If ∆ABC ∼ ∆DEF such that AB = 5 cm,
Question:

If $\triangle \mathrm{ABC} \sim \Delta \mathrm{DEF}$ such that $\mathrm{AB}=5 \mathrm{~cm}$, area $(\Delta \mathrm{ABC})=20 \mathrm{~cm}^{2}$ and area $(\Delta \mathrm{DEF})=45 \mathrm{~cm}^{2}$, determine $\mathrm{DE}$.

Solution:

Given: The area of two similar $\triangle \mathrm{ABC}=20 \mathrm{~cm}^{2}, \triangle \mathrm{DEF}=45 \mathrm{~cm}^{2}$ respectively and $\mathrm{AB}=5 \mathrm{~cm}$.

To find: measure of DE 

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\left(\frac{\mathrm{AB}}{\mathrm{DE}}\right)^{2}$

$\frac{20}{45}=\left(\frac{5}{\mathrm{DE}}\right)^{2}$

$\frac{20}{45}=\frac{25}{\mathrm{DE}^{2}}$

$\mathrm{DE}^{2}=\frac{25 \times 45}{20}$

$\mathrm{DE}^{2}=\frac{225}{4}$

$D E=7.5 \mathrm{~cm}$

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