Question

Find the area of quadrilateral ABCD in which AD = 24 cm, ∠BAD = 90° and ∆BCD is an equilateral triangle having each side equal to 26 cm. Also, find the perimeter of the quadrilateral. [Given: $\sqrt{3}=1.73]$]

Answer 1

We know that $△$BAD is a right-angled triangle.

∴ $AB=\sqrt{B{D}^2-A{D}^2}=\sqrt{{26}^2-{24}^2}=\sqrt{676-576}=\sqrt{100}=10\mathrm{cm}$

$$\begin{gathered} \mathrm{Now,} \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}BAD=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ =\frac{1}{2}\times AB\times AD \\ =\frac{1}{2}\times 10\times 24 \\ =120{\mathrm{cm}}^2 \end{gathered}$$

Also, we know that $△$BDC is an equilateral triangle.
$$\begin{gathered} \therefore \mathrm{Area}\mathrm{of}\mathrm{equilateral}\mathrm{triangle}=\frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 \\ =\frac{\sqrt{3}}{4}\times (26{)}^2 \\ =\frac{\sqrt{3}}{4}\times 676 \\ =169\sqrt{3} \\ =292.37{\mathrm{cm}}^2 \end{gathered}$$

Now,
Area of quadrilateral ABCD = Area of $△$ABD + Area of $△$BDC
= (120 + 292.37) cm² = 412.37 cm²