# RD Sharma Solutions for Class 9 Maths Chapter 15 Area of Parallelograms and Triangles Exercise 15.3

Get free RD Sharma Solutions for Class 9 Maths Chapter 15 exercise 15.3 here. In this exercise, students will learn figures on the same base and between the same parallels, mainly parallelograms and triangles. The step by step solution in RD Sharma Solution for Class 9 Mathematics offers invaluable help while preparing for exams.

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Question 1: In figure, compute the area of quadrilateral ABCD.

Solution:

A quadrilateral ABCD with DC = 17 cm, AD = 9 cm and BC = 8 cm (Given)

In right Î”ABD,

Using Pythagorean Theorem,

152 = AB2 + 92

AB2 = 225âˆ’81=144

AB = 12

Area of Î”ABD = 1/2(12Ã—9) cm2 = 54 cm2

In right Î”BCD:

Using Pythagorean Theorem,

CD2 = BD2 + BC2

172 = BD2 + 82

BD2 = 289 â€“ 64 = 225

or BD = 15

Area of Î”BCD = 1/2(8×17) cm2 = 68 cm2

Now, area of quadrilateral ABCD = Area of Î”ABD + Area of Î”BCD

= 54 cm2 + 68 cm2

= 112 cm2

Question 2: In figure, PQRS is a square and T and U are, respectively, the mid-points of PS and QR . Find the area of Î”OTS if PQ = 8 cm.

Solution:

T and U are mid points of PS and QR respectively (Given)

Therefore, TU||PQ => TO||PQ

In Î”PQS ,

T is the mid-point of PS and TO||PQ

So, TO = 1/2 PQ = 4 cm

(PQ = 8 cm given)

Also, TS = 1/2 PS = 4 cm

[PQ = PS, As PQRS is a square)

Now,

Area of Î”OTS = 1/2(TOÃ—TS)

= 1/2(4Ã—4) cm2

= 8cm2

Area of Î”OTS is 8 cm2.

Question 3: Compute the area of trapezium PQRS in figure.

Solution:

From figure,

Area of trapezium PQRS = Area of rectangle PSRT + Area of Î”QRT

= PT Ã— RT + 1/2 (QTÃ—RT)

= 8 Ã— RT + 1/2(8Ã—RT)

= 12 RT

In right Î”QRT,

Using Pythagorean Theorem,

QR2 = QT2 + RT2

RT2 = QR2 âˆ’ QT2

RT2 = 172âˆ’82 = 225

or RT = 15

Therefore, Area of trapezium = 12Ã—15 cm2 = 180 cm2

Question 4: In figure, âˆ AOB = 90o, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Î”AOB.

Solution:

Given: A triangle AOB, with âˆ AOB = 90o, AC = BC, OA = 12 cm and OC = 6.5 cm

As we know, the midpoint of the hypotenuse of a right triangle is equidistant from the vertices.

So, CB = CA = OC = 6.5 cm

AB = 2 CB = 2 x 6.5 cm = 13 cm

In right Î”OAB:

Using Pythagorean Theorem, we get

AB2 = OB2 + OA2

132 = OB2 + 122

OB2 = 169 â€“ 144 = 25

or OB = 5 cm

Now, Area of Î”AOB = Â½(Base x height) cm2 = 1/2(12 x 5) cm2 = 30cm2

Question 5: In figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.

Solution:

Given: ABCD is a trapezium, where AB = 7 cm, AD = BC = 5 cm, DC = x cm, and

Distance between AB and DC = 4 cm

Consider AL and BM are perpendiculars on DC, then

AL = BM = 4 cm and LM = 7 cm.

In right Î”BMC :

Using Pythagorean Theorem, we get

BC2 = BM2 + MC2

25 = 16 + MC2

MC2 = 25 â€“ 16 = 9

or MC = 3

Again,

Using Pythagorean Theorem, we get

25 = 16 + DL2

DL2 = 25 â€“ 16 = 9

or DL = 3

Therefore, x = DC = DL + LM + MC = 3 + 7 + 3 = 13

=> x = 13 cm

Now,

Area of trapezium ABCD = 1/2(AB + CD) AL

= 1/2(7+13)4

= 40

Area of trapezium ABCD is 40 cm2.

