Question

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)?

A

a​

Answer 1

Step-by-step explanation:

$Let,</p><p></p><p>The area of the rectangle be A1</p><p></p><p>The area of the white square be A2</p><p></p><p>The area of the triangle be A3</p><p></p><p>The area of the semi circle be A4</p><p></p><p>The area of the grey square be A5</p><p></p><p>Side of a square=\frac{a}{\sqrt{2}}=2a</p><p></p><p></p><p>A2 == A = (\frac{a}{\sqrt{2}})^2=(2a)2 =\frac{a^2}{2}=2a2</p><p></p><p>A1 =length×breadth = 2a.a = 2a2 = 4A</p><p></p><p></p><p>A3 = \frac{base×height}{2}2base×height = \frac{\frac{a}{\sqrt{2}}.\frac{a}{\sqrt{2}}}{2} = \frac{a^2}{4} = \frac{A}{2}22a.2a=4a2=2A</p><p></p><p></p><p>A4 = \frac{\prod.(Radius^2)}{2}=2∏.(Radius2)</p><p></p><p></p><p>= \frac{\prod.(\frac{a}{2})^2}{2}=2∏.(2a)2 = \frac{\prod.a^2}{8}=8∏.a2 = \frac{\prod.A}{4}=4∏.A</p><p></p><p>A5= A = (\frac{a}{\sqrt{2}})^2=(2a)2 =\frac{a^2}{2}=2a2</p><p></p><p></p><p>The surface of the grey area is A1-A2+A3+A4+2A5=A1+A2+A3+A4=4A+A+A/2+\frac{\prod.A}{4}4∏.A .</p><p></p><p>$