Area Of Similar Polygons

We can see that the ratio of the perimeters is $12:18=2:3$ . This means the ratio of the areas of the two similar polygons is equal to

$2^2:3^2=4:9$

We have

$4x+9x=65$

$\implies 13x=65$

$\implies x=5$

$\implies 9x=5\times 9 = \boxed{45}\ \blacksquare$

From the given information, the ratio of the sides of the polygons is $\frac{3}{2}$ , so the ratio of the areas is $\frac{9}{4}$ .

Let the area of the larger polygon be $x$ ; then we must have $\frac{4}{9}x+x=65\Rightarrow x=\boxed{45}$ .

First of all area is in unit squared, so u first square both 12 units and 18 units to get the total as 468 square units. Than we know that 468 squared units is therefore equal to 65 square units. Thus using proportions we divide 65 by 468 and then multiply it by 18 squared. This will give you your final result which is 45.

First: the ratio of the perimeters. We have $18:12 \rightarrow 3:2$ as the ratio. Now, the ratio of the areas. The shapes are 3 dimensional, so we square the ratio of the perimeters to get $4:9$ . W want the area of the larger polygon, so we multiply $65 \cdot \frac{9}{9+4} = 65 \cdot \frac{9}{13} = 5 \cdot 9 = \fbox{45}$ , and we are done.

12/18 = 2/3 So the corresponding areas E1/E2 = (2/3)^2 = 4/9. Since 4 + 9 =13 and 65/13 =5 we get: 9*5 =45 for the area of the larger polygon.

The areas of similar plane figures or similar surfaces have the same ratio as the squares of any two corresponding lines. Let $A$ and $B$ be the areas, then we have

$\dfrac{A}{B}=\dfrac{12^2}{18^2}$

However, $A=65-B$ , so

$\dfrac{65-B}{B}=\dfrac{144}{324}$

$21060-324B=144B$

$B=\boxed{45}$

Length factor 2:3, so area factor 4:9 65/13*9=45

We know that for similar polygons, ratio of corresponding sides of the similar triangles is equal to ratio of the perimeter of the corresponding polygons. Then again for similar polygons, we have the Area Theorem which states that---

$\text{(Ratio of area of two similar polygons)}=\text{(Ratio of corresponding sides)}^{2}=\text{(Ratio of their perimeters)}^{2}$

Let us take the area of one similar polygon as $x$ and then the other should have area $(65-x)$ as both the polygons have a total area of 65 sq. units. The perimeters of the polygons are $12$ units and $18$ units respectively.

$\frac{x}{65-x}=(\frac{12}{18})^{2}$

$\implies \frac{x}{65-x}=(\frac{2}{3})^{2}$

$\implies \frac{x}{65-x}=\frac{4}{9} \implies 9x=260-4x \implies 13x=260 \implies x=20$

Then the polygons have areas $x=20$ and $65-x=65-20=45$ units respectively where 45 units is the larger area.

So, the answer is $=\boxed{45}$