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

• Last Updated : 30 Apr, 2021

### 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,

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]

(8)2 = 4 x DC

DC = 64/4 = 16 cm

Hence, the length of side CD = 16 cm

### Question 6. In fig., âˆ ABC = 90o 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 = 90o

Required to find: BC

let âˆ BCD=x

âˆ ADB = âˆ CDB = 90o [right angled triangle]

âˆ ABD = âˆ CBD       [âˆ ABD = âˆ CBD = 90o – 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.

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]

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 = 90o   [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 = 90o

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 [each 90o]

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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