Question

A circle of radius $2\sqrt{10}$ cm is divided by a chord of length $12$ cm into two segments. Then maximum area of a rectangle that can be inscribed into the smaller circular segment is

• A. $2\sqrt{15}$
• B. $3\sqrt{15}$
• C. $4\sqrt{15}$
• D. $6\sqrt{15}$

Answer 1

The correct option is D $6\sqrt{15}$


Let $PC=l$
$AK=6$
$OK=\sqrt{O{A}^2-A{K}^2}=2$
$PQ=2\sqrt{(2\sqrt{10}{)}^2-(l+2{)}^2}$
Area of rectangle = $l\times 2\sqrt{36-l^2-4l}$
$\frac{dA}{dl}=2\{\frac{l(-l-2)}{\sqrt{36-l^2-4l}}+\sqrt{36-l^2-4l}\}=0$
$=\frac{2}{\sqrt{36-l^2-4l}}(-l^2-2l+36-l^2-4l)$
$=\frac{-(l+6)(l-3)\times 2}{\sqrt{36-l^2-4l}}$
- . + . -
-6 3
$A_{max.}\text{at}l=3$
$A_{max.}=3\times 2\sqrt{36-9-12}$
$=6\sqrt{15}$