Question

The area of the quadrilateral whose sides measure 9 cm, 40 cm, 28 cm and 15 cm and in which the angle between the first two sides is a right angle is

Answer 1

92 cm is the correct answer . 92 cm

Answer 2

$\sf \large \underline{ \color{aqua}SOLUTION:-}$

•Let ABCD be the given quadrilateral in which it is given,

$\sf BC=40cm,CD=28cm \\ \sf,DA=15cm \: and \angle ABC=90⁰$

$\sf \blue{ \underline{by \: Pythagoras \: theorum:}}$

$\sf AC= \sqrt{AB²+BC²} cm= \sqrt{(9)²+(40)²} cm$

$\sf \implies \sqrt{1681} cm=41cm$

$\sf \green{Area \: of \: \triangle ABC}={ \bigg(} \frac{1}{2} ×AB×BC {\bigg)}$

$\sf \implies{ \bigg(}\frac{1}{2} \times 9 \times 40 {\bigg)}$

$\sf \implies 180cm²</p><p></p><p>$

$\sf \red {In \triangle ACD}$

$\sf Let \: a=AC=41cm \\ \sf,b=CD=28cm \\ \sf ,c=DA=15cm$

$\sf \therefore s= \frac{a+b+c}{2} = \frac{41 + 28 + 15}{2} = 42cm$

$\sf \therefore(s-a)=1cm,(s-b)=14cm \: and \: (s-c)=27cm$

$\sf \therefore Area \: of \: \triangle ACD= \sqrt{s(s-a)(s-b)(s-c)}$

$\sf \implies\sqrt{42×1×14×27} cm²$

$\sf \implies (14×3×3)cm²=126cm²$

$\sf \purple{ Now,Area \: of \: quadrilateral \: ABCD=}$

$= \sf \green{area \: of \: \triangle ABC}+ \red{area \: of \: \triangle ACD}$

$= \sf (180+126)cm²=306cm²$