Question

The height of a trapezium is 68 cm, and the sum of its parallel sides is 75 cm. If the area of the trapezium is $\frac{6}{17}$ times of the area of a square, then the length of the diagonal of the square is: (Take $\sqrt{2}= 1.41$)

• A. 127.39 cm
• B. 183.49 cm
• C. 119.85 cm
• D. 102.39 cm

Answer 1

Answer:

C. 119.85 cm

Solution:

Given:

The height of a trapezium is 68 cm, and the sum of its parallel sides is 75 cm and the area of the trapezium is $\frac{6}{17}$ times of the area of a square.

Formula used:

Area of the trapezium $=\frac{(a+b)h}{2}$, where (a+b) is the sum of its parallel sides and h is height.

Area of the square = a² , where 'a' is the length of each side

Diagonal of a square $=\sqrt{2a}$

Calculation:

Area of the trapezium $=\frac{68\times{75}}{2}=2550$

According to the question,

The area of the trapezium $=\frac{6}{17}\times$the area of a square.

⇒ $2550 =\frac{6}{17}\times$ the area of a square.

Area of the square $=\frac{17}{6}\times{2550}=a^2$

Solving for a, 

$a=\sqrt{7225}=85cm$

∴ The length of the diagonal of the square is $1.45\times{85}$ ≈119.85cm