AB is the T.V. tower and CB is the,width of the river. D is a point which is 20 m away from C
Now angles of elevation from A to C and D are 60° and 30° respectively.

Question 28.
From the top of a 7 m high building, the – angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution:
Let AB be the building and CD be the cable tower
From the top of building, the angle of elevation of the top of tower is 60° and angle of depression of the foot of the tower is 45° Now AB = 7 m

Question 29.
As observed from the top of a 75 m tall lighthouse, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution:
Let LH is die light house and A, B are two ships From L, angles of depression of the ships A and B are 30° and 45° respectively

Question 30.
The angle of elevation of the top of the building from the foot of the tower is 30° and the angle of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution:
Let AB be the tower and CD is the building The angle of elevation of the top of the building from the foot of the tower is 30° and the angle of the top of the tower from the foot of the building is 60°

Question 31.
From a point on a bridge across a river the angles of depression of the banks onopposite side of the river are 30° and 45° respectively. If bridge is at the height of 30 m from the banks, find the width of the river.
Solution:
Let BR is the bridge with a height of 30 m and angles of depression from B the top of the bridge to two given points C and D on the opposite sides of the river are 30° and 45° respectively CD is the width of the river Let CR = x and DR = y

Question 32.
Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point between them on the road the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.
Solution:
AB and CD are two equal poles on either side of the road BD which is 80 m wide P is a point on the road such that the angles of elevation of the tops of the poles are 60° and 30° respectively

Let the length of each pole in h and let BP = x, then DP = 80 – x
In right ΔABP,

Question 33.
A man sitting at a height of 20 m on a tall tree on a small island in the middle of a river observes two poles directly opposite to each other on the two banks of the river and in line with the foot of tree. If the angles of depression of the feet of the poles from a point at which the man is sitting on the tree on either side of the river are 60° and 30° respectively. Find the width of the river.
Solution:
Let AB be the tree on which a man is sitting the tree is on a small island in the middle of the river. C and D are the foot of the poles on either bank of the river The angles of depression from A to the poles C and D are 60° and 30° respectively and AB = 20 m

Question 34.
A vertical tower stands on a horizontal plane and is surmounted by a flag-staff of height 7 m. From a point on the plane, the angle of elevation of the bottom of the flag-staff is 30° and that of the top of the flag-staff is 45°. Find the height of the tower.
Solution:
Let AB be a flag post on a building BC From a point P on the same ground, angle of elevations of A and B are 45° and 30° respectively

Question 35.
The length of the shadow of a tower standing on level plane is found to be 2x metres longer when the sun’s altitude is 30° than when it was 45°. Prove that the height of tower is x (\(\sqrt { 3 } \) +1) metres.
Solution:
Let PQ be the tower whose shadow is QA at the elevation of 45° and QB at the elevation of 30° such that AB = 2x

Question 36.
A tree breaks due to the storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with the ground. The distance from the foot of the tree to the point where the top touches the ground is 10 Find the height of the tree.
Solution:
Let TR be the tree arid it is broken at S and its broken parts touches the ground at P making ah angle of 30° elevation angle,
PR = 10 m
Let length of the tree TR = h and let TS = SP = x

Question 37.
A balloon is connected to a meteoro­logical ground station by a cable of length 215 m inclined at 60° to the horizontal. Determine the height of the balloon from the ground. Assume that there is no slack in the cable.
Solution:
Let B is the balloon which is connected by the cable BC which is 215 m long and makes an angle of elevation of 60° with the ground Let h be the height of the balloon

Question 38.
Two men on either side of the cliff 80 m high observes the angles of elevation of the top of the cliff to be 30° and 60° respectively. Find the distance between the two men.
Solution:
Let CL is the cliff and A and B are two men on either side of the cliff making angles of elevation with C as 30“ and 60“ respectively Height of cliff CL = 80 m

Question 39.
Find the angle of elevation of the sun (sun’s altitutde) when the length of the shadow of a vertical pole is equal to its height.
Solution:
Let the height of pole AB = h m
Then its shadow = h m

Question 40.
An aeroplane is flying at a height of 210 m. Flying at this height at some instant the angles of depression of two points in a line in opposite directions on both the banks of the river are 45° and 60°. Find the width of the river. (Use \(\sqrt { 3 } \)  = 1.73)          [CBSE 2015]
Solution:
Height of the aeroplane = 210 m
Let AB is the height of aeroplane and C and D are the points on the opposite banks of a river.

Question 41.
The angle of elevation of the top of a chimney from the top of a tower is 60° and the angle of depression of the foot of the chimney from the top of the tower is 30°. If the height of the tower is 40 m, find the height of the chimney. According to pollution control norms, the minimum height of a smoke emitting chimney should be 100 m. State if the height of the above mentioned chimriey meets the pollution norms. What value is discussed in this question?  [CBSE 2014]
Solution:
Let CD be the tower area AB be the chimney angle of elevation of the top of tower with the top of the chimney is 60° and foot of chimney with the top of tower is 30°
CD = 40 m

Question 42.
Two ships are there in the sea on either side of a light house in such a way that the ships and the light house are in the same straight line. The angle of depression of two ships are observed from the top of the light house are 60° and 45° respectively. If the height of the light house is 200 m, find the distance between the two ships. (Use \(\sqrt { 3 } \) = 1.73)
Solution:
Let AB be the light house and C and D are two ships which make angle of depression with the top A of the light house of 60° and 45°
AB = 20 m

Question 43.
The horizontal distance between two poles is 15 m. The angle of depression of the top of the first pole as seen from the top of the second pole is 30°. If the height of the second pole is 24 m, find the height of the first pole. (\(\sqrt { 3 } \) = 1.732)                [CBSE 2013]
Solution:
Let AB and CD are two poles and distance between them = 15 m, and AB = 24 m Angle of elevation from top of pole CD, to pole AB = 30°
From C, draw CE || DB

