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

Given:

A quadrilateral $ABCD$ in which $AB = 17\ cm, AD = 9\ cm, CD = 12\ cm, \angle ACB = 90^o$ and $AC = 15\ cm$.

To do:

We have to find the perimeter and area of the quadrilateral.

Solution:

In right angled triangle $ABC$,

$\angle ACB = 90^o$

This implies,

$AB^2 = AC^2 + BC^2$

$(17)^2 = (15)^2+ BC^2$

$289 = 225 + BC^2$

$\mathrm{BC}^{2}=289-225$

$=64$

$=(8)^{2}$

$\Rightarrow \mathrm{BC}=8 \mathrm{~cm}$

Area of $\triangle \mathrm{ABC}=\frac{1}{2}\ base \times\ altitude$

$=\frac{1}{2} \times 8 \times 15$

$=60 \mathrm{~cm}^{2}$

Area of $\triangle \mathrm{ADC}$,

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

$=\frac{15+12+9}{2}$

$=\frac{36}{2}$

$=18$

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

$=\sqrt{18(18-15)(18-12)(18-9)}$

$=\sqrt{18 \times 3 \times 6 \times 9}$

$=\sqrt{6 \times 3 \times 3 \times 6 \times 3 \times 3}$

$= 3 \times 3 \times 6$

$=54 \mathrm{~cm}^{2}$

Area of quadrilateral $\mathrm{ABCD}=60+54$

$=114 \mathrm{~cm}^{2}$

Perimeter of the quadrilateral $=A B+B C+C D+D A$

$=17+8+12+9$

$=46 \mathrm{~cm}$.

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Updated on: 10-Oct-2022

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