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 $\mathrm{ABCD}=30+84$

$=114 \mathrm{~m}^{2}$.

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Updated on: 10-Oct-2022

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