Find the area of quadrilateral ABCD in which AB=17 cm ,AD=9 cm,CD=12 cm,angle ACB=90 and AC=15 cm

COULD YU EXPLAIN ME THAT IN THESE TYPE OF QUESTIONS IN WHICH ORDER SHALL THE FIGURE MUST BE DRAWN

Question

Find the area of quadrilateral ABCD in which AB=17 cm ,AD=9
cm,CD=12 cm,angle ACB=90 and AC=15 cm
COULD YU EXPLAIN ME THAT IN THESE TYPE OF QUESTIONS IN WHICH
ORDER SHALL THE FIGURE MUST BE DRAWN

Shridhar Kaushik · Accepted Answer

Given that: AB = 17cm

AD = 9cm

CD = 12cm

AC = 15cm

and ∠ACB = 90°

In right angled triangle ACB,

${(BC)}^{2} = {(17)}^{2} - (15)^{2} = 289 - 225 = 64 \Rightarrow BC = 8 cm$

We have to find the area of quadrilateral ABCD.

Area of Quadrilateral ABCD = Area of ΔACB + Area of ΔACD ....... (1)

Now, Area of right angled triangle ACB

$= \frac{1}{2} \times Base \times height = \frac{1}{2} \times 15 \times 8 {cm}^{2} = 60 {cm}^{2} ........... (2)$

We will calculate the area of triangle ACD using Heron's Formula as :

$Area of Δ ACD = \sqrt{s (s - a) (s - b) (s - c)}$

Where a, b, c are the sides of triangle and S is the semi-perimeter of the triangle and is given by

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

Here, a = 15cm, b = 12cm, c = 9cm.

$∴ S = \frac{15 + 12 + 9}{2} cm = \frac{36}{2} cm = 18 cm$

$∴ Area of Δ ACD = \sqrt{18 (18 - 15) (18 - 12) (18 - 9)} {cm}^{2} = \sqrt{18 \times 3 \times 6 \times 9} {cm}^{2} = \sqrt{18 \times 18 \times 9} {cm}^{2} = 18 \times 3 {cm}^{2} = 54 {cm}^{2} ........... (3)$

∴ From (1),

Area of quadrilateral ABCD = (60 + 54) cm²

=114 cm²

Find the area of quadrilateral ABCD in which AB=17 cm ,AD=9 cm,CD=12 cm,angle ACB=90 and AC=15 cm COULD YU EXPLAIN ME THAT IN THESE TYPE OF QUESTIONS IN WHICH ORDER SHALL THE FIGURE MUST BE DRAWN

Find the area of quadrilateral ABCD in which AB=17 cm ,AD=9 cm,CD=12 cm,angle ACB=90 and AC=15 cm

COULD YU EXPLAIN ME THAT IN THESE TYPE OF QUESTIONS IN WHICH ORDER SHALL THE FIGURE MUST BE DRAWN