RD Sharma 2020 solution class 9 chapter 14 Area of Parallelograms and Triangles VSAQS

VSAQS

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Question 1:

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

Answer 1:

Given: (1) ΔABC is equilateral triangle.

(2) ΔBDE is equilateral triangle.

(3) D is the midpoint of BC.

To find:

PROOF : Let us draw the figure as per the instruction given in the question.

We know that area of equilateral triangle = , where a is the side of the triangle.

Let us assume that length of BC is a cm.

This means that length of BD is cm, Since D is the midpoint of BC.

------(1)

------(2)

Now, ar(ΔABC) : ar(ΔBDE) = (from 1 and 2)

=

Hence we get the result ar(ΔABC) : ar(ΔBDE) =

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Question 2:

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

Answer 2:

Given: (1) ABCD is a rectangle.

(2) CD = 6 cm

(3) AD = 8cm

To find: Area of rectangle CDEF.

Calculation: We know that,

Area of parallelogram = base × height

The Area of parallelogram and a rectangle on the same base and between the same parallels are equal in area.

Here we can see that rectangle ABCD and Parallelogram CDEF are between the same base and same parallels.

Hence,

Hence we get the result as Area of Rectangle CDEF =

Question 3:

In the given figure, find the area of ΔGEF.

Answer 3:

Given: (1) ABCD is a rectangle.

(2) CD = 6 cm

(3) AD = 8cm

To find: Area of ΔGEF.

Calculation: We know that,

Area of Parallelogram = base × height

If a triangle and a parallelogram are on the same base and between the same parallels , the area of the triangle is equal to half of the parallelogram

Here we can see that Parallelogram ABCD and triangle GEF are between the same base and same parallels.

Hence,

Hence we get the result as

Question 4:

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

Answer 4:

Given: (1) ABCD is a rectangle.

(2) AB = 10 cm

(3) AD = 5cm

To find: Area of ΔEGF.

Calculation: We know that,

Area of Rectangle = base × height

If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to half of the parallelogram

Here we can see that Rectangle ABCD and triangle GEF are between the same base and same parallels.

Hence,

Hence we get the result as Area of ΔGEF =

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 (ΔRAS)

Answer 5:

Given: Here from the given figure we get

(1) PQRS is a rectangle inscribed in a quadrant of a circle with radius 10cm,

(2) PS = 5cm

(3) PR = 13cm(radius of the quadrant)

To find: Area of ΔRAS.

Calculation: In right ΔPSR, (Using Pythagoras Theorem)

Hence we get the Area of ΔRAS =

Question 6:

In square ABCD, P and Q are mid-point of AB and CD respectively. If AB = 8cm and PQ and BD intersect at O, then find area of ΔOPB.

Answer 6:

Given: Here from the given question we get

(1) ABCD is a square,

(2) P is the midpoint of AB

(3) Q is the midpoint of CD

(4) PQ and BD intersect at O.

(5) AB = 8cm

To find : Area of ΔOPB

Calculation: Since P is the midpoint of AB,

BP = 4cm ……(1)

Hence we get the Area of ΔOBP =

 

Question 7:

ABC is a triangle in which D is the mid-point of BC. E and F are mid-points of DC and AE respectively. IF area of ΔABC is 16 cm2, find the area of ΔDEF.

Answer 7:

Given: Here from the given question we get

(1) ABC is a triangle

(2) D is the midpoint of BC

(3) E is the midpoint of CD

(4) F is the midpoint of A

Area of ΔABC = 16 cm2

To find : Area of ΔDEF

Calculation: We know that ,

The median divides a triangle in two triangles of equal area.

For ΔABC, AD is the median

For ΔADC , AE is the median .

Similarly, For ΔAED , DF is the median .

Hence we get Area of ΔDEF =

Question 8:

PQRS is a trapezium having PS and QR as parallel sides. A is any point on PQ and B is a point on SR such that AB || QR. If area of ΔPBQ is 17cm2, find the area of ΔASR.

Answer 8:

Given: Here from the given figure we get

(1) PQRS is a trapezium having PS||QR

(2) A is any point on PQ

(3) B is any point on SR

(4) AB||QR

(5) Area of ΔBPQ = 17 cm2

To find : Area of ΔASR.

Calculation: We know that ‘If a triangle and a parallelogram are on the same base and the same parallels, the area of the triangle is equal to half the area of the parallelogram’

Here we can see that:

Area (ΔAPB) = Area (ΔABS) …… (1)

And, Area (ΔAQR) = Area (ΔABR) …… (2)

Therefore,

Area (ΔASR) = Area (ΔABS) + Area (ΔABR)

From equation (1) and (2), we have,

Area (ΔASR) = Area (ΔAPB) + Area (ΔAQR)

⇒ Area (ΔASR) = Area (ΔBPQ) = 17 cm2

Hence, the area of the triangle ΔASR is 17 cm2.

Question 9:

ABCD is a parallelogram. P is the mid-point of AB. BD and CP intersect at Q such that CQ : QP = 3.1. If ar (ΔPBQ) = 10cm2, find the area of parallelogram ABCD.

Answer 9:

It is given that CQ : QP = 3: 1 and Area (PBQ) = 10 cm2

Let CQ = x and QP = 3x

We need to find area of the parallelogram ABCD.

From the figure,

Area (PBQ) =

And,

Area (BQC) =

Now, let H be the perpendicular distance between AP and CD. Therefore,

Area (PCB) = …… (1)

Thus the area of the parallelogram ABCD is,

Area (ABCD) = AB × H

⇒ Area (ABCD) = 2BP × H

From equation (1), we get

Area (ABCD) = 4 × 30 = 120 cm2

Hence, the area of the parallelogram ABCD is 120 cm2.

Question 10:

P is any point on base BC of ΔABC and D is the mid-point of BC. DE is drawn parallel to PA to meet AC at E. If ar (ΔABC) = 12 cm2, then find area of ΔEPC.

Answer 10:

Given: Area (ABC) = 12 cm2, D is midpoint of BC and AP is parallel to ED. We need to find area of the triangle EPC.

Since, AP||ED, and we know that the area of triangles between the same parallel and on the same base are equal. So,

Area (APE) = Area (APD)

⇒ Area (APM) + Area (AME) = Area (APM) + Area (PMD)

⇒ Area (AME) = Area (PMD) …… (1)

Since, median divide triangles into two equal parts. So,

Area (ADC) = Area (ABC) = = 6 cm2

⇒ Area (ADC) = Area (MDCE) + Area (AME)

⇒Area (ADC) = Area (MDCE) + Area (PMD) (from equation (1))

⇒ Area (ADC) = Area (PEC)

Therefore,

Area (PEC) = 6 cm2.

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