A geometry problem by rowegie lambojon

linear ratio is 2:3 and area ratio is 4:9, therefore the smaller figure has area 20 and the larger figure has area 65

$A_{1}+A_{2}=65$ or $A_{2}=65-A_{1}$

$\dfrac{A_{1}}{A_{2}}=\dfrac{12^2}{18^2}$

$\dfrac{A_{1}}{65-A_{1}}=\dfrac{144}{324}$

$324A_{1}=9360-144A_{1}$

$468A_{1}=9360$

$A_{1}=20$

It follows that $A_{2}=65-20=$ $\boxed{45}$

Note:

The areas of similar plane figures have the same ratio as the squares of any two corresponding lines. In the above problem, the two polygons are similar, so they have the same number of sides. Since the perimeter is given, we don't need to know the length of each side of the polygon.

Area of n sided polygon with side length s is given by A=1/4[ns^2cot(π/n)]
In this question polygons are similar i.e n is same for both. let side length of small polygon be s¹ and that of big polygon be s²

ns¹=12 and ns²=18 Use sum of areas applying above formula and then you will get,

1/4{[(ns¹)^2 + (ns²^2)][1/n(cotπ/n) ]}=65

1/4[144+324][1/n(cotπ/n) ]=65

[1/n(cotπ/n) ]=65/117

Now area of larger polygon will be 1/4[(ns²^2)(1/ncotπ/n)] =1/4(328)(65/117)=45