Question 44.
The angles of depression of two ships from the top of a light house and on the same side of it are found to be 45° and 30° respectively. If the ships are 200 m apart, find the height of the light house.
Solution:
Let PQ be the light house and A and B are two ships on the same side of the light house Angle of depression from top of the light house of the two ships are 30° and 45°

Question 45.
The angles of elevation of the top of a tower from two points at’a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution:
Let TR be the tower. A and B are two points which make the angled of elevation with top of tower as θ and 90° – θ (∵ angles are complementary)
Let height of tower TR = h and AR = 9 m, BR = 4m

Question 46.
From the top of a 50 m high tower, the angles of depression of the top and bottom of a pole are observed to be 45° and 60° respectively. Find the height of the pole.
Solution:
Let AB be the tower and CD is the pole. Angles of depression from the top A to the top and bottom of the pole are 45° and 60° respectively
AB = 50 m, let CD = h and BD = EC = x
∵ CE || DB
∴ EB = CD = h
and AE = 50 – h

Question 47.
The horizontal distance between two trees of different heights is 60 m. The angle of depression of the top of the first tree when seen from the top of the second tree is 45°. If the height of the second tree is 80 m, find the height of the first tree.
Solution:
Let AB and CD be the two trees
AB = 80 m, angle of depression from A of C is 45°. Draw CE || DB

Question 48.
A flag-staff stands on the top of a^5 m high tower. From a point on the ground, the angle of elevation of the top of the flag-staff is 60° and from the same point, the angle of elevation of the top of the tower is 45°. Find the height of the flag­staff.
Solution:
Let FT is the flag-staff situated on the top of the tower TR. A is any point on the same plane which makes angles of elevation with top of the flag-staff and top of the tower are 60° and 45° respectively.

Question 49.
The angle of elevation of the top of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation of the top is 45°. Calculate the height of the tower.
Solution:
Let TR is the tower From a point X, the angle of elevation of T is 60° and 40 m above X, from the point Y, the angle of elevation is 45°

Question 50.
As observed from the top of a 150 m tall light house, the angles of depression of two ships approaching it are 30° and 45°. If one ship is directly behind the other, find the distance between the two ships.
Solution:
Let LH be the light house, A apd B ate two ships making angles of elevation with the top of the light house as 30° and 45° respectively.
LH = 150 m.

Question 51.
The angles of elevation of the top of a rock from the top and foot of a 100 m high tower are respectively 30° and 45°. Find the height of the rock.
Solution:
Let RS is the rock and TP is the tower. The angles of elevation of the top of rock with the top and foot of the tower are 30° and 45° respectively

Height of TP = 100 m
Let height of rock RS = h
From T, draw TQ || PS                                 _
Then QS = TP = 100 m                               ‘
and RQ = h – 100
Let PS = TQ = x
Now in right ΔRPS

Question 52.
A straight highway leads to the foot of a tower of height 50 m. From the top of the tower, the angles of depression of two cars standing on the highway are 30° and 60° respectively. What is the distance between the two cars and how far is each car from the tower ?
Solution:
Let TR be the tower and A and B are two cars on the road making angles of elevation with T the top of tower as 30° and 60°
Height of the tower TR = 50 m
Let AR = x and BR = y
Now in right ΔTAR,

Question 53.
From the top of a building AB, 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find
(i) the horizontal distance between AB and CD.
(ii) the height of the lamp post
(iii) the difference between the heights of the building and the lamp post. [CBSE 2009]
Solution:
Let AB is the building and CD is the verticle lamp
From A, the top of the building angles of depression of C and D are 30° and 60° respectively
Height of building AB = 60 m

Let height of CD = h
Draw CE || DB || AX
∴ ∠ACE = ∠XAC = 30° and ∠ADB = ∠XAD = 60°
EB = CD = h and AE = 60 – h
Let DB = CE = x
Now in right ΔACE,

Question 54.
Two boats approach a light house in mid­sea from opposite directions. The angles of elevation of the top of the light house from two boats are 30° and 45° respectively. If the distance between two boats is 100 m, find the height of the light house. [CBSE 2014]
Solution:
Let LH is the light house and A and B are two boats on the opposite directions of the light house which are making angle of elevation of the top L of the light house as 30° and 45°

Question 55.
The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower is 50 m high, what is the height of the hill ?[CBSE 2006C, 2013]
Solution:
Let TR be the tower, HL is the hill and angles of elevation of top of the hill is 60° and top of the tower is 30

Question 56.
A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/h.   [CBSE 2017]
Solution:
Let the distance BC be x m and CD be y m.

Question 57.
From the fop of a 120 m ifigh tower, a man observes two cars on the opposite sides of the tower and in straightline with the base of tower with angles of depression as 60° and 45°. Find the distance between the cars. (Take \(\sqrt { 3 } \)  = 1.732) [CBSE 2017]
Solution:
Let BD be the tower and A and C be the two points on ground.

Then, BD, the height of the tower = 120m
∠BAD = 45°, ∠BCD = 60°

Question 58.
Two points A and B are on the same side of a tower and in the same straight line with its base. The angles of depression of these points from the top of the tower are 60° and 45° respectively. If the height of the tower is 15 m, then find the distance between these points.  [CBSE2017]
Solution:
Let TR be the tower and A, B are two objects which makes angles of elevation with top of the tower as 45° and 60° respectively

Question 59.
A fire in a building B is reported on telephone to two fire stations P and Q, 20 km apart from each other on a straight road. P observes that the fire is at an angle of 60° to the road and Q observes that it is at an angle of 45° to the road. Which station should send its team and how much will this team have to travel ?
Solution:
Let B is the building one fire and P and Q the fire stations which are 20 km apart i.e., PQ = 20 km.


