You visited us 1 times! Enjoying our articles? Unlock Full Access!

Sum of Infinite Terms of a GP

The maximum a...

Question

The maximum area of the rectangle that can be inscribed in a circle of radius $r,$ is

π r^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

r^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

π r^{2} / 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 r^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $2 r^{2}$

Let $A B C D$ be rectangle of Area $= a b$
As shown,
${(2 r)}^{2} = a^{2} + b^{2}$
$\Rightarrow b = \sqrt{{4 r}^{2} - a^{2}}$
Area, $A = a b = a {({4 r}^{2} - a^{2})}^{1 / 2}$
$\frac{d A}{d a} = \frac{a (- 2 a)}{2 {({4 r}^{2} - a^{2})}^{1 / 2}} + {({4 r}^{2} - a^{2})}^{1 / 2} = 0 \Rightarrow {4 r}^{2} - a^{2} = a^{2}$
$\Rightarrow a = r \sqrt{2} = b$
Area, $= a b = {(r \sqrt{2})}^{2} = {2 r}^{2}$ .

Suggest Corrections

Similar questions

If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then

r^{2} s q u n i t s

\frac{1}{2} r^{2} s q u n i t s

2 r^{2} s q u n i t s

\sqrt{2} r^{2} s q u n i t s

Let $c_{1} & c_{2}$ be the centers and $r_{1} & r_{2}$ be the radius of two circles. Then

Cases Conditions

p. 1. $| r_{1} - r_{2} | < c_{1} c_{2} < r_{1} + r_{2}$

q. 2. $| r_{1} - r_{2} | = c_{1} c_{2}$

r. 3. $c_{1} c_{2} < | r_{1} + r_{2} |$

4. $r_{1} + r_{2} < c_{1} r_{2}$

t. 5. $r_{1} + r_{2} = c_{1} r_{2}$

R_{1} + R_{2} = R

R_{1} + R_{2} < R

R_{1}^{2} + R_{2}^{2} < R^{2}

R_{1}^{2} + R_{2}^{2} = R^{2}

\frac{1}{2}

\sqrt{2}

Join BYJU'S Learning Program