Question

Find the perimeter and area of the quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and ∠CBD = 90°.

Answer 1

We know that $△$DBC is a right-angled triangle.
$$\begin{gathered} \therefore \mathrm{Area}\mathrm{of}\mathrm{triangle}DBC=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ =\frac{1}{2}\times BC\times BD \\ =\frac{1}{2}\times 20\times 21 \\ =210{\mathrm{cm}}^2 \end{gathered}$$

Now,
$$\begin{gathered} \mathrm{Let}: \\ a=42\mathrm{cm},b=34\mathrm{cm}\mathrm{and}c=20\mathrm{cm} \\ s=\frac{a+b+c}{2}=\frac{42+34+20}{2}=48\mathrm{cm} \\ \mathrm{By}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula},\mathrm{we}\mathrm{have}: \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}ABD=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{48(48-42)(48-34)(48-20)} \\ =\sqrt{48\times 6\times 14\times 28} \\ =\sqrt{4\times 2\times 6\times 6\times 7\times 2\times 7\times 4} \\ =4\times 2\times 6\times 7 \\ =336{\mathrm{cm}}^2 \end{gathered}$$

∴ Area of quadrilateral ABCD = Area of $△$DBC + Area of $△$ABD
=(210 + 336) cm² = 546 cm²
Also,
Perimeter of quadrilateral ABCD = AB + BC + CD + AD = 42 + 21 + 29 + 34 = 126 cm