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# Class 10 RD Sharma Solutions – Chapter 4 Triangles – Exercise 4.6 | Set 2

• Last Updated : 26 May, 2021

### Question 11. The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other.

Solution:

Let us consider âˆ†ABC and âˆ†DEF, AL and DM are the medians of âˆ†ABC and âˆ†DEF

It is given that the area of âˆ†ABC = 121 cm2 and area of âˆ†DEF = 64 cm2

AL = 12.1 cm

Let us assume DM = x cm

Given that, âˆ†ABC ~ âˆ†DEF

So,

ar(âˆ†ABC)/ar(âˆ†DEF) = AL2/DM2

= 121/64 = (12.1)2/x2

11/8 = 12.1/x

â‡’ x = (8 Ã— 12.1)/11 = 8.8

Hence, the median of the second triangle is 8.8cm

### Question 12. In âˆ†ABC ~ âˆ†DEF such that AB = 5 cm and (âˆ†ABC) = 20 cm2 and area (âˆ†DEF) = 45 cm2, determine DE.

Solution:

Given that,

area (âˆ†ABC) = 20 cmÂ²

area (âˆ†DEF) = 45 cmÂ²

AB = 5 cm

Let us consider DE = x cm

Also, given that âˆ†ABC ~ âˆ†DEF

ar(âˆ†ABC)/ar(âˆ†DEF) = AB2/DE2

â‡’20/45 = (5)2/x2

â‡’20/45 = 25/x

â‡’x2 = (25 Ã— 45)/20 = 225/4 = (15/2)2

x = 15/2 = 7.5

DE = 7.5cm

### Question 13. In âˆ†ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divide âˆ†ABC into two parts equal in area. Find BP/AB.

Solution:

It is given that, in âˆ†ABC, PQ || BC and line PQ divide the âˆ†ABC into two parts

âˆ†APQ and trap. BPQC equally

i.e., area âˆ†APQ = area BPQC

Now we have to find BP/AB.

As we know that PQ||BC

So, âˆ†APQ âˆ¼ âˆ†ABC

â‡’ ar.(âˆ†APQ)/ar.(âˆ†ABC) = AP2/AB2

â‡’ ar.(âˆ†ABC)/ar.(âˆ†APQ) = AB2/AP2

2/1 = AB2/AP2

{area âˆ†APQ = area trap. BPQC

Area âˆ†ABC = 2area (âˆ†APQ)}

â‡’ AB/AP = âˆš2/1

â‡’âˆš2 AP = AB = AP + PB

â‡’âˆš2AP – AP = PB

â‡’(âˆš2 – 1)AP = PB

BP/AP = (âˆš2 – 1)/1

### Question 14. The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR.

Solution:

Given that, area (âˆ†ABC) : area (âˆ†PQR) = 9 : 16

âˆ†ABC ~ âˆ†PQR

and BC = 4.5 cm

Let us considered QR = x cm

As we know that âˆ†ABC ~ âˆ†PQR

ar.(âˆ†ABC)/ar.(âˆ†PQR) = BC2/QR2 â‡’ 9/16 = (4.5)2/x2

â‡’ (3/4)2 = (4.5/x)2 â‡’ 4.5/x = 3/4

x = (4.5 Ã— 4)/3 = 60

Hence, the length of QR is 6cm

### Question 15. ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of âˆ†APQ is one sixteenth of the area of âˆ†ABC.

Solution:

Given that, in âˆ†ABC, P and Q are two points on line AB and AC

AP = 1 cm, PB = 3 cm, AQ = 1.5 cm and QC = 4.5 cm

Now, AP/PB = 1/3 and AQ/QC = 1.5/4.5 = 1/3

In âˆ†APQ and âˆ†ABC

AP/PB = AQ/QC

PQ||BC

Hence, âˆ†APQ âˆ¼ âˆ†ABC

So, ar.(âˆ†APQ)/ar.(âˆ†ABC) = AP2/PB2 = AP2/(AP + PB)2

ar.(âˆ†APQ)/ar.(âˆ†ABC) = 12/(1 + 3)2 = 1/16

Hence, area of âˆ†APQ = 1/16 of area of âˆ†ABC

### Question 16. If D is a point on the side AB of âˆ†ABC such that AD : DB = 3 : 2 and E is a point on BC such that DE || AC. Find the ratio of areas of âˆ†ABC and âˆ†BDE.

Solution:

Given that in âˆ†ABC, D is a point on AB such that AD : DB = 3 : 2

DE||AC

In âˆ†BDE and âˆ†ABC

âˆ BDE = âˆ A

âˆ DBE = âˆ ABC

So, by AA, âˆ†BED âˆ¼ âˆ†ABC

Therefore, ar.(âˆ†ABC)/ar.(âˆ†BDE) = AB2/BD2 = (BD + AD)2/BD2

= (2 + 3)2/22 = 52/22 = 25/4

Hence, the ratio of areas of âˆ†ABC and âˆ†BDE is 25:4

### Question 17. If âˆ†ABC and âˆ†BDE are equilateral triangles, where D is the mid point of BC, find the ratio of areas of âˆ†ABC and âˆ†BDE.

