# 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

CD^{2} = BD^{2} + BD^{2}

⇒ 17^{2} = BD^{2} + 8^{2}

⇒ BD^{2} = 17^{2} – 8^{2}

⇒ BD^{2} = 289 – 64

⇒ BD^{2} = 225

⇒ BD = 15 cm

since area of triangle

So , Area of triangle ΔBCD

= 60 cm^{2}

In right angled triangle ABD

BD^{2} = AB^{2} + AD^{2}

15^{2} = AB^{2} + 9^{2}

= 12 cm

Area of right triangle

= 54 cm^{2}

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

= 54 + 60

= 114 cm^{2}

Hence we get Area of quadrilateral ABCD = 114 cm^{2}

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

(3) U is the mid-point of PS which means

(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)