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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.
Given:
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.
To do:
We have to find its area.
Solution:
Let in quadrilateral $ABCD$,
$AB = 5\ m, BC = 12\ m, CD = 14\ m, DA = 15\ m$ and $\angle ABC = 90^o$
Join $AC$.
In right angled triangle $ABC$,
By Pythagoras theorem,
$AC^2 = AB^2 + BC^2$
$= 5^2 + (12)^2$
$= 25 + 144$
$= 169$
$= (13)^2$
$\Rightarrow AC = 13\ m$
Area of right angled triangle $ABC=\frac{1}{2} \times base \times height$
$=\frac{1}{2} \times \mathrm{AB} \times \mathrm{BC}$
$=\frac{1}{2} \times 5 \times 12$
$=30 \mathrm{~m}^{2}$
In $\triangle \mathrm{ACD}$
$a=13 \mathrm{~m}, b=14 \mathrm{~m}, c=15 \mathrm{~m}$
$s=\frac{a+b+c}{2}$
$=\frac{13+14+15}{2}$
$=\frac{42}{2}$
$=21 \mathrm{~m}$
Area of triangle $\mathrm{ACD}=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{21(21-13)(21-14)(21-15)}$
$=\sqrt{21 \times 8 \times 7 \times 6}$
$=\sqrt{3 \times 7 \times 2 \times 2 \times 2 \times 7 \times 2 \times 3}$
$=7 \times 3 \times 2 \times 2$
$=84 \mathrm{~m}^{2}$
Area of quadrilateral $\mathrm{ABCD}=30+84$
$=114 \mathrm{~m}^{2}$.