There are several criteria for proving the similarity of triangles. All these criteria are discussed in this exercise. Students who want to understand these criteria in detail can refer to the RD Sharma Solutions Class 10 for securing high or full marks in the examinations. Also, make use of the RD Sharma Solutions for Class 10 Maths Chapter 4 Triangles Exercise 4.5 PDF provided below for further assistance.

## RD Sharma Solutions for Class 10 Maths Chapter 4 Triangles Exercise 4.5 Download PDF

### Access RD Sharma Solutions for Class 10 Maths Chapter 4 Triangles Exercise 4.5

**1. In fig. 4.136,Â Î”ACB âˆ¼ Î”APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ.**

**Solution: **

Given,

Î”ACB âˆ¼ Î”APQ

BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm

Required to find: CA and AQ

We know that,

Î”ACB âˆ¼ Î”APQ [given]

BA/ AQ = CA/ AP = BC/ PQ [Corresponding Parts of Similar Triangles]

So,

6.5/ AQ = 8/ 4

AQ =Â (6.5 x 4)/ 8

AQ = 3.25 cm

Similarly, as

CA/ AP = BC/ PQ

CA/ 2.8 = 8/ 4

CA = 2.8 x 2

CA = 5.6 cm

Hence, CA = 5.6 cm and AQ = 3.25 cm.

**2. In fig.4.137, **AB âˆ¥ QR**, find the length of PB.**

**Solution: **

Given,

Î”PQR, AB âˆ¥ QR and

AB = 3 cm, QR = 9 cm and PR = 6 cm

Required to find: PB

In Î”PAB and Î”PQR

We have,

âˆ PÂ = âˆ PÂ [Common]

âˆ PAB =Â âˆ PQR [Corresponding angles as AB||QR with PQ as the transversal]

â‡’ Î”PAB **âˆ¼** Î”PQR [By AA similarity criteria]

Hence,

AB/ QR = PB/ PR [Corresponding Parts of Similar Triangles are propositional]

â‡’ 3/ 9 = PB/6

PB = 6/3

Therefore, PB = 2 cm

**3. In fig.Â 4.138 given,Â **XYâˆ¥BC**. Find the length of XY.**

**Solution: **

Given,

XYâˆ¥BC

AX = 1 cm, XB = 3 cm and BC = 6 cm

Required to find: XY

In Î”AXY and Î”ABC

We have,

âˆ AÂ = âˆ AÂ [Common]

âˆ AXY =Â âˆ ABC [Corresponding angles as AB||QR with PQ as the transversal]

â‡’ Î”AXY **âˆ¼** Î”ABC [By AA similarity criteria]

Hence,

XY/ BC = AX/ AB [Corresponding Parts of Similar Triangles are propositional]

We know that,

(AB = AX + XBÂ = 1 + 3 = 4)

XY/6 = 1/4

XY/1 = 6/4

Therefore, XY = 1.5 cm

**4. In a right-angled triangle with sides a and b and hypotenuse c, the altitude drawn on the hypotenuse is x. Prove that ab = cx.**

**Solution: **

ConsiderÂ Î”ABCÂ to be a right angle triangle having sides a and b and hypotenuse c. Let BD be the altitude drawn on the hypotenuse AC.

Required to prove: ab = cx

We know that,

In Î”ACB and Î”CDB

âˆ BÂ = âˆ BÂ [Common]

âˆ ACBÂ = âˆ CDB = 90^{o}

â‡’ Î”ACB **âˆ¼** Î”CDB [By AA similarity criteria]

Hence,

AB/ BD = AC/ BC [Corresponding Parts of Similar Triangles are propositional]

a/ x = c/ b

â‡’ xc = ab

Therefore, ab = cx

**5. In fig. 4.139,Â **âˆ ABC**Â = 90 andÂ **BDâŠ¥AC**. If BD = 8 cm, and AD = 4 cm, find CD.**

**Solution: **

Given,

âˆ ABCÂ = 90^{o} andÂ BDâŠ¥AC

BD = 8 cm

AD = 4 cm

Required to find: CD.

We know that,

ABC is a right angled triangle andÂ BDâŠ¥AC.

Then,Â Î”DBAâˆ¼Î”DCBÂ [By AA similarity]

BD/ CD = AD/ BD

BD^{2}Â = AD x DC

(8)^{2}Â = 4 x DC

DC = 64/4 = 16 cm

Therefore, CD = 16 cm

**6. In fig.4.140,Â âˆ ABCÂ = 90 ^{o} andÂ BD âŠ¥ AC. If AC = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, Find BC.**

**Solution: **

Given:

BD âŠ¥ AC

AC = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm

âˆ ABCÂ = 90^{o}

Required to find: BC

We know that,

Î”ABC âˆ¼ Î”BDC [By AA similarity]

âˆ BCAÂ = âˆ DCA = 90^{o}

âˆ AXY =Â âˆ ABC [Common]

Thus,

AB/ BD = BC/ CD [Corresponding Parts of Similar Triangles are propositional]

5.7/ 3.8 = BC/ 5.4

BC = (5.7 x 5.4)/ 3.8 = 8.1

Therefore, BC = 8.1 cm

**7. In the fig.4.141 given,Â **DE âˆ¥ BC**Â such that AE = (1/4)AC. If AB = 6 cm, find AD.**

**Solution: **

Given:

DEâˆ¥BC

AE = (1/4)AC

AB = 6 cm.

