CTET Class I to V 2015 Feb Paper

Perimeter of square $=44\mathit{c}\mathit{m}$
$\Rightarrow 4\times$ side of square $=44$
$\therefore$ Side of square $=11\mathit{c}\mathit{m}$
$\therefore$ Length of rectangle $=11+5=16\mathit{c}\mathit{m}$
Given that The perimeter of a rectangle $=$ The perimeter of the square.
$\Rightarrow 2($ length $+$ breadth $)=44$
$\Rightarrow 2(16+$ breadth $)=44$
$\therefore$ Breadth $=6$
Area of Rectangle $=16\times 6=96{\mathit{c}\mathit{m}}^2$
Area of square $={\left(11\right)}^2=121{\mathit{c}\mathit{m}}^2$
Sum of area of square and area of rectangle $=96+121=217{\mathit{c}\mathit{m}}^2$