P and Q observes the angles with B, as 60° and 45° respectively.
Draw BA ⊥ PQ
Let AB = h Now in right ΔBAQ

Question 60.
A man on the deck of a ship is 10 m above the water level. He observes that the angle of elevation of the top of a cliff is 45° and the angle of depression of the base is 30°. Calculate the distance of the cliff from the ship and the height of the cliff.
Solution:
Let M is a man on the deck MN such that MN = 10 m, AB is the cliff
From M the angle of elevation of A is 45°
and angle of depression of B is 30°

Question 61.
A man standing on the deck of a ship, which is 8 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship and the height of the hill.
Solution:
Let M is the man on the deck MN such that MN = 8 m. AB is the hill From M, the angle of elevation of A is 60° and angle of depression of B is 30°

Question 62.
There are two temples, one on each bank of a river, just opposite to each other. One temple is 50 m high. From the top of this temple, the angles of depression of the top and the foot of the other temple are 30° and 60° respectively. Find the width of the river and the height of the other temple.
Solution:
Let AB and CD are two temples on the banks of the river.
AB = 50 m
From A, the angles of depression of the top and botttom of the other temple are 30° and 60° respectively.

Question 63.
The angle of elevation of an aeroplane from a point on the ground is 45°. After a flight of 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 3000 metres, find the speed of the aeroplane.
Solution:
Let A is the plane flying in the sky at is height of 3000 m i.e., AB = 3000 m
P is a point on the ground which from an angle of elevation of 45° at A and then after a flight of 15 seconds at A’ the angle of elevation because 30°

Question 64.
An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the the speed of the aeroplane in km/hr.
Solution:
Let A be the aeroplane and AB is the height which 1 km and make an angle of elevations of 60° from a point P on the ground After moving 10 second’s flight, the angle of elevation becomes 30° from P

A’B’=AB =1 km = 1000 m
Let PB = y and BB’ = x
Now in right ΔAPB,

Question 65.
A tree standing on a horizontal plane is leaning towards east. At two points situated at distance a and b exactly due west on it, the angles of elevation of the top are respectively a and p. Prove that the height of the top from the ground is

Solution:
Let CD is the tree which is leaning towardsEast and A and B are two points on the West making angles of elevation with top C of the tree as α and β
A and B are at the distance of a and b from the foot of the tree CD, then AD = a, BD = b

Question 66.
The angle of elevation of a stationery cloud from a point 2500 m above a lake is 15° and the angle of depression of its reflection in the lake is 45°. What is the height of the cloud above the lake level? (Use tan 15° = 0.268)
Solution:
Let C is the cloud over a lake LK
From a point M which, is 2500 m above the lake level, angle of elevation of C is 15° and angle of depression of the reflection of C in the lake which is R is 45°

Question 67.
If the angle of elevation of a cloud from a point h metres above a lake is a and the angle of depression of its reflection in the lake be p, prove that the distance of the cloud from the point of \(\frac { 2h\sec { \alpha } }{ \tan { \beta -\tan { \alpha } } }\)  [CBSE 2004]
Solution:
Let C be the cloud and from a point M which h m is above the lake level angle of elevation is α and angle of reflection of the cloud C, is β

Question 68.
From an aeroplane vertically above a straight horizontal road, the angles of depression of two consecutive mile stones on opposite sides of the aeroplane are observed to be a and p. Show that the height in miles of aeroplane above the road is given by \(\frac { \tan { \alpha } \tan { \beta } }{ \tan { \alpha } +\tan { \beta } }\)   [CBSE 2004]
Solution:
Let A is aeroplane and C and D are two such points that the angles of depression from A are a and p respectively and CD = 1 km
Let height of the plane be h

Question 69.
PQ is a post of given height a and AB is a tower at some distance. If a and p are the angles of elevation of B, the top of the tower, at P and Q respectively. Find the height of the tower and its distance from the post.
Solution:
Let PQ is post and AB is the tower Angles of elevation of B, from P and Q are a and P respectively

Question 70.
A ladder rests against a wail at an angle a to the horizontal. Its foot is pulled away from the wall through a distance a, so that it slides a distance b down the wall making an angle p with the horizontal.
Show that \(\frac { a }{ b }\)  = \(\frac { \cos { \alpha } -\cos { \beta } }{ \sin { \beta } -\sin { \alpha } }\)
Solution:
In the figure, AC and ED is the same stair, so AC = ED

Question 71.
A tower subtends an angle a at a point A in the plane of its base and the angle of depression of the foot of the tower at a point b metres just above A is p. Prove that the height of the tower is b tan α cot β.
Solution:
Let TR is the tower which subtends angle α at a point A on the same plane
AB = b and angle of depression of R from B is β

Question 72.
An observer, 1.5 m tall, is 28.5 m away from a tower 30 m high. Determine theangle of elevation of the top of the tower from his eye.
Solution:
Let TR is the tower and CD is the observer who is 28.5 m away from the tower TR
Height of the tower TR = 30 m
and height of observer CD = 1.5 m

Question 73.
A carpenter makes stools for electricians with a square top of side 0.5 m and at a height of 1.5 m above the ground. Also, each leg is inclined at an angle of 60° to the ground. Find the length of each leg and also the lengths of two steps to be put at equal distances.
Solution:
Let AC be the leg of stool whose top in a square shaped of side AB = 0.5 m
Height of stool AL = 1.5 m, and angle of inclination by the leg of the stool = 60°

Question 74.
A boy is standing on the ground and flying a kite with 100 m of string at an elevation of 30°. Another boy is standing on the roof of a 10 m high building and is flying his kite at an elevation of 45°. Both the boys are on opposite sides of both the kites. Find the length of the string that the second boy must have so that the two kites meet.
Solution:
Let K be the kite. A and B are two boys flying kites. Boy B is standing on a building 10 m high
The string AK of kite of boy A is 100 m Let h be the height of the kite and x is the length of string of kite of second boy B

