Let
$A_1=\left\{(x,y):\left|x\right|\le y^2, \left|x\right|+2y\le 8\right\}$ and $A_2=\left\{\right(x,y):|x|+|y|\le k\}$. If 27 (Area $\left.{\mathrm{A}}_1\right)=5$

(Area ${\mathrm{A}}_2$ ), then $\mathrm{k}$ is equal to :

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

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

$=2\left[{\left(\frac{y^3}{3}\right)}_0^2+{\left(8y-y^2\right)}_2^4\right]$

 $\begin{array}{l}\mathrm{area} \left(A_1\right)=2\times \frac{20}{3}=\frac{40}{3}\\ Area \left({\mathrm{A}}_2\right)=4\times \frac{1}{2}{k}^2\\ \mathrm{Area} \left({\mathrm{A}}_2\right)=2k^2\\ Now\\ 27\left( \right.Area \left.A_1\right)=5\left( \right.Area\left.A_2\right)\\ 9\times 4=k^2\\ \mathrm{k}=6\end{array}$