The area of an isosceles trapezium is 176 , c | vedclass

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(D) Let the parallel sides be $AD = 4x \, cm$ and $BC = 7x \, cm$.The height $h = \frac{2}{11} 	imes (AD + BC) = \frac{2}{11} 	imes (4x + 7x) = \fra

Vedclass · Accepted Answer

$2\sqrt{137}$

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(D) Let the parallel sides be $AD = 4x \, cm$ and $BC = 7x \, cm$.The height $h = \frac{2}{11} 	imes (AD + BC) = \frac{2}{11} 	imes (4x + 7x) = \frac{2}{11} 	imes 11x = 2x \, cm$.The area of the trapezium is $\frac{1}{2} 	imes (AD + BC) 	imes h = 176$.Substituting the values: $\frac{1}{2} 	imes (4x + 7x) 	imes 2x = 176$.$11x^2 = 176 \implies x^2 = 16 \implies x = 4$.Thus,$AD = 16 \, cm$ and $BC = 28 \, cm$. The height $h = 2 	imes 4 = 8 \, cm$.In an isosceles trapezium,the projection of the non-parallel side on the base is $\frac{BC - AD}{2} = \frac{28 - 16}{2} = 6 \, cm$.Considering the right-angled triangle formed by the height,the projection,and the diagonal,the base of the triangle for the diagonal is $AD + 6 = 16 + 6 = 22 \, cm$.The diagonal $d = \sqrt{h^2 + (AD + 6)^2} = \sqrt{8^2 + 22^2} = \sqrt{64 + 484} = \sqrt{548} = \sqrt{4 	imes 137} = 2\sqrt{137} \, cm$.

The area of an isosceles trapezium is $176 \, cm^2$ and the height is $2/11$ of the sum of its parallel sides. If the ratio of the length of the parallel sides is $4:7$,then the length of a diagonal (in $cm$) is

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