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

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:
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 1
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 2
Solution:
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 3
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 4
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 5
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 6

Question 2.
What values of x will make DE || AB In the figure
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 7
Solution:
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 8

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 9

Question 4.
In the figure, DE || CB. Determine AC and AE.
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 10
Solution:
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 11
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 12

Question 5.
In the figure, given that ∆ABC ~ ∆PQR and quad ABCD ~ quad PQRS. Determine the values of x, y, z in each case.
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 13
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 14
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 15
Solution:
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 16
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 17

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 18
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 19

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.
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 20
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 21
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 22
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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 23
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 24
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 25
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 26
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 27
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 28
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 29

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 30

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 31
8x = 24
=> x = 3
AQ = 3 cm

Question 10.
In the figure, ∆AMB ~ ∆CMD; determine MD in terms of x, y and z.
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 32
Solution:
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 33

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 34
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 35

Question 12.
In the figure, l || m
(i) Name three pairs of similar triangles with proper correspondence; write similarities.
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 36
Solution:
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 37
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 38

Question 13.
In the figure, AB || DC
(i) ∆DMU ~ ∆BMV
(ii) DM x BV = BM x DU
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 39
Solution:
Given : In the figure,
ABCD is a trapezium in which
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 40

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 41

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 42

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.
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 43
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 44
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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 45
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 46

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 47

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 48

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.
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 49
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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 50
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.
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 51
Solution:
(i) In figure (i)
In ∆ABC, ∠B = 90°
BD ⊥ AC
∆ABD ~ ∆CBD
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 52
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 53
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 54
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 55

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 56

Question 23.
In ∆ABC, AD and BE are altitudes. Prove that :
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 57
Solution:
Given : In ∆ABC,
AD ⊥ BC and BE ⊥ AC
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 58
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 59

Question 24.
The diagonals of quadrilateral ABCD intersect at O. Prove that
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 60
Solution:
Given : ABCD is quadrilateral in which diagonals AC and BD intersect each other atO
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 61
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 62
= \(\frac { BO }{ DO }\) {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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 63
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 64
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 65

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.
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 66
Solution:
There are 4 steps in staircase AB
Taking first step,
In ∆ALP
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 67
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 68

Question 27.
In ∆ABC, ∠A = 60°. Prove that BC² = AB² + AC² – AB.AC.
Solution:
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 69
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 70
= 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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 71

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 72
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 73

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 74

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 75

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 76
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 77

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 78
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 79

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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 80

Question 35.
In ∆ABC, ∠ABC = 135°. Prove that: AC² = AB² + BC² + 4 ar (∆ABC).
Solution:
Given : In ∆ABC, ∠ABC = 135°,
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 81
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 82
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
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 83
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 84

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°
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 85
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 86
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 87

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.
RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise 88
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.

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