Question

Calculate the area of quadrilateral $ABCD$, in which $\angle A={90}^{\circ }$, $△BCD$ is an equilateral triangle of side $24cm$ and $AD=26cm$.

• A. $369.41c{m}^2$
• B. $349.41c{m}^2$
• C. $269.41c{m}^2$
• D. $249.41c{m}^2$

Answer 1

The correct option is A $369.41c{m}^2$

Given,
ABD is a right angled triangle and
BCD is an equilateral triangle.
$AB=A{D}^2-B{D}^2$
$=>AB={26}^{-}{24}^2$
$=>AB=676-576$
$=>AB=\sqrt{100}$
$=>AB=10$
Area of $△ABC=\frac{1}{2}\times AB\times BD$
$=\frac{1}{2}\times 10\times 24$
$=120c{m}^2$
Area of $△BCD=\frac{\sqrt{3}}{4}B{D}^2$
=$\frac{\sqrt{3}}{4}\times 24\times 24$
=$144\sqrt{3}c{m}^2$
Area of ABCD=Area of $△ABC$+Area of $△BCD$
=$120+144\sqrt{3}c{m}^2$
=$369.12c{m}^2$