The sides of a quadrangular field taken in order are $26\ m, 27\ m, 7\ m$ and $24\ m$ respectively. The angle contained by the last two sides is a right angle. Find its area.

Given:

The sides of a quadrangular field taken in order are $26\ m, 27\ m, 7\ m$ and $24\ m$ respectively. The angle contained by the last two sides is a right angle.

To do:

We have to find its area.

Solution:

Let in a quadrilateral $ABCD$,

$AB = 26\ m, BC = 27\ m CD = 7\ m, DA = 24\ m, \angle CDA = 90^o$

Join $AC$,

In $\triangle \mathrm{ACD}$,

$\mathrm{AC}=\sqrt{\mathrm{CD}^{2}+\mathrm{AD}^{2}}$

$=\sqrt{(7)^{2}+(24)^{2}}$

$=\sqrt{49+576}$

$=\sqrt{625}$

$=25 \mathrm{~m}$

Area of the triangle $ACD=\frac{1}{2} C D \times A D$

$=\frac{1}{2} \times 7 \times 24$

$=84 \mathrm{~m}^{2}$

In $\triangle \mathrm{ABC}$,

$s=\frac{a+b+c}{2}$

$=\frac{25+26+27}{2}$

$=\frac{78}{2}$

$=39$

Area of triangle $ABC=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{39(39-25)(39-26)(39-27)}$

$=\sqrt{39 \times 14 \times 13 \times 12}$

$=\sqrt{13 \times 3 \times 2 \times 7 \times 13 \times 2 \times 2 \times 3}$

$=13 \times 3 \times 2 \times \sqrt{7 \times 2}$

$=78 \times \sqrt{14}$

$=78 \times 3.74$

$=291.72 \mathrm{~m}^{2}$

Total area of quadrilateral $\mathrm{ABCD}=84 \times 291.72$ 

$=291.72+84$

$=375.72 \mathrm{~m}^{2}$.

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Updated on: 10-Oct-2022

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