Question

Sides of a quadrilateral are 9 cm, 12 cm, 5cm and 8 cm. Angle made by the sides 12 cm and 5 cm is 90°. Find the area of quadrilateral.​

Answer 1

The sides of a quadrilateral ABCD are 6cm,8cm,12cm and 14cm (taken in order).

i.e. ∠ABC=90

o

Now, let us join the points A & C.

So, we get two triangles namely △ABC (a right angled triangle) and △ACD.

Applying Pythagoras theorem in △ABC, we get

AC=

AB

2

+BC

2

=

6

2

+8

2

=

36+64

=

100

=10cm

So, the area of △ABC=

2

1

×base×height

=

2

1

×AB×BC

=

2

1

×6×8=24

Now, in △ACD, we have

AC=10cm,CD=12cm,AD=14cm.

According to Heron's formula the area of triangle (A)=

[s(s−a)(s−b)(s−c)]

where, 2s=(a+b+c).

Here, a=10cm,b=12cm,c=14cm

s=

2

(10+12+14)

=

2

36

=18

Area of △ACD=

[18×(18−10)(18−12)(18−14)]

=

(18×8×6×4)

=

(2×3×3×2×2×2×2×3×2×2)

=

[(2×2×2×2×2×2×3×3)×2×3]

=2×2×2×3×

6

=24

6

So, total area of quadrilateral ABCD =△ABC+△ACD

=24+24

6

=24(

6

+1)