Question

Let $A_1=\left\{(x,y):|x|\le y^2,|x|+2y\le 8\right\}$ and $A_2=\{(x,y):|x|+|y|\le k\}$. If 27 (Area $A_1)=5$ (Area $A_2$ ), then $k$ is equal to:

Answer 1

Answer: 6

$A_1=\left\{(x,y):|x|\le y^2,|x|+2y\le 8\right\}$ and
$A_2=\{(x,y):|x|+|y|\le k\}$

$\text{area}\left(A_1\right)=2\left[\underset{0}{\overset{2}{\int }}{y}^{2}dy+\underset{2}{\overset{4}{\int }}(8-2y)dy\right]$
$=2\left[{\left(\frac{y^3}{3}\right)}_0^2+{\left(8y-y^2\right)}_2^4\right]$

$\text{area}\left(A_1\right)=2\times \frac{20}{3}=\frac{40}{3}$
$\text{Area}\left(A_2\right)=4\times \frac{1}{2}{k}^2$
Area $\left(A_2\right)=2k^2$
Now
$27($ Area $A_1)=5($ Area $A_2)$
$9\times 4=k^2$
$k=6$