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Rd sharma solution 2019 class 9 chapter 14 Areas of paralllelograms and triangles

Exercise 14.3


QUESTION 1

In the given f‌igure, compute the area of quadrilateral ABCD.

Sol :

Given : Here from the given figure we gwt

(1) ABCD is a quadrilateral with base AB .

(2) ΔABD is a right angled triangle.

(3) ΔBCD is a right angled triangle with base BC right angled at B

To find : Area of quadrilateral ABCD

Calculation :

In right triangle ΔBCD , by using Pythagoreans theorem

CD2 = BD2 + BD2

⇒ 172 = BD2 + 82

⇒ BD2 = 172 – 82

⇒ BD2 = 289 – 64

⇒ BD2 = 225

⇒ BD = 15 cm

since area of triangle =\dfrac{1}{2}\times \text{base}\times \text{height}

So , Area of triangle ΔBCD =\dfrac{1}{2}\times \text{BC}\times \text{BD}

=\dfrac{1}{2}\times 8 \times 15

= 60 cm2

 

In right angled triangle ABD

BD2 = AB2 + AD2

152 = AB2 + 92

\text{AB}=\sqrt{225-81}

=\sqrt{144}

= 12 cm

Area of right triangle ΔABD=\dfrac{1}{2} \times \text{AB} \times \text{AD}

=\dfrac{1}{2}\times 12 \times 9

= 54 cm2

Area of parallelogram ABCD = area ( ΔABD) + area ( ΔBCD)

= 54 + 60

= 114 cm2

Hence we get Area of quadrilateral ABCD = 114 cm2


QUESTION 2

In the given 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

Sol :

Given : Here from the given figure we get

(1) PQRS is a square .

(2) T is a mid-point of PS which means \text{TS}=\dfrac{1}{2}\text{PS}

(3) U is the mid-point of PS which means \text{QU}=\dfrac{1}{2}\text{QR}

(4) QU = 8 cm

To find : Area of ΔOTS

Calculation: 

Since it is given that PQ = 8 cm . So

PS = SR =  QR = 8 cm (side of square are equal)

 

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