If the lengths of three sides of a trapezium | vedclass

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(A) Let the trapezium be $ABCD$ where $AD = DC = CB = 10 \ cm$. Draw perpendiculars $DP$ and $CQ$ on the base $AB$. Let $AP = x \ cm$. Since the trape

Vedclass · Accepted Answer

$75\sqrt{3} \ cm^2$

Vedclass · Answer

(A) Let the trapezium be $ABCD$ where $AD = DC = CB = 10 \ cm$. Draw perpendiculars $DP$ and $CQ$ on the base $AB$. Let $AP = x \ cm$. Since the trapezium is isosceles,$QB = x \ cm$. The length $PQ = DC = 10 \ cm$. Thus,the base $AB = x + 10 + x = 2x + 10 \ cm$.By the Pythagorean theorem in $\Delta APD$,the height $h = DP = \sqrt{10^2 - x^2} = \sqrt{100 - x^2}$.The area $A$ of the trapezium is given by $A = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$.$A(x) = \frac{1}{2} (AB + DC) 	imes h = \frac{1}{2} (2x + 10 + 10) \sqrt{100 - x^2} = (x + 10) \sqrt{100 - x^2}$.To find the maximum area,we find the derivative $A'(x)$:$A'(x) = (1) \sqrt{100 - x^2} + (x + 10) \frac{-2x}{2\sqrt{100 - x^2}} = \frac{100 - x^2 - x^2 - 10x}{\sqrt{100 - x^2}} = \frac{100 - 10x - 2x^2}{\sqrt{100 - x^2}}$.Setting $A'(x) = 0$,we get $2x^2 + 10x - 100 = 0 \implies x^2 + 5x - 50 = 0 \implies (x + 10)(x - 5) = 0$.Since $x$ is a length,$x = 5 \ cm$.The second derivative test confirms this is a maximum.Substituting $x = 5$ into the area formula:$A(5) = (5 + 10) \sqrt{100 - 5^2} = 15 \sqrt{75} = 15 	imes 5\sqrt{3} = 75\sqrt{3} \ cm^2$.

If the lengths of three sides of a trapezium other than the base are equal to $10 \ cm$,find the maximum area of the trapezium.

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