Question

Calculate the area of quadrilateral ABCD in which AB=32cm , AD=24cm angle A = 90° and BC=CD =52cm

Answer 1

Refer attachment please ...

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Answer 2

Find the area of quadrilateral by adding the area's of ∆ABD , ∆BCD

$\underline{ \rm{Area \: of \: \: \triangle ABD}}$

$\rm{Area \: of \: \: \triangle ABD = \frac{1}{2} \times AB \times AD}$

$\rm{ = \frac{1}{2} \times 32 \times 24}$

$\rm \fbox{Area \: of \: \: \triangle ABD = 384 {cm}^{2} }$

Now find the length of BD by Pythagoras Theorem

$\implies\rm{BD}^{2} = {AB }^{2}+ {AD}^{2}$

$\implies\rm{BD}^{2} = {32 }^{2}+ {24}^{2}$

$\implies \rm{BD}^{2} = 1024+ 576$

$\implies\rm{BD}^{2} = 1600$

$\implies\rm{BD}= \sqrt{ 1600}$

$\fbox{\rm{BD}= 40cm}$

$\underline{ \rm{Area \: of \: \: isosceles \: \triangle BCD}}$

$\rm{Area \: of \: \: isosceles \: \triangle BCD = \frac{1}{4} b \sqrt{4 {a}^{2} - {b}^{2} } }$

a = equal side , b = base

$\implies \rm{ \frac{1}{4} \times 40 \sqrt{4 ({52})^{2} - {40}^{2} } }$

$\implies \rm{ 10 \sqrt{4 ({2704})- {1600} } }$

$\implies \rm{ 10 \sqrt{10816 - 1600 } }$

$\implies \rm{ 10 \sqrt{9216} }$

$\implies \rm{ 10 \times 96 }$

$\fbox{\rm{Area \: of \: \: isosceles \: \triangle BCD = 960 {cm}^{2} }}$

$\rm Area \: of \: quadrilateral = Area \: of \: \triangle \: ABD + Area \: of \: \triangle BCD</p><p>$

$\implies \rm \: 384 + 960$

$\fbox{ \therefore\rm Area \: of \: quadrilateral =1344 {cm}^{2} }$