Question

Find the perimeter and area of the quadrilateral ABCD in which AB = 21 cm, $\angle BAC={90}^{\circ },$ AC = 20 cm, CD = 42 cm and AD = 34 cm.

Answer 1

ANSWER:
In right angled ∆ ABC,
$B{C}^2=A{B}^2+A{C}^2$
(Pythagoras Theorem)
⇒ $B{C}^2={21}^2+{20}^2$
⇒ $B{C}^2$= 441 + 400
⇒ $B{C}^2$ = 841
⇒ BC = 29 cm
Area of ∆ ABC = $\frac{1}{2}\times AB\times AC$
= $\frac{1}{2}\times 21\times 20$
= 210 $c{m}^2$
....(1)
In ∆ ACD,
The sides of the triangle are of length 20 cm, 34 cm and 42 cm.
∴ Semi-perimeter of the triangle is
s=20+34+422=962=48 cm
∴ By Heron's formula,
Area of ∆ACD=$\sqrt{s(s-a)(s-b)(s-c)}$

= $\sqrt{48(48-20)(48-34)(48-42)}$

= $\sqrt{48\ast 28\ast 14\ast 6}$

=336 $c{m}^2$
Thus,
Area of quadrilateral ABCD = Area of ∆ ABC + Area of ∆ ACD
= (210 + 336) $c{m}^2$
= 546 $c{m}^2$
Also,
Perimeter of quadrilateral ABCD = (34 + 42 + 29 + 21) cm
= 126 cm
Hence, the perimeter and area of quadrilateral ABCD is 126 cm and 546 $c{m}^2$ , respectively.