Question

Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm AD = 9cm, CD = 12cm, $\angle ACB={90}^0$and AC = 15cm.

Answer 1

In right $\Delta ABC,\angle ACB={90}^0$
$A{B}^2=A{C}^2+B{C}^2$

$(17{)}^2=(15{)}^2+B{C}^2$

$\Rightarrow B{C}^2=289-225=64=(8{)}^2$
$\therefore$ BC= 8 cm
Now Area of $\Delta ABC$,
= $\frac{1}{2}base\times altitude$
=$\frac{1}{2}\times 8\times 15=60c{m}^2$
and area of $\Delta$ ADC,
$s=\frac{a+b+c}{2}=\frac{15+12+9}{2}=\frac{36}{2}=18$
$\therefore$ Area= $\sqrt{s(s-a)(s-b)(s-c)}$
=$\sqrt{18(18-15)(18-12)(18-9)}$
= $\sqrt{18\times 3\times 6\times 9}$= $\sqrt{6\times 3\times 3\times 6\times 3\times 3}$
= $3\times 3\times 6=54c{m}^2$
$\therefore$ Area of quadrilateral ABCD = 60+54
=$114c{m}^2$
and perimeter = AB+BC+CD+DA
= 17+8+12+9=46 cm