Required to find: AD.

In Î”ADE and Î”ABC

We have,

âˆ AÂ = âˆ AÂ [Common]

âˆ ADE =Â âˆ ABC [Corresponding angles as AB||QR with PQ as the transversal]

â‡’ Î”ADE **âˆ¼** Î”ABC [By AA similarity criteria]

Then,

AD/AB = AE/ AC [Corresponding Parts of Similar Triangles are propositional]

AD/6 = 1/4

4 x AD = 6

AD = 6/4

Therefore, AD = 1.5 cm

**8. In the fig.4.142 given, if AB âŠ¥ BC,Â DC âŠ¥ BC, andÂ DE âŠ¥ AC, prove thatÂ Î”CED âˆ¼ Î”ABC**

**Solution: **

Given:

AB âŠ¥ BC,

DC âŠ¥ BC,

DE âŠ¥ AC

Required to prove:Â Î”CEDâˆ¼Î”ABC

We know that,

FromÂ Î”ABC andÂ Î”CED

âˆ BÂ =Â âˆ EÂ = 90^{o}Â Â [given]

âˆ BACÂ =Â âˆ ECDÂ Â Â Â [alternate angles since, AB || CD with BC as transversal]

Therefore,Â Î”CEDâˆ¼Î”ABCÂ [AA similarity]

**9. Diagonals AC and BD of a trapezium ABCD withÂ **AB âˆ¥ DC**Â intersect each other at the point O. Using similarity criterion for two triangles, show thatÂ **OA/ OC = OB/ OD**Â Â **

**Solution: **

Given: OC is the point of intersection of AC and BD in the trapezium ABCD, withÂ AB âˆ¥ DC.

Required to prove:Â OA/ OC = OB/ OD

We know that,

InÂ Î”AOB andÂ Î”COD

âˆ AOBÂ =Â âˆ COD [Vertically Opposite Angles]

âˆ OABÂ =Â âˆ OCDÂ Â Â [Alternate angles]

Then,Â Î”AOB âˆ¼ Î”COD

Therefore,Â OA/ OC = OB/ ODÂ Â [Corresponding sides are proportional]

**10. IfÂ Î” ABC andÂ Î” AMP are two right triangles, right angled at B and M, respectively such thatÂ âˆ MAPÂ =Â âˆ BAC. Prove that**

**(i)Â Î”ABC âˆ¼ Î”AMP**

**(ii)Â CA/ PA = BC/ MP**

**Solution: **

(i)Â Given:

Î”Â ABC andÂ Î”Â AMP are the two right triangles.

We know that,

âˆ AMPÂ =Â âˆ BÂ = 90^{o}

âˆ MAPÂ =Â âˆ BAC [Vertically Opposite Angles]

â‡’ Î”ABCâˆ¼Î”AMPÂ Â Â [AA similarity]

(ii)Â Since,Â Î”ABCâˆ¼Î”AMP

CA/ PA = BC/ MPÂ [Corresponding sides are proportional]

Hence proved.

**11. A vertical stick 10 cm long casts a shadow 8 cm long. At the same time, a tower casts a shadow 30 m long. Determine the height of the tower.**

**Solution: **

Given:

Length of stick = 10cm

Length of the stickâ€™s shadow = 8cm

Length of the towerâ€™s shadow = 30m = 3000cm

Required to find: the height of the tower = PQ.

In Î”ABC âˆ¼ Î”PQR

âˆ ABCÂ =Â âˆ PQR = 90^{o}

âˆ ACBÂ =Â âˆ PRQ [Angular Elevation of Sun is same for a particular instant of time]

â‡’ Î”ABC âˆ¼ Î”PQRÂ Â [By AA similarity]

So, we have

AB/BC = PQ/QR [Corresponding sides are proportional]

10/8 = PQ/ 3000

PQ =Â (3000×10)/ 8

PQ =Â 30000/8

PQ =Â 3750/100

Therefore, PQ = 37.5 m

**12. In fig.4.143,Â âˆ AÂ =Â âˆ CED, prove thatÂ Î”CAB âˆ¼ Î”CED. Also find the value of x.**

**Solution: **

Given:

âˆ A**Â =Â **âˆ CED

Required to prove: Î”CAB âˆ¼ Î”CED

In Î”CAB âˆ¼ Î”CED

âˆ C **= **âˆ C**Â **[Common]

âˆ A**Â =Â **âˆ CED [Given]

â‡’ Î”CAB âˆ¼ Î”CED [By AA similarity]

Hence, we have

CA/ CE = AB/ EDÂ [Corresponding sides are proportional]

15/10 = 9/x

x = (9 x 10)/ 15

Therefore, x = 6 cm