Question

The sides of ABCD are 6 cm, 8 cm, 12 cm and 14 cm ( take in order ) respectively
and the angle between first two sides is a right angle. Find its area.
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Answer 1

Given :-

Length of side A = 6 cm

Length of side B = 8 cm

Length of side C = 12 cm

Length of side D = 14 cm

The angle between first two sides is a right angle.

To Find :-

The area.

Solution :-

Given that,

ABCD is a quadrilateral having sides AB = 6 cm, BC = 8 cm, CD = 12 cm and DA = 14 cm. Now, join AC.

(Please refer the attachment for the figure)

According to the question,

ABC is a right angled triangle at B

Using Pythagoras theorem,

$\underline{\boxed{\sf AC^{2}=AB^{2}+BC^{2}}}$

Substituting their values,

$\sf =6^{2}+8^{2}=36+64=100$

Taking positive root square,

$\sf AC=10 \ cm$

Area of the quadrilateral ABCD = Area of ΔABC + Area of ΔACD

$\underline{\boxed{\sf Area \ of \ a \ triangle=\dfrac{1}{2} (Base \times Height)}}$

Now,

$\sf \Delta ABC=\dfrac{1}{2} \times AB \times BC$

$\sf =\dfrac{1}{2} \times 6 \times 8=24 \ cm^{2}$

In ΔACD,

AC = a = 10 cm

CD = b = 12 cm

DA = c = 14 cm

Next, finding the semi-perimeter

$\underline{\boxed{\sf s =\dfrac{a+b+c}{2} }}$

Substituting their values,

$\sf \dfrac{10+12+14}{2}$

$\sf \dfrac{36}{2} =18 \ cm$

Therefore, the semi perimeter is 18 cm

By Heron's formula,

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

Substituting their values,

Area = $\sf \sqrt{18(18-10)(18-12)(18-14)}$

$\sf =\sqrt{18 \times 8 \times 6 \times 4} =\sqrt{(3)^{2} \times 2 \times 4 \times 2 \times 3 \times 2 \times 4}$

$\sf =3 \times 4 \times 2\sqrt{3 \times 2}$

$=\sf 24\sqrt{6} \ cm^{2}$

Therefore, the area is 24√6 cm²