## 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.

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