## 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.

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