Question

Calculate the area of quadrilateral ABCD in which angle ABD=90° , triangle BCD is an equilateral triangle of side 24cm and AD=26cm

Answer 1

Answer:

$\text{Area of ABCD}\:=(12 \times 12 \sqrt{3})+ \frac{1}{2}\times 10 \times 24\\</p><p>=24(6\sqrt{3}+5) unit^2$

Step-by-step explanation:

Add areas of the right angles triangle ABD and equilateral triangle BCD.

Answer 2

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

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

$\rm \: Area \: of \: equilateral \: \triangle BCD = \frac{ \sqrt{3} }{4} {a}^{2}$

$\rm \implies \: \frac{ \sqrt{3} }{4} \times 24 \times 24$

$\rm \implies \: { \sqrt{3} } \times 6\times 24$

$\rm \fbox{\rm \: Area \: of \: equilateral \: \triangle BCD = 249.408 {cm}^{2} }$

Now find the length of AB by Pythagoras Theorem

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

$\implies\rm {26}^{2}={AB}^{2}+{24}^{2}$

$\implies\rm {AB}^{2} = {26}^{2} - {24}^{2}$

$\implies\rm {AB}^{2} = 100$

$\implies\rm {AB} = \sqrt{100}$

$\fbox{\rm {AB} = 10}$

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

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

$\rm \: \implies \frac{1}{2} \times \rm \: 10 \times \rm \: 24$

$\fbox{\rm \: Area \: of \: \triangle \: ABD = 120 {cm}^{2} }$

$\rm \: Area \: of \: quadrilateral \: ABCD = Area \: of \: \triangle ABD + Area \: of \: \triangle BCD$

$\rm \implies \: 249.408 + 120$

$\rm \implies \: 369.408$

$\fbox{ \rm \: Area \: of \: quadrilateral \: ABCD =369.41}$