Question

Problem E

The white square in the drawing below 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 a of the square. What is the surface of the grey area (without the white square)?

Image is found on the website below

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

Answer 1

Let,

The area of the rectangle be A1

The area of the white square be A2

The area of the triangle be A3

The area of the semi circle be A4

The area of the grey square be A5

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

A₂ $=$ A $= (\frac{a}{\sqrt{2}})^2$ $=\frac{a^2}{2}$

A₁ =length×breadth = 2a.a = 2a² = 4A

A₃ = $\frac{base×height}{2}$ = $\frac{\frac{a}{\sqrt{2}}.\frac{a}{\sqrt{2}}}{2} = \frac{a^2}{4} = \frac{A}{2}$

A₄ $= \frac{\prod.(Radius^2)}{2}$

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

A₅= 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+$\frac{\prod.A}{4}$ .