Question

Assertion: If the length of three sides of a trapezium other than base are equal to $10cm$, then the area of trapezium when it is maximum, is $75\sqrt{3}c{m}^2$.
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 $AP=xcm$. Note that $△APD\sim \Delta BQC$.

Therefore, $QB=xcm.$
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 )
$=\frac{1}{2}(2x+10+10)\left(\sqrt{100-x^2}\right)=(x+10)\left(\sqrt{100-x^2}\right)$
or $A^{\prime }(x)=(x+10)\frac{(-2x)}{2\sqrt{100-x^2}}+\sqrt{100-x^2}$
$=\frac{-2x^2-10x+100}{\sqrt{100-x^2}}$
Now, $A^{\prime }(x)=0$ gives
$2x^2+10x-100x=0$
i.e., $x=5$ and $x=-10.$
Since, $x$ represents distance, it cannot be negative.
So, $x=5$, Now
$A^{\prime \prime }\left(x\right)=\frac{\left(\sqrt{100-x^2}\left(-4x-10\right)-\left(-2x^2-10x+100\right)\frac{\left(-2x\right)}{2\sqrt{100-x^2}}\right)}{100-x^2}$
$=\frac{2x^3-300x-1000}{{\left(100-x^2\right)}^{3/2}}$ (on simplification)
or $A^{\prime \prime }(5)=\frac{2(5{)}^3-300(5)-1000}{{\left(100-(5{)}^2\right)}^{\frac{3}{2}}}$
$=\frac{-2250}{75\sqrt{75}}=\frac{-30}{\sqrt{75}}<0$
Thus, area of trapezium is maximum at $x=5$ and the area is given by
$A(5)=(5+10)\sqrt{100-(5{)}^2}$
$=15\sqrt{75}=75\sqrt{3}c{m}^2$