CTET 2 Math and Science Sep 2016 Paper

Let a and b be the sides of the larger and smaller equilateral triangles respectively.
Since, the ratio of area of both the triangles = 16 : 9Then,
$\frac{\sqrt{3}}{4}{\mathit{a}}^2:\frac{\sqrt{3}}{4}{\mathit{b}}^2=16:9$
${\mathit{a}}^2:{\mathit{b}}^2=16:9$
$\frac{{\mathit{a}}^2}{{\mathit{b}}^2}=\frac{16}{9}$
${\mathit{a}}^2=\frac{16}{9}{\mathit{b}}^2$
$\mathit{a}=\frac{4}{3}\mathit{b}$.......(1)
Since, the perimeter of the smaller triangle = 63 m
Then, $3\mathit{b}=63$
$\mathit{b}=21$
Then, from the equation (1)
$\mathit{a}=\frac{4}{3}\mathit{b}$
$=\frac{4}{3}\times 21$
$=28$
Now,
$\mathit{a}-\mathit{b}=28-21=7$
Hence, the side of the larger triangle is 7 times largerthan the side of the smaller triangle.