Let A= (x, y) ∈ R 2: y ≥ 0,2 x ≤ y ≤ √4-(x-1)2 and B= (x, y) ∈ R × R: 0 ≤ y ≤ min 2 x, √4-(x-1)2 . Then the ratio of the area of A to the area of B is

Question

Let A= (x, y) ∈ R 2: y ≥ 0,2 x ≤ y ≤ √4-(x-1)2 and B= (x, y) ∈ R × R: 0 ≤ y ≤ min 2 x, √4-(x-1)2 . Then the ratio of the area of A to the area of B is (A) (π/π+1) (B) (π-1/π+1) (C) (π+1/π-1) (D) (π/π-1)

Tardigrade Admin · Accepted Answer

y2+(x-1)2=4

shaded portion = text circular ( OABC ) - operatornameAr( triangle OAB ) =(π(4)/4)-(1/2)(2)(1) A =(π-1)

Area B = operatornameAr( triangle AOB )+ Area of arc of circle ( ABC ) =(1/2)(1)(2)+(π(2)2/4)=π+1 ( A / B )=(π-1/π+1)

Q. Let $A = {(x, y) \in R^{2} : y \geq 0, 2 x \leq y \leq 4 - (x - 1)^{2}}$ and
$B = {(x, y) \in R \times R : 0 \leq y \leq min {2 x, 4 - (x - 1)^{2}}} .$
Then the ratio of the area of $A$ to the area of $B$ is

$\frac{π}{π + 1}$

$\frac{π - 1}{π + 1}$

$\frac{π + 1}{π - 1}$

$\frac{π}{π - 1}$

$y^{2} + (x - 1)^{2} = 4$

shaded portion $= circular (O A BC)$
$- Ar (△ O A B)$
$= \frac{π ( 4 )}{4} - \frac{1}{2} (2) (1)$
$A = (π - 1)$

Area $B = Ar (△ A OB) +$ Area of arc of circle $(A BC)$
$= \frac{1}{2} (1) (2) + \frac{π ( 2 ) ^{2}}{4} = π + 1$
$\frac{A}{B} = \frac{π - 1}{π + 1}$

Q. Let A={(x,y)∈R2:y≥0,2x≤y≤4−(x−1)2​} and B={(x,y)∈R×R:0≤y≤min{2x,4−(x−1)2​}}. Then the ratio of the area of A to the area of B is

π+1π​

π+1π−1​

π−1π+1​

π−1π​