Question

The perimeter of an isosceles trapezium is $16cm$ and the shorter parallel side is one third of the longer parallel side. If the sum of the lengths of the opposite sides of the trapezium are equal, find the area (in ${cm}^2$) of the trapezium?

• A. $8\sqrt{3}$
• B. $4\sqrt{3}$
• C. $12\sqrt{3}$
• D. $16\sqrt{3}$

Answer 1

The correct option is A $8\sqrt{3}$

REF.Image.

In isosceles trapezium, AD = BC

Let DC = 3a cm

$\Rightarrow$ AB = a cm

As sum of opposite sides is equal,

$\Rightarrow AD+BC=4a\Rightarrow 2AD=4a\Rightarrow AD=2a$

Now perimeter = 16 cm $\Rightarrow 8a=16\Rightarrow a=2cm$

In $△ALD,A{L}^2=A{D}^2-D{L}^2$

(Pythagoras th.)$\Rightarrow A{L}^2=16-4=12$

$\Rightarrow AL=\sqrt{12}=2\sqrt{3}cm$

Now, Area of trapezium $=\frac{1}{2}$(Sum of parallel sides)Height

$=\frac{1}{2}\left(4a\right)\left(2\sqrt{3}\right)$

$=2a\times 2\sqrt{3}$

$=4\times 2\sqrt{3}=8\sqrt{3}c{m}^2$