Question 75.
From the top of a light house, the angles of depression of two ships on the opposite sides of it are observed to be α and β. If the height of the light house be h metres and the line joining the ships passes through the foot of the light house, show that the distance between the ship is \(\frac { h(\tan { \alpha } +\tan { \beta } ) }{ \tan { \alpha } \tan { \beta } }\)   meters.
Solution:
Let LH be the light house and A and B are two ships which make angles of elevation with L are a and p respectively

Question 76.
From the top of a tower hmetre high, the angles of depression of two objects, which are in the line with the foot of the tower are a and β (β > α). Find the distance between the two objects. [NCERT Exemplar]
Solution:
Let the distance between two objects is x m
and CD = y m
Given that, ∠BAX = α = ∠ABD                              [alternate angle]
∠CAY = β = ∠ACD                                                   [alternate angle]
and the height of tower, AD = h m Now, in ΔACD,

Question 77.
A window of a house is h metre above the ground. From the window, the angles of elevation and depression of the top and bottom of another house situated on the opposite side of the lane are found to be α and β respectively. Prove that the height of the house is h( 1 + tan α tan β) metres. [NCERT Exemplar]
Solution:
Let the height of the other house = OQ = H
and OB = MW = x m
Given that, height of the first house = WB = h = MO
and ∠QWM = α, ∠OWM =β= ∠WOB                                 [alternate angle]
Now, in ΔWOB,

Question 78.
The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30° respectively. Find the height of the balloon above the ground.[NCERT Exemplar]
Solution:
Let the height of the balloon from above the ground is H.
A and OP = w₂R = w₁Q = x
Given that, height of lower window from above the ground = w₂P = 2 m = OR
Height of upper window from above the lower window = w₁w₂ = 4 m = QR
∴ BQ = OB – (QR + RO)
BQ = H – (4 + 2)
BQ = H – 6

Hope given RD Sharma Class 10 Solutions Chapter 12 Heights and Distances Ex 12.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5

Other Exercises

• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.2
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.3
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7
• RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise
• RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS
• RD Sharma Class 10 Solutions Chapter 7 Triangles MCQS

Question 1.
In the figure, ∆ACB ~ ∆APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ. (C.B.S.E. 1991)
Solution:
In the figure,
∆ACB ~ ∆APQ
BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm

Question 2.
In the figure, AB || QR. Find the length of PB. (C.B.S.E. 1994)

Solution:
In the figure,
In ∆PQR, AB || QR
AB = 3 cm, QR =9 cm, PR = 6 cm
In ∆PAB and ∆PQR
∠P = ∠P (common)
∠PAB = ∠PQR (corresponding angles)
∠PBA = ∠PRQ (corresponding angles)
∠PAB = ∠PQR (AAA axiom)

Question 3.
In the figure, XY || BC. Find the length of XY. (C.B.S.E. 1994C)
Solution:
In the figure
In ∆ABC XY || BC
AX = 1 cm, BC = 6 cm

Question 4.
In a right angled triangle with sides a and b and hypotenuse c, the altitude drawn on hypotenuse is x. Prove that ab = cx.
Solution:
Given : In right ∆ABC, ∠B is right angle BD ⊥ AC
Now AB = a, BC = b, AC = c and BD = x

Question 5.
In the figure, ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm and AD = 4 cm, find CD.
Solution:

Question 6.
In the figure, ∠ABC = 90° and BD ⊥ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.
Solution:
In right ∆ABC, ∠B = 90°
BD ⊥ AC
AB = 5.7 cm, BD = 3.8 and CD = 5.4 cm

Question 7.
In the figure, DE || BC such that AE = \(\frac { 1 }{ 4 }\) AC. If AB = 6 cm, find AD.
Solution:
In the figure, in ∆ABC, DE || BC
AE = \(\frac { 1 }{ 4 }\) AC, AB = 6 cm

Question 8.
In the figure, if AB ⊥ BC, DC ⊥ BC and DE ⊥ AC, prove that ∆CED ~ ∆ABC.

Solution:
Given : In the figure AB ⊥ BC, DC ⊥ BC and DE ⊥ AC
To prove : ∆CED ~ ∆ABC.
Proof: AB ⊥ BC
∠B = 90°
and ∠A + ∠ACB = 90° ….(i)
DC ⊥ BC
∠DCB = 90°
=> ∠ACB + ∠DCA = 90° ….(ii)
From (i) and (ii)
∠A = ∠DCA
Now in ∆CED and ∆ABC,
∠E = ∠B (each 90°)
∠DEA or ∠DCE = ∠A (proved)
∆CED ~ ∆ABC (AA axiom)
Hence proved.

Question 9.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using similarity criterion
for two triangles, show that \(\frac { OA }{ OC }\) = \(\frac { OB }{ OD }\)
Solution:
Given : ABCD is a trapezium in which AB || DC and diagonals AC and BD intersect each other at O
To Prove : \(\frac { OA }{ OC }\) = \(\frac { OB }{ OD }\)

Question 10.
In ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively such that ∠MAP = ∠BAC. Prove that
(i) ∆ABC ~ ∆AMP
(ii) \(\frac { CA }{ PA }\) = \(\frac { BC }{ MP }\).
Solution:
Given : In ∆ABC and ∆AMP,
∠B = ∠M = 90°
∠MAP = ∠BAC

Question 11.
A vertical stick 10 cm long casts a shadow of 8 cm long. At the same time a tower casts a shadow 30 m long. Determine the height of the tower. (CB.S.E. 1991)
Solution:
The shadows are casted by a vertical stick and a tower at the same time
Their angles will be equal

Question 12.
In the figure, A = CED, prove that ∠CAB ~ ∠CED. Also find the value of x.

