Name: Area of a Quadrilateral and Rhombus
Uploaded: 2019-07-19
Duration: 8 min 51 s

Question

Find the perimeter and area of a quadrilateral $ABCD$ in which $BC = 12 cm, CD = 9 cm, BD = 15 cm, DA = 17 cm$ and $∠ ABD = 90 °$ .

Accepted Answer

The given figure is

Consider $∆ ABD$ ,

Using the Pythagoras theorem, we have,

${AD}^{2} = {AB}^{2} + {BD}^{2}$

By substituting the values

$17^{2} = {AB}^{2} + 15^{2}$

On further calculation

${AB}^{2} = 289 - 225$

$\Rightarrow {AB}^{2} = 64$

$\Rightarrow AB = \sqrt{64}$

$\Rightarrow AB = 8 cm$

Perimeter of the quadrilateral $ABCD = AB + BC + CD + AD$ .

By substituting the values

Perimeter $= 8 + 12 + 9 + 17$

$= 46 cm$

We know that area of $∆ ABD = \frac{1}{2} \times base \times height$

It can be written as

Area of $∆ ABD = \frac{1}{2} \times AB \times BD$

By substituting the values, we get,

Area of $∆ ABD = \frac{1}{2} \times 8 \times 15$

Thus, area of $∆ ABD = 60 {cm}^{2}$

Consider $∆ BCD$

We know that $BC = 12 cm, CD = 9 cm$ and $BD = 15 cm$

It can be written as $a = 12 cm, b = 9 cm$ and $c = 15 cm$

So we get

Semi-perimeter, $s = \frac{a + b + c}{2}$

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

$s = 18 cm$

By Heron's formula, we know that

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

By substituting the values

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

So we get,

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

It can be written as

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

On further calculation

Area $= 6 \times 3 \times 3$

Thus, Area of $∆ BCD$ $= 54 {cm}^{2}$

So the area of quadrilateral $ABCD =$ Area of $∆ ABD +$ Area of $∆ BCD$

By substituting the values, we get:

Area of quadrilateral $ABCD = 60 + 54 = 114 {cm}^{2}$

Hence, the perimeter is $46 cm$ and the area is $114 {cm}^{2}$ .

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