Question

Assertion: If the length of three sides of a trapezium other than base are equal to $10 \,cm$, then the area of trapezium when it is maximum, is $75 \sqrt{3} \,cm ^{2}$.
Reason: Area of trapezium is maximum at $x =5$.

• A. Assertion is correct, reason is correct; reason is a correct explanation for assertion.
• B. Assertion is correct, reason is correct; reason is not a correct explanation for assertion
• C. Assertion is correct, reason is incorrect
• D. Assertion is incorrect, reason is correct.

Answer 1

Answer: (A)

The required trapezium is as given in figure. Draw perpendicular DP and $CQ$ on $AB$. Let $AP = x\, cm$. Note that $\triangle APD \sim \Delta BQC$.

Therefore, $QB = x \,cm .$
Also, by Pythagoras theorem,
$DP = QC -\sqrt{100- x ^{2}}$.
Let $A$ be the area of the trapezium. Then,
$ A = A ( x )=\frac{1}{2}$ (Sum of parallel sides ) $ \times $ (Height )
$=\frac{1}{2}(2 x +10+10)\left(\sqrt{100- x ^{2}}\right)=( x +10)\left(\sqrt{100- x ^{2}}\right) $
or $ A'( x )=( x +10) \frac{(-2 x )}{2 \sqrt{100- x ^{2}}}+\sqrt{100- x ^{2}} $
$=\frac{-2 x ^{2}-10 x +100}{\sqrt{100- x ^{2}}}$
Now, $ A '( x )=0$ gives
$ 2 x ^{2}+10 x -100 x =0 $
i.e., $ x =5 $ and $ x =-10.$
Since, $x$ represents distance, it cannot be negative.
So, $x=5$, Now
$A''\left(x\right) = \frac{\left(\sqrt{100 -x^{2}}\left(-4x-10\right)-\left(-2x^{2} - 10x + 100\right) \frac{\left(-2x\right)}{2\sqrt{100 - x^{2}}}\right)}{100 - x^{2}}$
$=\frac{2 x^{3}-300 x-1000}{\left(100-x^{2}\right)^{3 / 2}} $ (on simplification)
or $ A''(5)=\frac{2(5)^{3}-300(5)-1000}{\left(100-(5)^{2}\right)^{\frac{3}{2}}}$
$=\frac{-2250}{75 \sqrt{75}}=\frac{-30}{\sqrt{75}} < 0$
Thus, area of trapezium is maximum at $x =5$ and the area is given by
$A (5)=(5+10) \sqrt{100-(5)^{2}}$
$=15 \sqrt{75}=75 \sqrt{3} cm ^{2}$