Question

The side of quadrangular field taken in order are 26 m, 24 m, 7 m, 27 m, respectively. The angle contained by the first two sides is a right angle. Find its area

Answer 1

Given:- Sides of quadrilateral are 26m, 24m, 7m and 27m and the angle contained by the first two sides is 90°.

To Find:- Area of quadrilateral.

Solution :-

Area of quadrilateral ABCD = Area of ΔADC + Area of ∆ABC

✏ Area of ∆ADC,

$\sf \: = \dfrac{1}{2} \times 29 \times 7$

$\sf \: \boxed{ \sf{= 84m {}^{2}} }$

✏ We've to find AC,

By using Pythagoras theorem,

AC = √(24)² + (7)²m

AC = √625m

AC = 25m

✏ By Heron's Formula, In ∆ABC,

$\sf \: S = \dfrac{Sum \: of \: sides}{2} = \dfrac{24 + 27 + 25}{2} = 39$

Area of ∆ABC = √s( s-a )( s-b )( s-c )

ㅤㅤㅤ= √[39(39-24)(39-27)(39-25)]

ㅤㅤㅤ= √[√39(13)(12)(14)]

ㅤㅤㅤ= √85176

ㅤㅤㅤ=291.85m²

∴ Area of ABCD:-

= 291.85 + 84 m²

=375.85m²