# Class 10 RD Sharma Solutions – Chapter 4 Triangles – Exercise 4.5 | Set 1

**Question 1. In fig., 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

To find: CA and AQ

We know that,

Î”ACB âˆ¼ Î”APQ [given]

BA/ AQ = CA/ AP = BC/ PQ [sides are proportional of Similar Triangles]

Now,

6.5/ AQ = 8/ 4

AQ = (6.5 x 4)/ 8

AQ = 3.25 cm

Similarly,

CA/ AP = BC/ PQ

CA/ 2.8 = 8/ 4

CA = 2.8 Ã— 2

CA = 5.6 cm

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

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

**Solution:**

Given,

Î”PQR, AB âˆ¥ QR

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

To find: length of PB

In Î”PAB and Î”PQR

We have,

âˆ P = âˆ P [Common angle ]

âˆ PAB = âˆ PQR [Corresponding angles]

âˆ PBA = âˆ PRQ [Corresponding angles]

Î”PAB âˆ¼ Î”PQR [By AAA similarity criteria]

AB/ QR = PB/ PR [sides are proportional of Similar Triangles]

â‡’ 3/ 9 = PB/6

PB = 6/3

Hence the length of PB = 2 cm

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

**Solution:**

Given,

XYâˆ¥BC

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

To find: length of XY

In Î”AXY and Î”ABC

We have,

âˆ A = âˆ A [Common angle]

âˆ AXY = âˆ ABC [Corresponding angles]

âˆ AYX = âˆ ACB [Corresponding angles]

Î”AXY âˆ¼ Î”ABC [By AAA similarity criteria]

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

Now,

(AB = AX + XB = 1 + 3 = 4)

XY/6 = 1/4

XY/1 = 6/4

Hence, length of XY = 1.5 cm

**Question 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:**

Let us consider Î”ABC to be a right angle triangle with sides a,b and c as hypotenuse. let BD be the altitude drawn on the hypotenuse AC.

To prove: ab = cx

Now,

In Î”ABC and Î”ADB

âˆ BAC = âˆ DAB [Common]

âˆ ACB = âˆ ABC = 90 [right angled triangle]

Î”ABC âˆ¼ Î”ADB [By AA similarity criteria]

So,

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

c/ a = b/ x

â‡’ xc = ab

ab = cx

Hence, proved.

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

**Solution:**

Given,

ABC is a right-angled triangle and BDâŠ¥AC.

BD = 8 cm, and AD = 4 cm

To find: CD.

Now in triangle Î”ABD and Î”CBD,

âˆ BDC=âˆ BDA [each 90]

âˆ ABD=âˆ CBD [BDâŠ¥AC]

Î”DBAâˆ¼Î”DCB [By AA similarity]

BD/ CD = AD/ BD

BD

^{2}= AD x DC(8)

^{2}= 4 x DCDC = 64/4 = 16 cm

Hence, the length of side CD = 16 cm

**Question 6. In fig., âˆ ABC = 90**^{o} and BD âŠ¥ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, Find BC.

^{o}and BD âŠ¥ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, Find BC.

**Solution:**

Given:

BD âŠ¥ AC

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

âˆ ABC = 90

^{o}Required to find: BC

let âˆ BCD=x

Now In Î”ADB and Î”CDB,

âˆ ADB = âˆ CDB = 90

^{o }[right angled triangle]âˆ ABD = âˆ CBD [âˆ ABD = âˆ CBD = 90

^{o }– x]Î”ABC âˆ¼ Î”BDC [By AA similarity]

so,

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

5.7/ 3.8 = BC/ 5.4

BC = (5.7 Ã— 5.4)/ 3.8 = 8.1

Hence, length of side BC = 8.1 cm

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

**Solution:**

Given:

DEâˆ¥BC

AE = (1/4)AC

AB = 6 cm.

To find: AD.

Now In Î”ADE and Î”ABC,

âˆ A = âˆ A [Common angle]

âˆ ADE = âˆ ABC [Corresponding angles]

Î”ADE âˆ¼ Î”ABC [By AA similarity criteria]

so,

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

AD/6 = 1/4

4 x AD = 6

AD = 6/4

AD=1.5 cm

Hence length of AD = 1.5 cm

**Question 8. In the fig. given, if AB âŠ¥ BC, DC âŠ¥ BC, and DE âŠ¥ AC, prove that Î”CED âˆ¼ Î”ABC**

**Solution:**

Given:

AB âŠ¥ BC,

DC âŠ¥ BC,

DE âŠ¥ AC

To prove: Î”CEDâˆ¼Î”ABC

Now In Î”ABC and Î”CED,

âˆ B = âˆ E = 90

^{o}[given]âˆ BAC = âˆ ECD [alternate angles]

Î”CEDâˆ¼Î”ABC [AA similarity]

Hence Proved

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

**Solution:**

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

To prove: OA/ OC = OB/ OD

Now In Î”AOB and Î”COD,

âˆ AOB = âˆ COD [Vertically Opposite Angles]

âˆ OAB = âˆ OCD [Alternate angles]

Î”AOB âˆ¼ Î”COD

so,

OA/ OC = OB/ OD [Corresponding sides are proportional]

Hence Proved

**Question 10. If Î” ABC and Î” AMP are two right triangles, right angled at B and M, respectively such that âˆ MAP = âˆ BAC. Prove that**

**(i) Î”ABC âˆ¼ Î”AMP**

**(ii) CA/ PA = BC/ MP**

**Solution:**

(i) Given:

Î” ABC and Î” AMP are the two right triangles.

Now In Î”ABC and Î”AMP,

âˆ AMP = âˆ B = 90

^{o}âˆ MAP = âˆ BAC [Vertically Opposite Angles]

Î”ABCâˆ¼Î”AMP [AA similarity]

(ii) Since, Î”ABCâˆ¼Î”AMP

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

Hence, proved.

**Question 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

To find: the height of the tower = PQ.

Now In Î”ABC and Î”PQR,

âˆ ABC = âˆ PQR = 90

^{o }[each 90^{o}]âˆ ACB = âˆ PRQ [angle are made at the same time]

Î”ABC âˆ¼ Î”PQR [By AA similarity]

So,

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

10/8 = PQ/ 3000

PQ = (3000Ã—10)/ 8

PQ = 30000/8

PQ = 3750/100

Hence the length of PQ = 37.5 m

**Question 12. In fig., âˆ A = âˆ CED, prove that Î”CAB âˆ¼ Î”CED. Also**,** find the value of x.**

**Solution:**

Given:

âˆ A = âˆ CED

To prove: Î”CAB âˆ¼ Î”CED

In Î”CAB and Î”CED

âˆ C = âˆ C [Common]

âˆ A = âˆ CED [Given]

Î”CAB âˆ¼ Î”CED [By AA similarity]

so,

CA/ CE = AB/ ED [by Corresponding sides are proportional]

15/10 = 9/x

x = (9 Ã— 10)/ 15

Hence, value for x = 6 cm

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