Question

The sides of a quadrilateral taken in order are 5, 12, 14 and 15 metres respectively, and the angle contained by the first two sides is a right angle. Find its area.

Answer 1

In quad. ABCD, AB=5 m, BC= 12 m CD= 14m, DA = 15m and $\angle ABC={90}^0$ Join AC,
Now in right $\Delta ABC$,
$A{C}^2=A{B}^2+B{C}^2=(5{)}^2+(12{)}^2$
= $25+144=169=(13{)}^2$
$\therefore AC=13m$
Now area of right $\Delta ABC$
= $\frac{1}{2}\times base\times height$
= $\frac{1}{2}\times AB\times BC$
=$\frac{1}{2}\times 5\times 12m^2=30m^2$
and in $\Delta ACD$, sides are 13 m, 14 m, 15 m
$\therefore s=\frac{a+b+c}{2}=\frac{13+14+15}{2}=\frac{42}{2}=21m$
Now area of $\Delta ACD$
= $\sqrt{s(s-a)(s-b)(s-c)}$
= $\sqrt{21(21-13)(21-14)(21-15)}$
= $\sqrt{21\times 8\times 7\times 6}$.
= $\sqrt{3\times 7\times 2\times 2\times 2\times 7\times 2\times 3}$.
= $7\times 3\times 2\times 2=84m^2$
$\therefore$ Area of quad. ABCD = 30+84= $114m^2$