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 $\sqrt{2}=1.41$ )
Option 1: 127.39 cm
Option 2: 183.49 cm
Option 3: 119.85 cm
Option 4: 102.39 cm
Correct Answer: 119.85 cm
Solution :
Given,
The height of a trapezium is 68 cm, the sum of its parallel sides is 75 cm and the area of the trapezium is $\frac{6}{17}$ times the area of a square.
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 = $a^2$, where $a$ is the length of each side.
Diagonal of a square = $\sqrt2a$
So, the area of the trapezium = $\frac{68×75}{2} = 2550$
According to the question,
The area of the trapezium = $\frac{6}{17}$ × the area of a square
⇒ 2550 = $\frac{6}{17}$ × the area of a square
⇒ Area of the square = $\frac{17}{6}×2550=a^2$
⇒ $a=\sqrt{7225}$
⇒ $a=85$ cm
⇒ The length of the diagonal of the square is $1.41×85=119.85$ cm
Hence, the correct answer is 119.85 cm.
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