$$\triangle$$ABC is similar to $$\triangle$$DEF . The area of $$\triangle$$ABC is 100 cm$$^2$$ and the area of $$\triangle$$DEF is 49 cm$$^2$$ If the altitude of $$\triangle$$ABC = 5 cm, then the corresponding altitude of $$\triangle$$DEF is:

Given that,

$$\triangle$$ABC is similar to $$\triangle$$DEF

The area of $$\triangle$$ABC is 100 cm$$^2$$

the area of $$\triangle$$DEF is 49 cm$$^2$$

$$\triangle$$ABC = 5 cm

We know that from the property of the similar triangle,

$$\Rightarrow \dfrac{ar(\triangle ABC)}{ar(\triangle DEF)}=(\dfrac{AB}{DE})^2=(\dfrac{BC}{EF})^2=(\dfrac{AC}{DF})^2=(\dfrac{AX}{DY})^2$$

Now substituting the values in the above,

$$\Rightarrow \dfrac{ar(\triangle ABC)}{ar(\triangle DEF)}=(\dfrac{AX}{DY})^2$$

$$\Rightarrow \dfrac{100C}{49}=(\dfrac{5}{DY})^2$$

Now, taking the square root of both side,

$$\Rightarrow \dfrac{10}{7}=\dfrac{5}{DY}$$

$$\Rightarrow DY=\dfrac{5\times 7}{10}=\dfrac{7}{2}=3.5cm$$