Solution:
Given : In ∆ABC,
∠CED = ∠A
AB = 9 cm, BE = 2 cm, EC = 10 cm, AD = 7 cm and DC = 8 cm
To prove :
(i) ∆CAB ~ ∆CED
(ii) Find the value of x
Proof: BC = BE + EC = 2 + 10 = 12 cm
AC = AD + DC = 7 + 8 = 15 cm
(i) Now in ∆CAB and ∆CED,
∠A = ∠CED (given)
∠C = ∠C (common)
∆CAB ~ ∆CED (AA axiom)

Question 13.
The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of first triangle is 9 cm, what is the corresponding side of the other triangle? (C.B.S.E. 2002C)
Solution:
Let perimeter of ∆ABC = 25 cm
and perimeter of ∆DEF = 15 cm
and side BC of ∆ABC = 9 cm
Now we have to find the side EF of ∆DEF
∆ABC ~ ∆DEF (given)

Question 14.
In ∆ABC and ∆DEF, it is being given that: AB = 5 cm, BC = 4 cm and CA = 4.2 cm; DE = 10 cm, EF = 8 cm and FD = 8.4 cm. If AL ⊥ BC and DM ⊥ EF, find AL : DM.
Solution:
In ∆ABC and ∆DEF,
AB = 5 cm, BC = 4 cm, CA = 4.2 cm, DE = 10 cm, EF = 8 cm and FD = 8.4 cm
AL ⊥ BC and DM ⊥ EF

Question 15.
D and E are the points on the sides AB and AC respectively of a ∆ABC such that: AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm. Prove that BC = \(\frac { 5 }{ 2 }\) DE.
Solution:
Given : In ∆ABC, points D and E are on the sides AB and AC respectively
and AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm


Hence proved

Question 16.
D is the mid-point of side BC of a ∆ABC. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE : EX = 3 : 1.
Solution:
Given : In ∆ABC, D is mid point of BC, and E is mid point of AD
BE is joined and produced to meet AC at X
To prove : BE : EX = 3 : 1
Construction : From D, draw DY || BX meeting AC at Y


Hence proved.

Question 17.
ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q. Prove that the rectangle obtained by BP and DQ is equal to the rectangle contained AB and BC.
Solution:
Given : ABCD is a parallelogram.
APQ is a straight line which meets BC at P and DC on producing at Q

Question 18.
In ∆ABC, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that :
(i) ∆OMA ~ ∆OLC
(ii) \(\frac { OA }{ OC }\) = \(\frac { OM }{ OL }\)
Solution:
Given : In ∆ABC, AL ⊥ BC, CM ⊥ AB which intersect each other at O

(ii) \(\frac { OA }{ OC }\) = \(\frac { OM }{ OL }\)
Hence proved.

Question 19.
ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the mid-points of AB, AC, CD and BD respectively, show that PQRS is a rhombus.
Solution:
Given : In quadrilateral ABCD, AD = BC
P, Q, R and S are the mid points of AB, AC, CD and AD respectively
PQ, QR, RS, SP are joined

Question 20.
In an isosceles ∆ABC, the base AB is produced both the ways to P and Q such that AP x BQ = AC². Prove that ∆APC ~ ∆BCQ.
Solution:
Given : In ∆ABC, AC = BC
Base AB is produced to both sides and points P and Q are taken in such a way that
AP x BQ = AC²
CP and CQ are joined
To prove : ∆APC ~ ∆BCQ

Question 21.
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp is 3.6 m above ground, find the length of her shadow after 4 seconds.
Solution:
Let AB be the lamp post and CD be the girl and AB = 3.6 m, CD = \(\frac { 90 }{ 100 }\) = 9 m
Distance covered in 4 seconds = 1.2 m x 4 = 4.8 m
BD = 4.8 m

x = 1.6
Length of her shadow = 1.6 m

Question 22.
A vertical stick of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
Let AB be stick and DE be tower.
A stick 6 m long casts a shadow of 4 m i.e., AB = 6 m and BC = 4 m
Let DE casts shadow at the same time which is EF = 28 m
Let height of tower DE = x
Now in ∆ABC and ∆DEF,
∠B = ∠E (each 90°)
∠C = ∠F (shadows at the same time)
∆ABC ~ ∆DEF (AA criterion)

Question 23.
In the figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE.

Solution:
Given: In the figure, ∆ABC is a right angled triangle right angle at C.
DE ⊥ AB
To prove:
(i) ∆ABC ~ ∆ADE
(ii) Find the length of AE and DE
Proof: In ∆ABC and ∆ADE,
∠ACB = ∠AED (each 90°)
∠BAC = ∠DAE (common)
∆ABC ~ ∆ADE (AA axiom)

Question 24.
In the figure, PA, QB and RC are each perpendicular to AC. Prove that

Solution:
Given : In the figure, PA, QB and RC are perpendicular on AC and PA = x, QB = z and RC = y



Hence proved.

Question 25.
In the figure, we have AB || CD || EF. If AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm and DE = y cm, calculate the values of x and y.

Solution:
In the figure, AB || CD || EF
AB = 6 cm, EF = 10 cm, BD = 4 cm, CD = x cm and DE = y cm
In ∆ABE, CE || AB
∆CED ~ ∆AEB

Hope given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4

Other Exercises

• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.2
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.3
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7
• RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise
• RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS
• RD Sharma Class 10 Solutions Chapter 7 Triangles MCQS

Question 1.
(i) In figure, if AB || CD, find the value of x.

(ii) In figure, if AB || CD, find the value of x.

