A geometry problem by Chung Kevin

Method 1:

let $A$ be the area of the oval-shaped region, $x$ be the area of the quarter circle (yellow region) and $y$ be the area of the white region

then,

$x=\dfrac{1}{4}(\pi)(1^2)=\dfrac{\pi}{4}$

it follows that,

$y=1^2-\dfrac{\pi}{4}=\dfrac{4-\pi}{4}$

Finally,

$A=1^2-2(y)=1-2(\dfrac{4-\pi}{4})=1-(\dfrac{4-\pi}{2})=1-2+\dfrac{\pi}{2}=-1+\dfrac{\pi}{2}=$ $\color{#3D99F6}\boxed{\dfrac{\pi}{2}-1}$ $\boxed{\text{answer}}$

Method 2:

let $A$ be the area of the oval-shaped region, $x$ be the area of the quarter circle (yellow region) and $y$ be the area of the white region

then,

$x=\dfrac{1}{4}(\pi)(1^2)=\dfrac{\pi}{4}$

it follows that,

$y=1^2-\dfrac{\pi}{4}=\dfrac{4-\pi}{4}$

Finally,

$A=x-y$

$A=\dfrac{\pi}{4}-\dfrac{4-\pi}{4}$

$A=\dfrac{\pi-(4-\pi)}{4}$

$A=\dfrac{\pi-4+\pi}{4}$

$A=\dfrac{2\pi-4}{4}$

$A=\dfrac{2\pi}{4}-\dfrac{4}{4}$

$A=$ $\color{#3D99F6}\boxed{\dfrac{\pi}{2}-1}$ $\boxed{\text{answer}}$

Method 3:

let $A$ be the area of the oval-shaped region, $x$ be the area of the quarter circle (yellow region) and $z$ be the area of the triangle (green region)

then,

$x=\dfrac{1}{4}(\pi)(1^2)=\dfrac{\pi}{4}$

it follows that,

$z=\dfrac{1}{2}(1)(1)=\dfrac{1}{2}$

then,

$\dfrac{A}{2}=x-z$

$\dfrac{A}{2}=\dfrac{\pi}{4}-\dfrac{1}{2}$

Finally,

$A=2(\dfrac{A}{2})=2(\dfrac{\pi}{4}-\dfrac{1}{2})=$ $\color{#3D99F6}\boxed{\dfrac{\pi}{2}-1}$ $\boxed{\text{answer}}$

Two quarter circles cover the whole area once and overlap in the blue area covering it twice.

An area of two quarter circles minus an area of a square = area of half a circle minus a square = $\frac{\pi}{2}-1$ .

the area of the quarter circle is π/4 minus 1/2 the area of the square will give us half the blue area so π/4 - 1/2. the blue area is twice that or π/2 - 1

You could just eyeball it. Option 1 is 2 - 1.57, so .43, Option 2 is .14. Option 3 is 1.57-1, or .57, and Option 4 is a number larger than 1. Since the area is 1 and the blue area is clearly about half, you can eliminate options 2 and 4. Then the question is is it closer to 60% blue or 40% blue. It is closer to 60%, so option 3.

By sight, the shaded oval is a bit more than half the area of the square which is 1. So the only answer that is a bit more than 0.5 is (pi/2)-1.

The blue area is equal to 2× [Area of the quarter circle - (1/2) area of the square] =

= The area of a semi-circle - The area of the square

= [(1/2)π(1)^(2)]-(1*1)

= (π/2)-1

Therefore the answer is (π/2) - 1

The area of the quarter circle is $\frac{\pi}{4}$ .

Thus the area of the part of the square - quarter of circle is = $1-\frac{\pi}{4}$ , since there are two of these areas,

then the unwanted area is $2-\frac{\pi}{2}\implies$ the area of the oval part is $1-(2-\frac{\pi}{2})=\frac{\pi}{2}-1$

Firstly divide the oval region shaped into two halves then calculate the area of the triangle = 1/2(1)(1)=1/2 Then calculate the area of quadrant = (πr*2)/4 which is equal to = π/4 then Subtract the both area =(π/4 - 1/2 )
This is the area of 1/2 Oval shaped and multiplied by 2 == (π/2-1)

Technically, that shape is called a vesica piscis - or you could just say the football-shaped region if you want to use a more familiar term. Oval is not a well-defined term, so fine, call I guess you can call it an oval. But I don't think this shape matches the common usage of the term.