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# Area and Perimeter of Similar Triangles

#### What is true about the ratio of the area of similar triangles?

Answer: If 2 triangles are similar, their areas are the square of that similarity ratio (scale factor)

For instance if the similarity ratio of 2 triangles is $$\frac 3 4$$ , then their areas have a ratio of $$\frac {3^2}{ 4^2} = \frac {9}{16}$$

Let's look at the two similar triangles below to see this rule in action.

##### Example 1

It's easiest to see that this is true if you look at some specific examples of real similar triangles.

Triangle 1

$$Area = \frac{1}{2}\cdot{12}\cdot{4} \\ Area = 24$$

Triangle 2

$$Area = \frac{1}{2}\cdot{24}\cdot{8} \\ Area = 96$$

Notice: $$\frac{24}{96} = \frac{1}{4}$$

Therefore, if you know the similarity ratio, all that you have to do is square it to determine ratio of the triangle's areas.

##### Example 2
Triangle 1
Perimeter of Triangle #1

Perimeter $$= 6 + 8 + 10 = 24$$

Triangle 2
Perimeter of Triangle #2

Perimeter $$= 5 + 3 +4 = 12$$

The ratio of the perimeter's is exactly the same as the similarity ratio!

$\frac{\text{perimeter #1}}{\text{perimeter #2}} = \frac{24}{12} = \frac{2}{1}$

### Practice Problems

##### Problem 1

$\text{Ratio of areas} = (\text{similarity ratio})^2 \\ = \Big(\frac{3}{2}\Big)^2 \\ = \frac{9}{4}$

##### Problem 2

$\text{ratio of perimeters} = \text{similarity ratio} \\ \text{similarity ratio} = \frac{11}{5} \\ \text{ratio of areas} = (\text{similarity ratio})^2 \\ = \Big(\frac{11}{5}\Big)^2 \\ \text{ratio of areas} = \frac{121}{25}$

##### Problem 3

$\text{ratio of areas} = (\text{similarity ratio})^2 \\ (\text{similarity ratio})^2 = \text{ratio of areas} \\ \text{similarity ratio} = \sqrt{\text{ratio of areas} } \\ = \sqrt{\Big(\frac{36}{17} \Big) } \\ = \frac{\sqrt{36}}{\sqrt{17 } } \\ = \frac {6 }{\sqrt{17 } }$

##### Problem 4

We need to find the similarity ratio first, since that ratio gives us a proportion between corresponding sides.

$\text{ratio of areas} = (\text{similarity ratio})^2 \\ (\text{similarity ratio})^2 = \text{ratio of areas} \\ \text{similarity ratio} = \sqrt{\text{ratio of areas} } \\ = \sqrt{\Big(\frac{25}{16} \Big) } \\ \text{similarity ratio} = \frac{5}{4 }$

$\frac{5}{4 } = \frac{HI}{XY} \\ \frac{5}{4 } = \frac{HI}{40} \\ \frac{40 \cdot 5}{4 } = HI \\ HI = 50$