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Given a square ABCD of sides 10 cm, and using the corners as centres, construct four quadrants with radius 10 cm each inside the square. The four arcs intersect at P, Q, R and S. Find the area enclosed by PQRS.

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A white cross is placed symmetrically in a red disc with the central square of side length sqrt 2 and the arms of the cross of length 1 unit. What is the area of the disc still showing?

Bound to Be

Age 14 to 16
Challenge Level

This problem attracted several correct answers anonymously submitted by students at Madras College . This solution is included because of its brevity:

$AD = 1$ unit, which is the radius of the circle centre $A$.

The coordinates of $P$ are $(AD \cos 60^{\circ} , AD \sin 60^{\circ} ) = ({1\over2}, {\sqrt{3}\over2})$

Similarly coordinates of $Q$ are $(AD \cos 30^{\circ} , AD \sin 30^{\circ} ) = ( {\sqrt{3}\over2},{1\over2})$ and angle $PAQ$ is $30^{\circ}$.

Area of square $PQRS = ({1\over2} - {\sqrt{3}\over2})^2 + ({\sqrt{3}\over2} - {1\over2})^2 = 2 - \sqrt{3}$

Area of a segment $PQ$ = [${1\over12}$ area of circle, centre $A$ radius $1$ unit] - [Area of Triangle $APQ$ ]

Using the formula ${1\over12}\pi r^2$ for the $30$ degree sector and ${1\over2} ab\sin C$ for the area of the triangle we get:

Area of a segment $PQ = {1\over12} \pi.1.1 - {1\over2}.1.1.{1\over2} = {\pi\over12} - {1\over4}$ \par Bounded area = area of square $PQRS$ + area of 4 equal segments

Bounded area $= 2 - \sqrt{3} + 4 ({\pi\over12} - {1\over4}) = 1 + {\pi\over3} - \sqrt{3}$