RD Sharma Class 9 Solutions Chapter 15 VSAQS (Updated for 2024)

RD Sharma Class 9 Solutions Chapter 15 VSAQS: These solutions are created by our experts in a detailed manner. This exercise helps students to revise all the concepts they have studied in this chapter. They will help you during the class 9 exam preparation. You can get direct access to RD Sharma Class 9 Solutions Chapter 15 VSAQS from the below article. 

• RD Sharma Class 9 Maths Solutions

Access RD Sharma Class 9 Solutions Chapter 15 VSAQS

Question 1: If ABC and BDE are two equilateral triangles such that D is the mid-point of BC, then find ar(△ABC): ar(△BDE).

Solution:

Given: ABC and BDE are two equilateral triangles.

We know, the area of an equilateral triangle = √3/4 (side)²

Let “a” be the side measure of the given triangle.

Find ar(△ABC):

ar(△ABC) = √3/4 (a) ²

Find ar(△BDE):

ar(△BDE) = √3/4 (a/2) ²

(D is the mid-point of BC)

Now,

ar(△ABC) : ar(△BDE)

or √3/4 (a) ² : √3/4 (a/2) ²

or 1 : 1/4

or 4:1

This implies, ar(△ABC) : ar(△BDE) = 4:1

Question 2: In the figure, ABCD is a rectangle in which CD = 6 cm, AD = 8 cm. Find the area of parallelogram CDEF.

Solution:

ABCD is a rectangle, where CD = 6 cm and AD = 8 cm (Given)

From figure: Parallelogram CDEF and rectangle ABCD on the same base and between the same parallels, which means both have equal areas.

Area of parallelogram CDEF = Area of rectangle ABCD ….(1)

Area of rectangle ABCD = CD x AD = 6 x 8 cm² = 48 cm²

Equation (1) => Area of parallelogram CDEF = 48 cm².

Question 3: In the figure, find the area of ΔGEF.

Solution:

From figure:

Parallelogram CDEF and rectangle ABCD are on the same base and between the same parallels, which means both have equal areas.

Area of CDEF = Area of ABCD = 8 x 6 cm²= 48 cm²

Again,

Parallelogram CDEF and triangle EFG are on the same base and between the same parallels, then

Area of a triangle = ½(Area of parallelogram)

In this case,

Area of a triangle EFG = ½(Area of parallelogram CDEF) = 1/2(48 cm²) = 24 cm².

Question 4: In the figure, ABCD is a rectangle with sides AB = 10 cm and AD = 5 cm. Find the area of ΔEFG.

Solution:

From figure:

Parallelogram ABEF and rectangle ABCD are on the same base and between the same parallels, which means both have equal areas.

Area of ABEF = Area of ABCD = 10 x 5 cm²= 50 cm²

Again,

Parallelogram ABEF and triangle EFG are on the same base and between the same parallels, then

Area of a triangle = ½(Area of parallelogram)

In this case,

Area of a triangle EFG = ½(Area of parallelogram ABEF) = 1/2(50 cm²) = 25 cm².

Question 5: PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then find ar(△RAS).

Solution:

PQRS is a rectangle with PS = 5 cm and PR = 13 cm (Given)

In △PSR:

Using Pythagoras’ theorem,

SR² = PR² – PS² = (13) ² – (5) ² = 169 – 25 = 114

or SR = 12

Now,

Area of △RAS = 1/2 x SR x PS

= 1/2 x 12 x 5

= 30

Therefore, the Area of △RAS is 30 cm².

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