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 solutions in RD Sharma Solutions for Class 9 Mathematics offer invaluable help while preparing for the exams.

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

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Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 15 Area of Parallelograms and Triangles Exercise 15.3 Page number 15.40

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 ΔABD,

Using Pythagorean Theorem,

AB² + AD² = BD²

15² = AB² + 9²

AB² = 225−81=144

AB = 12

Area of ΔABD = 1/2(12×9) cm² = 54 cm²

In right ΔBCD:

Using Pythagorean Theorem,

CD² = BD² + BC²

17² = BD² + 8²

BD² = 289 – 64 = 225

or BD = 15

Area of ΔBCD = 1/2(8×17) cm² = 68 cm²

Now, area of quadrilateral ABCD = Area of ΔABD + Area of ΔBCD

= 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 ΔOTS if PQ = 8 cm.

Solution:

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

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

In ΔPQS ,

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

Now,

Area of ΔOTS = 1/2(TO×TS)

= 1/2(4×4) cm²

= 8cm²

Area of ΔOTS 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 ΔQRT

= PT × RT + 1/2 (QT×RT)

= 8 × RT + 1/2(8×RT)

= 12 RT

In right ΔQRT,

Using Pythagorean Theorem,

QR² = QT² + RT²

RT² = QR² − QT²

RT² = 17²−8² = 225

or RT = 15

Therefore, Area of trapezium = 12×15 cm² = 180 cm²

Question 4: In figure, ∠AOB = 90^o, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of ΔAOB.

Solution:

Given: A triangle AOB, with ∠AOB = 90^o, 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 ΔOAB:

Using Pythagorean Theorem, we get

AB² = OB² + OA²

13² = OB² + 12²

OB² = 169 – 144 = 25

or OB = 5 cm

Now, Area of ΔAOB = ½(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 ΔBMC :

Using Pythagorean Theorem, we get

BC² = BM² + MC²

25 = 16 + MC²

MC² = 25 – 16 = 9

or MC = 3

Again,

In right Δ ADL :

Using Pythagorean Theorem, we get

AD² = AL² + DL²

25 = 16 + DL²

DL² = 25 – 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 = 2√5 cm, find the area of the rectangle.

Solution:

From given:

Radius = OD = 10 cm and OE = 2√5 cm

In right ΔDEO,

By Pythagoras theorem

OD² = OE² + DE²

(10)² = (2√5 )² + DE²

100 – 20 = DE²

DE = 4√5

Now,

Area of rectangle OCDE = Length x Breadth = OE x DE = 2√5 x 4√5 = 40

Area of rectangle is 40 cm².

Question 7: In figure, ABCD is a trapezium in which AB || DC. Prove that ar(ΔAOD) = ar(ΔBOC)

Solution:

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

To Prove: ar(ΔAOD) = ar(ΔBOC)

Proof:

From figure, we can observe that ΔADC and ΔBDC are sharing common base i.e. DC and between same parallels AB and DC.

Then, ar(ΔADC) = ar(ΔBDC) ……(1)

ΔADC is the combination of triangles, ΔAOD and ΔDOC. Similarly, ΔBDC is the combination of ΔBOC and ΔDOC triangles.

Equation (1) => ar(ΔAOD) + ar(ΔDOC) = ar(ΔBOC) + ar(ΔDOC)

or ar(ΔAOD) = ar(ΔBOC)

Hence Proved.

Question 8: In figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(ΔADE) = ar(ΔBCF).

Solution:

Here, ABCD, CDEF and ABFE are parallelograms:

Which implies:

AD = BC

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

ΔADE ≅ ΔBCF

Since both the triangles are congruent, then ar(ΔADE) = ar(ΔBCF).

Hence Proved,

Question 9: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar(ΔAPB) x ar(ΔCPD) = ar(ΔAPD) x ar(ΔBPC).

Solution:

Consider: BQ and DR are two perpendiculars on AC.

To prove: ar(ΔAPB) x ar(ΔCPD) = ar(ΔAPD) x ar(ΔBPC).

Now,

L.H.S. = ar(ΔAPB) x ar(ΔCDP)

= (1/2 x AP × BQ) × (1/2 × PC × DR)

= (1/2 x PC × BQ) × (1/2 × AP × DR)

= ar(ΔAPD) x ar(ΔBPC)

= 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(ΔABC) = ar(ΔABD).

Solution:

Draw two perpendiculars CP and DQ on AB.

Now,

ar(ΔABC) = 1/2×AB×CP ⋅⋅⋅⋅⋅⋅⋅(1)

ar(ΔABD) = 1/2×AB×DQ ⋅⋅⋅⋅⋅⋅⋅(2)

To prove the result, ar(ΔABC) = ar(ΔABD), we have to show that CP = DQ.

In right angled triangles, ΔCPO and ΔDQO

∠CPO = ∠DQO = 90^o

CO = OD (Given)

∠COP = ∠DOQ (Vertically opposite angles)

By AAS condition: ΔCP0 ≅ ΔDQO

So, CP = DQ …………..(3)

(By CPCT)

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

ar(ΔABC) = ar(ΔABD)

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.