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If length of three sides of a trapezium other than base are equal to 10 cm, then find the area of the trapezium when it is maximum.

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The required trapezium is as given in figure.

Draw perpendiculars DP and CQ on AB.

Let AP = x cm.

Note that Δ APD ≅ ΔBQC.


QB = x cm.


By Pythagoras theorem,

DP = QC = \(\sqrt{100-x^2}.\)

Let A be the area of the trapezium.


A = A(x) = \(\frac{1}{2}\)

(sum of parallel sides) x (height)


A'(x) = 0 gives

2x2 + 10x - 100 = 0,

i.e., x = 5 and x = - 10

Since x represents distance, it cannot be negative.

So, x = 5.


Area of trapezium is maximum at x = 5 and the maximum area is given by


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