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The maximum area of the trapezium whose three non-base sides are each $10$ cm (maximum at $x=5$) is
A$50\sqrt{3}$ cm$^2$
B$75\sqrt{3}$ cm$^2$
C$100\sqrt{3}$ cm$^2$
D$60\sqrt{3}$ cm$^2$
Answer & Solution
Correct answer: B. $75\sqrt{3}$ cm$^2$
1. Area $A(x)=(x+10)\sqrt{100-x^2}$, maximum at $x=5$.
2. $A(5)=(5+10)\sqrt{100-25}$.
3. $=15\sqrt{75}=15\cdot 5\sqrt{3}=75\sqrt{3}$.
_Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.22_
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