(iii) In figure, AB || CD. If OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4, find x. (C.B.S.E. 2000C)

Solution:
(i) In the figure, AB || CD
The diagonals of a trapezium divides each other proportionally




=> (x – 8)(x – 11) = 0
Either x – 8 = 0, then x = 8 or x – 11 =0, then x = 11
x = 8 or 11

Hope given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

Mark the correct alternative in each of the following :
Question 1.
The distance between the points (cosθ, sinθ) and (sinθ, -cosθ) is
(a) √3
(b) √2
(c) 2
(d) 1
Solution:
(b) Distance between (cosθ, sinθ) and (sinθ, -cosθ)

Question 2.
The distance between the points (a cos 25°, 0) and (0, a cos 65°) is
(a) a
(b) 2a
(c) 3a
(d) None of these
Solution:
(a) Distance between (a cos 25°, 0) and (0, a cos 65°)

Question 3.
If x is a positive integer such that the distance between points P (x, 2) and Q (3, -6) is 10 units, then x =
(a) 3
(b) -3
(c) 9
(d) -9
Solution:
(c) Distance between P (x, 2) and Q (3, -6) = 10 units

=> x (x – 9) + 3 (x – 9) = 0
(x – 9) (x + 3) = 0
Either x – 9 = 0, then x = 9 or x + 3 = 0, then x = -3
x is positive integer
Hence x = 9

Question 4.
The distance between the points (a cosθ + b sinθ, 0) and (0, a sinθ – b cosθ) is
(a) a² + b²
(b) a + b
(c) a² – b²
(d) √(a²+b²)
Solution:
(d) Distance between (a cosθ + b sinθ, 0) and (0, a sinθ – b cosθ)

Question 5.
If the distance between the points (4, p) and (1, 0) is 5, then p =
(a) ±4
(b) 4
(c) -4
(d) 0
Solution:
(a) Distance between (4, p) and (1, 0) = 5

Question 6.
A line segment is of length 10 units. If the coordinates of its one end are (2, -3) and the abscissa of the other end is 10, then its ordinate is
(a) 9, 6
(b) 3, -9
(c) -3, 9
(d) 9, -6
Solution:
(b) Let the ordinate of other end = y
then distance between (2, -3) and (10, y) = 10 units

Question 7.
The perimeter of the triangle formed by the points (0, 0), (1, 0) and (0, 1) is
(a) 1 ± √2
(b) √2 + 1
(c) 3
(d) 2 + √2
Solution:
(d) Let the vertices of ∆ABC be A (0, 0), B(1, 0) and C (0, 1)

Question 8.
If A (2, 2), B (-4, -4) and C (5, -8) are the vertices of a triangle, then the length of the median through vertices C is
(a) √65
(b) √117
(c) √85
(d) √113
Solution:
(c) Let mid point of A (2, 2), B (-4, -4) be D whose coordinates will be

Question 9.
If three points (0, 0), (3, √3) and (3, λ) form an equilateral triangle, then λ =
(a) 2
(b) -3
(c) -4
(d) None of these
Solution:
(d) Let the points (0, 0), (3, √3) and (3, λ) from an equilateral triangle
AB = BC = CA
=> AB² = BC² = CA²
Now, AB² = (x₂ – x₁)² + (y₂ – y₁)²

Question 10.
If the points (k, 2k), (3k, 3k) and (3, 1) are collinear, then k
(a) \(\frac { 1 }{ 3 }\)
(b) \(\frac { -1 }{ 3 }\)
(c) \(\frac { 2 }{ 3 }\)
(d) \(\frac { -2 }{ 3 }\)
Solution:
(b) Let the points A (k, 2k), B (3k, 3k) and C (3, 1) be the vertices of a ∆ABC

Question 11.
The coordinates of the point on x-axis which are equidistant from the points (-3, 4) and (2, 5) are
(a) (20, 0)
(b) (-23, 0)
(c) (\(\frac { 4 }{ 5 }\) , 0)
(d) None of these
Solution:
(d)

Question 12.
If (-1, 2), (2, -1) and (3, 1) are any three vertices of a parallelogram, then
(a) a = 2, b = 0
(b) a = -2, b = 0
(c) a = -2, b = 6
(d) a = 0, b = 4
Solution:
(d) In ||gm ABCD, diagonals AC and AD bisect each other at O
O is mid-point of AC

Question 13.
If A (5, 3), B (11, -5) and P (12, y) are the vertices of a right triangle right angled at P, then y =
(a) -2, 4
(b) -2, 4
(c) 2, -4
(d) 2, 4
Solution:
(c) A (5, 3), B (11, -5) and P (12, y) are the vertices of a right triangle, right angle at P

Question 14.
The area of the triangle formed by (a, b + c), (b, c + a) and (c, a + b) is
(a) a + b + c
(b) abc
(c) (a + b + c)²
(d) 0
Solution:
(d) Vertices of a triangle are (a, b + c), (b, c + a) and (c, a + b)

Question 15.
If (x, 2), (-3, -4) and (7, -5) are coliinear, then x =
(a) 60
(b) 63
(c) -63
(d) -60
Solution:
(c) Area of triangle whose vertices are (x, 2), (-3, -4) and (7, -5)

Question 16.
If points (t, 2t), (-2, 6) and (3, 1) are collinear, then t =
(a) \(\frac { 3 }{ 4 }\)
(b) \(\frac { 4 }{ 3 }\)
(c) \(\frac { 5 }{ 3 }\)
(d) \(\frac { 3 }{ 5 }\)
Solution:
(b) The area of triangle whose vertices are (t, 2t), (-2, 6) and (3, 1)

Question 17.
If the area of the triangle formed by the points (x, 2x), (-2, 6) and (3, 1) is 5 square units, then x =
(a) \(\frac { 2 }{ 3 }\)
(b) \(\frac { 3 }{ 5 }\)
(c) 2
(d) 5
Solution:
(c) Area of triangle whose vertices are (x, 2x), (-2, 6) and (3, 1)

