Question

Diagonals $\overline{AC}$ and $\overline{BD}$ of quadrilateral $ABCD$ are perpendicular. $AD=DC=8,AC=BC=6,m\angle ADC={60}^o$. The area of $ABCD$ is

• A. $4\sqrt{5}+8\sqrt{3}$
• B. $16\sqrt{3}$
• C. $32\sqrt{3}$
• D. $8\sqrt{5}+16\sqrt{3}$
• E. $48$

Answer 1

Answer: (D)

$8\sqrt{5}+16\sqrt{3}$

The altitude to the base of an isosceles triangle also bisects the vertex angle, so $m\angle ADE=30$.

With the hypotenuse of the triangle having a length of $8$, $AE=4$ and $DE=4\sqrt{3}$.

$△AEC$ is a right angle with leg $4$ and hypotenuse $6$.

Use the pythagorean theorem to determine that

$BE=\sqrt{6^2-4^2}=\sqrt{36-16}=\sqrt{20}=2\sqrt{5}$

The area of the quadrilateral with perpendicular diagonals is equal to half the product of the diagonals, so the area of $ABCD$ is:

$A=\frac{1}{2}\times 8\times (2\sqrt{5}+4\sqrt{3})=4(2\sqrt{5}+4\sqrt{3})=8\sqrt{5}+16\sqrt{3}$