Question

Problem E- The white square in the drawing is located in the centre of the grey rectangle and has a surface area of A. The width of the rectangle is twice the width of a of the square. What is the surface of the grey area(without the white square)?

Image- https://iymc.info/docs/IYMC_Qualification_Round_2020.pdf

Answer 1

Let,

The area of the rectangle be $A_1$

The area of the white square be $A_2$

The area of the triangle be $A_3$

The area of the semi circle be $A_4$

The area of the grey square be $A_5$

Side of a square $=\frac{a}{\sqrt{2}}$

$A_1 =length×breadth = 2a.a = 2a^2 = 4A$

 $A_2 = A= (\frac{a}{\sqrt{2}})^2$ $=\frac{a^2}{2}$

 $A_3=\frac{base×height}{2}$ = $\frac{\frac{a}{\sqrt{2}}.\frac{a}{\sqrt{2}}}{2} = \frac{a^2}{4} = \frac{A}{2}$

$A_4$  $= \frac{\prod.(Radius^2)}{2}$ $= \frac{\prod.(\frac{a}{2})^2}{2}$ $= \frac{\prod.a^2}{8}$ $= \frac{\prod.A}{4}$

$A_5$ = A $= (\frac{a}{\sqrt{2}})^2$ $=\frac{a^2}{2}$

The surface of the grey area is

$A1-A2+A3+A4+2A5=A1+A2+A3+A4=4A+A+A/2+\frac{\prod.A}{4} 4 ∏.A$