Question 18.
If points (a, 0), (0, b) and (1, 1) are collinear, then \(\frac { 1 }{ a }\) + \(\frac { 1 }{ b }\) =
(a) 1
(b) 2
(c) 0
(d) -1
Solution:
(a) The area of triangle whose vertices are (a, 0), (0, b) and (1, 1)

Question 19.
If the centroid of a triangle is (1, 4) and two of its vertices are (4, -3) and (-9, 7), then the area of the triangle is
(a) 183 sq. units
(b) \(\frac { 183 }{ 2 }\) sq. units
(c) 366 sq. units
(d) \(\frac { 183 }{ 4 }\) sq. units
Solution:
(b) Centroid of a triangle = (1, 4)
and two vertices of the triangle are (4, -3) and (-9, 7)
Let the third vertex be (x, y), then

= \(\frac { 183 }{ 2 }\) sq. units

Question 20.
The line segment joining points (-3, -4) and (1, -2) is divided by y-axis in the ratio
(a) 1 : 3
(b) 2 : 3
(c) 3 : 1
(d) 2 : 3
Solution:
(c) The point lies on y-axis
Its abscissa will be zero
Let the point divides the line segment joining the points (-3, -4) and (1, -2) in the ratio m : n

Question 21.
The ratio in which (4, 5) divides the join of (2, 3) and (7, 8) is
(a) -2 : 3
(b) -3 : 2
(c) 3 : 2
(d) 2 : 3
Solution:
(d) Let the point (4, 5) divides the line segment joining the points (2, 3) and (7, 8) in the ratio m : n

Question 22.
The ratio in which the X-axis divides the segment joining (3, 6) and (12, -3) is
(a) 2 : 1
(b) 1 : 2
(c) -2 : 1
(d) 1 : -2
Solution:
(a) The point lies on x-axis
Its ordinate is zero
Let this point divides the line segment joining the points (3, 6) and (12, -3) in the ratio m : n

Question 23.
If the centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is at the origin, then a³ + b³ + c³ =
(a) abc
(b) 0
(c) a + b + c
(d) 3 abc
Solution:
(d) Centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is origin (0, 0)

Question 24.
If points (1, 2), (-5, 6) and (a, -2) are collinear, then a =
(a) -3
(b) 7
(c) 2
(d) -2
Solution:
(b) The area of a triangle whose vertices are (1, 2), (-5, 6) and (a, -2)

Question 25.
If the centroid of the triangle formed by (7, x), (y, -6) and (9, 10) is at (6, 3), then (x, y) =
(a) (4, 5)
(b) (5, 4)
(c) (-5, -2)
(d) (5, 2)
Solution:
(d) Centroid of (7, x), (y, -6) and (9, 10) is (6, 3)

Question 26.
The distance of the point (4, 7) from the x-axis is
(a) 4
(b) 7
(c) 11
(d) √65
Solution:
(b) The distance of the point (4, 7) from x-axis = 7

Question 27.
The distance of the point (4, 7) from the y-axis is
(a) 4
(b) 7
(c) 11
(d) √65
Solution:
(a) The distance of the point (4, 7) from y-axis = 4

Question 28.
If P is a point on x-axis such that its distance from the origin is 3 units, then the coordinates of a point Q on OY such that OP = OQ, are
(a) (0, 3)
(b) (3, 0)
(c) (0, 0)
(d) (0, -3)
Solution:
(a) P is a point on x-axis and its distance from 0 is 3
Co-ordinates of P will be (3, 0)
Q is a point on OY such that OP = OQ
Co-ordinates of Q will be (0, 3)

Question 29.
If the point (x, 4) lies on a circle whose centre is at the origin and radius is 5, then x =
(a) ±5
(b) ±3
(c) 0
(d) ±4
Solution:
(b) Point A (x, 4) is on a circle with centre O (0, 0) and radius = 5

Question 30.
If the point P (x, y) is equidistant from A (5, 1) and B (-1, 5), then
(a) 5x = y
(b) x = 5y
(c) 3x = 2y
(d) 2x = 3y
Solution:
(c)

Question 31.
If points A (5, p), B (1, 5), C (2, 1) and D (6, 2) form a square ABCD, then p =
(a) 7
(b) 3
(c) 6
(d) 8
Solution:
(c) Vertices of a square are A (5, p), B (1, 5), C (2, 1) and D (6, 2)
The diagonals bisect each other at O
O is the mid-point of AC and BD
O is mid-point of BD, then

Question 32.
The coordinates of the circumcentre of the triangle formed by the points O (0, 0), A (a, 0) and B (0, b) are
(a) (a, b)
(b) (\(\frac { a }{ 2 }\) , \(\frac { b }{ 2 }\))
(c) (\(\frac { b }{ 2 }\) , \(\frac { a }{ 2 }\))
(d) (b, a)
Solution:
(b) Let co-ordinates of C be (x, y) which is the centre of the circumcircle of ∆OAB
Radii of a circle are equal

Question 33.
The coordinates of a point on x-axis which lies on the perpendicular bisector of the line segment joining the points (7, 6) and (-3, 4) are
(a) (0, 2)
(b) (3, 0)
(b) (0, 3)
(d) (2, 0)
Solution:
(d) The given point P lies on x-axis
Let the co-ordinates of P be (x, 0)
The point P lies on the perpendicular bisector of of the line segment joining the points A (7, 6), B (-3, 4)

Question 34.
If the centroid of the triangle formed by the points (3, -5), (-7, 4), (10, -k) is at the point (k, -1), then k =
(a) 3
(b) 1
(c) 2
(d) 4
Solution:
(c) O (k, -1) is the centroid of triangle whose vertices are

Question 35.
If (-2, 1) is the centroid of the triangle having its vertices at (x, 0), (5, -2), (-8, y), then x, y satisfy the relation
(a) 3x + 8y = 0
(b) 3x – 8y = 0
(c) 8x + 3y = 0
(d) 8x = 3y
Solution:
(-2, 1) is the centroid of triangle whose vertices are (x, 0), (5, -2), (-8, y)

