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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 $=A B+B C+C D+D A$
$=17+8+12+9$
$=46 \mathrm{~cm}$.