Question

Let $ABCD$ be a square. An arc of a circle with $A$ as center and $AB$ as radius is drawn inside the square joining the points $B$ and $D$. Points $P$ on $AB$, $S$ on $AD$, $Q$ and $R$ on arc $BD$ are taken such that PQRS is a square. Further suppose that PQ and RS are parallel to AC. Then $\frac{\text{area}\left(PQRS\right)}{\text{area}\left(ABCD\right)}$is

• A. $\frac{1}{8}$
• B. $\frac{1}{5}$
• C. $\frac{1}{4}$
• D. $\frac{2}{5}$

Answer 1

The correct option is D $\frac{2}{5}$


Let the length of $AB=a$
the equation of the arc be $x^2+y^2=a^2$
the length of $PQ=k$
and the cordinates of $P=(x_1,0);S=(0,y_1)$
$\therefore Q=(x_1+k\cos {45}^{\circ },0+k\sin {45}^{\circ })\Rightarrow Q=(x_1+\frac{k}{\sqrt{2}},0+\frac{k}{\sqrt{2}})\therefore R=(0+k\cos {45}^{\circ },y_1+k\sin {45}^{\circ })\Rightarrow R=(0+\frac{k}{\sqrt{2}},y_1+\frac{k}{\sqrt{2}})$
$Q$ and $R$ lies on the arc so,
$\therefore {(x_1+\frac{k}{\sqrt{2}})}^2+\frac{k^2}{2}=a^2$.........(i)
Now we know that $△PAS$ is a right angled at $A$ so,
${x_1}^2+{y_1}^2=k^2$......(ii)
Slope of $AC=1$
$\therefore slopeofPS=-1$
$\Rightarrow \frac{-y_1}{x_1}=-1\Rightarrow x_1=y_1$
using equation (ii)
$x_1=\frac{k}{\sqrt{2}}$
using equation (i)
$2k^2+\frac{k^2}{2}=a^2\Rightarrow \frac{k^2}{a^2}=\frac{2}{5}$

Hence the ratio $\frac{\text{area}PQRS}{\text{area}ABCD}=\frac{k^2}{a^2}=\frac{2}{5}$

Alternate solution:


Let the length of the side of square $PQRS$ be $2a$.
Diagonal $PR=2\sqrt{2}a$
Now we know that,
$AT=PTAT=TSPT=TS=a\therefore AP=\sqrt{2}a$
In $△ARP\Rightarrow A{R}^2=A{P}^2+R{P}^2\Rightarrow A{R}^2=(\sqrt{2}a{)}^2+(2\sqrt{2}a{)}^2\Rightarrow AR=\sqrt{10}a\Rightarrow AB=AR=\sqrt{10}a$
Hence
$\frac{\text{area}PQRS}{\text{area}ABCD}=\frac{(2a{)}^2}{(\sqrt{10}a{)}^2}=\frac{2}{5}$