Question

Find the area and the perimeter of quadrilateral ABCD. If $AB=8$ cm, $AD=10$ cm, $BD=12$ cm, $DC=13$ cm and $\angle DBC={90}^{\circ }$.

• A. $69.7c{m}^2$ and $36cm$
• B. $69.7c{m}^2$ and $38.97cm$
• C. $88.63c{m}^2$ and $36cm$
• D. None of these

Answer 1

The correct option is A $69.7c{m}^2$ and $36cm$

Given,
$AB=8cm$ and $AD=10cm$

$BD=12cm$ and $CD=13cm$

$\angle DBC=90$

$△DBC$ is a right angle $△$.

$\therefore D{C}^2=D{B}^2+B{C}^2$

$=>B{C}^2=D{C}^2-D{B}^2$

$=>B{C}^2={13}^2-{12}^2$

$=>B{C}^2=169-144$

$=>B{C}^2=25$

$=>BC=5cm$

$\therefore$ Area of $△ABC=\frac{1}{2}\times BD\times BC$

$=\frac{1}{2}\times 13\times 5$

$=30c{m}^2$

In $△ABD,S=\frac{AB+BD+AD}{2}$

$=\frac{8+12+10}{2}$

$=15cm$

Area of $△ABD=\sqrt{S(S-AC)(S-BD)(S-AD)}$

$=\sqrt{15(15-8)(15-12)(15-10)}$

$=\sqrt{15\left(7\right)\left(3\right)\left(5\right)}$

$=39.68c{m}^2$

$\therefore$ Area of Quadrilateral ABCD= Area of $△ABD$+Area of $△DBC$
$=(39.68+30)c{m}^2$

$=69.68c{m}^2$

$\simeq 69.7c{m}^2$

Perimeter of Quadrilateral $ABCD=AB+BC+CD+AD$

$=(8+5+13+10)cm$

$=36cm$