Question 6: In figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 2âˆš5 cm, find the area of the rectangle.

Solution:

From given:

Radius = OD = 10 cm and OE = 2âˆš5 cm

In right Î”DEO,

By Pythagoras theorem

OD2 = OE2 + DE2

(10)2 = (2âˆš5 )2 + DE2

100 â€“ 20 = DE2

DE = 4âˆš5

Now,

Area of rectangle OCDE = Length x Breadth = OE x DE = 2âˆš5 x 4âˆš5 = 40

Area of rectangle is 40 cm2.

Question 7: In figure, ABCD is a trapezium in which AB || DC. Prove that ar(Î”AOD) = ar(Î”BOC)

Solution:

ABCD is a trapezium in which AB || DC (Given)

To Prove: ar(Î”AOD) = ar(Î”BOC)

Proof:

From figure, we can observe that Î”ADC and Î”BDC are sharing common base i.e. DC and between same parallels AB and DC.

Î”ADC is the combination of triangles, Î”AOD and Î”DOC. Similarly, Î”BDC is the combination of Î”BOC and Î”DOC triangles.

Equation (1) => ar(Î”AOD) + ar(Î”DOC) = ar(Î”BOC) + ar(Î”DOC)

or ar(Î”AOD) = ar(Î”BOC)

Hence Proved.

Question 8: In figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(Î”ADE) = ar(Î”BCF).

Solution:

Here, ABCD, CDEF and ABFE are parallelograms:

Which implies:

DE = CF and

AE = BF

Again, from triangles ADE and BCF:

AD = BC, DE = CF and AE = BF

By SSS criterion of congruence, we have

Since both the triangles are congruent, then ar(Î”ADE) = ar(Î”BCF).

Hence Proved,

Question 9: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar(Î”APB) x ar(Î”CPD) = ar(Î”APD) x ar(Î”BPC).

Solution:

Consider: BQ and DR are two perpendiculars on AC.

To prove: ar(Î”APB) x ar(Î”CPD) = ar(Î”APD) x ar(Î”BPC).

Now,

L.H.S. = ar(Î”APB) x ar(Î”CDP)

= (1/2 x AP Ã— BQ) Ã— (1/2 Ã— PC Ã— DR)

= (1/2 x PC Ã— BQ) Ã— (1/2 Ã— AP Ã— DR)

= ar(Î”APD) x ar(Î”BPC)

= R.H.S.

Hence proved.

Question 10: In figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar(Î”ABC) = ar(Î”ABD).

Solution:

Draw two perpendiculars CP and DQ on AB.

Now,

ar(Î”ABC) = 1/2Ã—ABÃ—CP â‹…â‹…â‹…â‹…â‹…â‹…â‹…(1)

ar(Î”ABD) = 1/2Ã—ABÃ—DQ â‹…â‹…â‹…â‹…â‹…â‹…â‹…(2)

To prove the result, ar(Î”ABC) = ar(Î”ABD), we have to show that CP = DQ.

In right angled triangles, Î”CPO and Î”DQO

âˆ CPO = âˆ DQO = 90o

CO = OD (Given)

âˆ COP = âˆ DOQ (Vertically opposite angles)

By AAS condition: Î”CP0 â‰… Î”DQO

So, CP = DQ â€¦â€¦â€¦â€¦..(3)

(By CPCT)

From equations (1), (2) and (3), we have

ar(Î”ABC) = ar(Î”ABD)

Hence proved.

## RD Sharma Solutions for Class 9 Maths Chapter 15 Area of Parallelograms and Triangles Exercise 15.3

RD Sharma Solutions Class 9 Maths Chapter 15 Area of Parallelograms and Triangles Exercise 15.3 is based on the topic – Parallelogram on the same base and between the same parallels. In this exercise, students will learn how to compute the area of any quadrilateral which contains triangles.