Question

Find the perimeter and area of a quadrilateral ABCD in which BC = 12 cm, CD = 9 cm, BD = 15 cm, DA = 17 cm and ∠ABD = 90°.

Answer 1

We know that $△$ABD is a right-angled triangle.
∴ $A{B}^2=\sqrt{A{D}^2-D{B}^2}=\sqrt{{17}^2-{15}^2}=\sqrt{289-225}=\sqrt{64}=8\mathrm{cm}$
$$\begin{gathered} \mathrm{Now,} \\ Area\mathrm{of}\mathrm{triangle}ABD=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ =\frac{1}{2}\times AB\times BD \\ =\frac{1}{2}\times 8\times 15 \\ =60{\mathrm{cm}}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{Let}: \\ a=9\mathrm{cm},b=15\mathrm{cm}\mathrm{and}c=12\mathrm{cm} \\ s=\frac{a+b+c}{2}=\frac{9+15+12}{2}=18\mathrm{cm} \\ \mathrm{By}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula},\mathrm{we}\mathrm{have}: \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}DBC=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{18(18-9)(18-15)(18-12)} \\ =\sqrt{18\times 9\times 3\times 6} \\ =\sqrt{6\times 3\times 3\times 3\times 3\times 6} \\ =6\times 3\times 3 \\ =54{\mathrm{cm}}^2 \end{gathered}$$

Now,
Area of quadrilateral ABCD = Area of $△$ABD + Area of $△$BCD
= (60 + 54) cm² =114 cm²
And,
Perimeter of quadrilateral ABCD = AB + BC + CD + AD = 17 + 8 + 12 + 9 = 46 cm