Question 36.
The coordinates of the fourth vertex of the rectangle formed by the points (0, 0), (2, 0), (0, 3) are
(a) (3, 0)
(b) (0, 2)
(c) (-2, 3)
(d) (3, 2)
Solution:
(c) Three vertices of a rectangle are A (0, 0), B (2, 0), C (0, 3)
Let fourth vertex be D (x, y)
The diagonals of a rectangle bisect eachother at O
O is the mid-point of AC, then
Coordinates of O will be (\(\frac { 0+0 }{ 2 }\) , \(\frac { 0+3 }{ 2 }\))

Question 37.
The length of a line segment joining A (2, -3) and B is 10 units. If the abscissa of B is 10 units, then its ordinates can be
(a) 3 or -9
(b) -3 or 9
(c) 6 or 27
(d) -6 or-27
Solution:
(a) Abscissa of B is 10 and co-ordinates of A are (2, -3)
Let ordinates of B be y, then

Question 38.
The ratio in which the line segment joining P(x₁, y₁) and Q (x₂, y₂) is divided by x-axis is
(a) y₁ : y₂
(b) -y₁ : y₂
(c) x₁ : x₂
(d) -x₁ : x₂
Solution:
(b) Let a point A on x-axis divides the line segment joining the points P (x₁, y₁), Q (x₂, y₂) in the ratio m₁ : m₂ and
let co-ordinates of A be (x, 0)

Question 39.
The ratio in which the line segment joining points A (a₁, b₁) and B (a₂, b₂) is divided by y-axis is
(a) -a₁ : a₂
(b) a₁ : a₂
(c) b₁ : b₂
(d) -b₁ : b₂
Solution:
(a) Let the point P on y-axis, divides the line segment joining the point A (a₁, b₁) and B (a₂, b₂) is the ratio m₁ : m₂ and
let the co-ordinates of P be (0, y), then

Question 40.
If the line segment joining the points (3, -4) and (1, 2) is trisected at points P

Solution:
(b)

Question 41.
If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (-2, 5), then the coordinates of the other end of the diameter are [CBSE 2012]
(a) (-6, 7)
(b) (6, -7)
(c) (6, 7)
(d) (-6, -7)
Solution:
(a) Let AB be the diameter of a circle with centre O

Question 42.
The coordinates of the point P dividing the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1 are
(a) (2, 4)
(b) (3, 5)
(c) (4, 2)
(d) (5, 3) [CBSE 2012]
Solution:
(b) Point P divides the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1
Let coordinates of P be (x, y), then

Question 43.
In the figure, the area of ∆ABC (in square units) is [CBSE 2013]
(a) 15
(b) 10
(c) 7.5
(d) 2.5

Solution:
(c)

Question 44.
The point on the x-axis which is equidistant from points (-1, 0) and (5, 0) is
(a) (0, 2)
(b) (2, 0)
(c) (3, 0)
(d) (0, 3) [CBSE 2013]
Solution:
(c) Let the point P (x, 0) is equidistant from the points A (-1, 0), B (5, 0)

Question 45.
If A (4, 9), B (2, 3) and C (6, 5) are the vertices of ∆ABC, then the length of median through C is
(a) 5 untis
(b) √10 units
(c) 25 units
(d) 10 units [CBSE 2014]
Solution:
(b) A (4, 9), B (2, 3) and C (6, 5) are the vertices of ∆ABC
Let median CD has been drawn C (6, 5)

Question 46.
If P (2, 4), Q (0, 3), R (3, 6) and S (5, y) are the vertices of a paralelogram PQRS, then the value of y is
(a) 7
(b) 5
(c) -7
(d) -8 [CBSE 2014]
Solution:
(a) P (2, 4), Q (0, 3), R (3, 6) and S (5, y) are the vertices of a ||gm PQRS

Question 47.
If A (x, 2), B (-3, -4) and C (7, -5) are collinear, then the value of x is
(a) -63
(b) 63
(c) 60
(d) -60 [CBSE 2014]
Solution:
(a) A (x, 2), B (-3, -4) and C (7, -5) are collinear, then area ∆ABC = 0
Now area of ∆ABC

Question 48.
The perimeter of a triangle with vertices (0, 4) and (0, 0) and (3, 0) is
(a) 7 + √5
(b) 5
(c) 10
(d) 12 [CBSE 2014]
Solution:
(d) A (0, 4) and B (0, 0) and C (3, 0) are the vertices of ∆ABC

Question 49.
If the point P (2, 1) lies on the line segment joining points A (4, 2) and B (8, 4), then
(a) AP = \(\frac { 1 }{ 3 }\) AB
(b) AP = BP
(c) BP = \(\frac { 1 }{ 3 }\) AB
(d) AP = \(\frac { 1 }{ 2 }\) AB
Solution:
(d) Given that, the point P (2, 1) lies on the line segment joining the points (4, 2) and B (8, 4), which shows in the figure below:

Question 50.
A line intersects the y-axis and x-axis at P and Q, respectively. If (2, -5) is the mid-point of PQ, then the coordinates of P and Q are, respectively
(a) (0, -5) and (2, 0)
(b) (0, 10) and (-4, 0)
(c) (0, 4) and (-10, 0)
(d) (0, -10) and (4, 0)
Solution:
(d) Let the coordinates of P and Q (0, y) and (x, 0), respectively.
So, the mid-point of P (0, y) and Q (x, 0) is

2 = \(\frac { x + 0 }{ 2 }\) and -5 = \(\frac { y + 0 }{ 2 }\)
=> 4 = x and -10 = y
=> x = 4 and y = -10
So, the coordinates of P and Q are (0, -10) and (4, 0).

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS are helpful to complete your math homework.

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