Solution:

Given that âˆ†ABC and âˆ†DBE are equilateral triangles, where D is mid point of BC

So, BD = 1/2BC

Now area of âˆ†ABC

âˆš3/4(side)2 = âˆš3/4BC2

and area of âˆ†DBE

âˆš3/4(side)2 = âˆš3/4BD2

âˆš3/4(side)2 = âˆš3/4(1/2BC)2

âˆš3/4(side)2 = âˆš3/16(BC)2

So, the ratio between areas is

= area(âˆ†ABC)/area(âˆ†DBE) =

Hence, the ratio of areas of âˆ†ABC and âˆ†BDE is 4:1

### Question 18. Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights.

Solution:

Let us consider two triangles, âˆ†ABC and âˆ†XYZ and these triangles have equal vertical angle, i.e., âˆ A and âˆ X

And AD and XO is the heights of these triangles.

So, âˆ†ABC/âˆ†XYZ = AB/AC = XY/XZ

In âˆ†ABC and âˆ†XYZ

âˆ A = âˆ X

AB/AC = XY/XZ

So, by SAS

âˆ†ABC ~ âˆ†XYZ

As we know that ar(âˆ†ABC)/ar(âˆ†XYZ) = 36/25

So,

Hence, the ratio of their corresponding heights is 6:5

### Question 19. In the figure, âˆ†ABC and âˆ†DBC are on the same base BC. If AD and BC intersect at O, prove that

Solution:

Given that two âˆ†ABC and âˆ†DBC are on the same base BC as shown in the given figure

AC and BD intersect each other at O

Now, Draw AL âŠ¥ BC and DM âŠ¥BC

Prove:

Proof:

In âˆ†ALO and âˆ†DMO,

âˆ L =âˆ M = 90Â°

âˆ AOL = âˆ DOM [Vertically opposite angles]

So, by AA, âˆ†ALO âˆ¼ âˆ†DMO

So, AL/DM = AO/DO

Now

But AL/DM = AO/DO      (Proved above)

So,

Hence proved

### (a)ar(âˆ†AOB)/ar(âˆ†COD)        (b)ar(âˆ†AOD)/ar(âˆ†COD)

Solution:

Given that ABCD is a trapezium in which AB || CD and the diagonals AC and BD intersect at O

Now, in the figure from point D, draw DLâŠ¥AC

(i) In âˆ†AOB and âˆ†COD

âˆ AOB =âˆ COD            [Vertically opposite angles]

âˆ OAB =âˆ OCD           [Alternate angles]

So, by AA criterion

âˆ†AOB ~ âˆ†COD

(ii) Given that OA = 6 cm, OC = 8 cm

As we know that âˆ†AOB ~ âˆ†COD

So, OA/OC = OB/OD = AB/CD

(a) ar(âˆ†AOB)/ar(âˆ†COD) = AO2/OC2

= 62/82 = 36/64 = 9/16

Therefore, ar(âˆ†AOB)/ar(âˆ†COD) = 9/16

(b) As we know that âˆ†AOD and âˆ†COD have their bases on the same line and their vertex A is common

Therefore,  ar(âˆ†AOD)/ar(âˆ†COD) = AO/OC = 6/8 = 3/4

### Question 21. In âˆ†ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of âˆ†APQ and trapezium BPQC.

Solution:

Given that, ABC is a triangle, in which P divides the side AB such that

AP : PB = 1 : 2. Q is a point in AC such that PQ || BC

In âˆ†APQ and âˆ†ABC

âˆ APQ = âˆ B

âˆ PAQ = âˆ BAC

So, by AA criterion

âˆ†APQ âˆ¼ âˆ†ABC

So,

ar(âˆ†APQ)/ar(âˆ†ABC) = (AP)2/(AB)2

ar(âˆ†APQ)/ar(âˆ†ABC) = (1)2/(1 + 2)2 = (1)2/(3)2 = 1/9

9 ar(âˆ†APQ) = ar(âˆ†ABC)

9 ar(âˆ†APQ) = ar(âˆ†APQ) + ar(trap. BPQC)

9 ar(âˆ†APQ) = ar(trap BPQC)

ar(âˆ†APQ)/ar(trap BPQC) = 1/9

Hence, the ratio of the areas of âˆ†APQ and trapezium BPQC is 1:9

### Question 22. AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed. Prove that Area (âˆ†ADE) : Area (âˆ†ABC) = 3 : 4.

Solution:

Given that AD is an altitude of an equilateral triangle ABC.

Prove: Area (âˆ†ADE) : Area (âˆ†ABC) = 3 : 4

Proof:

Area of âˆ†ABC = âˆš3/4 BC2