Question

find the area of a quadrilateral ABCD whose sides are 9m ,40 m, 28 m, and 15 m respectively and the angle between the first two sides is a right angle.​

Answer 1

Answer

Refer to the attachment

Answer 2

$\huge \underline\textbf{Solution}$

Let ABCD be the given quadrilateral such that

∠$\green{\bold{ABC \:=\:90dgree}}$

$\green{\bold{AB \:=\:9\:m}}$

$\green{\bold{BC\:=\:40\:m}}$

$\green{\bold{CD\:=\:28\:m}}$

$\green{\bold{AD\:=\:m}}$

In △ ABC, we have

$\blue{\bold{ {AC}^{2}\:=\: {AB}^{2}\:+\: {BC}^{2}}}$

$\blue{\bold{ {AC}^{2}\:=\: {9}^{2}\:+\: {40}^{2}\:=\:1681}}$

$\blue{\bold{AC\:=\:41\:m}}$

Now,

$\blue{\bold{Area\:of\:ABC\:=\: \frac{1}{2}(base\:×\:Height)}}$

$\blue{\bold{Area\:of\:ABC\:=\:\frac{1}{2}(AB\:×\:BC)}}$

$\blue{\bold{Area\:of\:ABC\:=\:\frac{1}{2}(9\:×\:40)\: {m}^{2}\:=\:180\: {m}^{2}}}$

In △ACD, we have

$\green{\bold{AC\:=\:41\:m}}$

$\green{\bold{CD\:=\:28\:m}}$

$\green{\bold{DA\:=\:15\:m}}$

$\blue{\bold{Let\:a\:=\:AC\:=\:41\:m}}$

$\blue{\bold{b\:=\:CD\:=\:28\:m}}$

$\blue{\bold{c\:=\:DA\:=\:15\:m}}$

Then,

$\blue{\bold{s\:=\: \frac{1}{2}(41\:+\:28\:+\:15)\:=\:42}}$

$\blue{\bold{Area\:of\:ACD\:=\: \sqrt{s(s-a)(s-b)(s-c)}}}$

$\blue{\bold{Area\:of\:ACD\:=\: \sqrt{42 (42-41)(42-28)(42-15)}}}$

$\blue{\bold{Area\:of\:ACD\:=\: \sqrt{42×1×14×27}}}$

$\blue{\bold{ \sqrt{14×3×14×27}}}$

$\blue{\bold{14\:×\:9\:=\: 126\:{m}^{2}}}$

Area of quadrilateral ABCD = (Area of △ABC) + (Area of △ACD)

$\sf\huge\underline\pink{(180\:+\:126)\:{m}^{2}\:=\:306\:{m}^{2}}$