Question

Prove that the areas of similar triangles have the same ratio as the squares of the corresponding altitudes.

Answer 1

Hint: Consider two similar triangles $\Delta AMT\text{ and }\Delta AHE$ . Now use the property that the area of a triangle is half of the product of the length of two sides of the triangle with the sine of the angle between these two sides.

Complete step-by-step answer:
To start with the question, let us consider two similar triangles $\Delta AMT\text{ and }\Delta AHE$ . And we let the ratio of sides of these similar triangles be k. We will also let the heights of the two triangles be ${{h}_{1}}\text{ and }{{h}_{2}}$ , respectively. Let us draw a diagram of the assumed situation for better understanding.

It is assumed that $\Delta AMT\sim \Delta AHE$ , and we know, two triangles are similar if their corresponding angles are equal and the corresponding sides are in proportion.
$\therefore \dfrac{MA}{HA}=\dfrac{AT}{AE}=\dfrac{TM}{EH}=k............(i)$
Also, $\angle AMT=\angle AHE.............(ii)$ .
Now we know that the area of a triangle is half of the product of the length of two sides of the triangle with the sine of the angle between these two sides.
$\therefore ar\left( \Delta AMT \right)=\dfrac{1}{2}\times MA\times TM\times \sin \left( \angle AMT \right)$
$\therefore ar\left( \Delta AHE \right)=\dfrac{1}{2}\times HA\times EH\times \sin \left( \angle AHE \right)$
Now we will find the ratio of the areas of the two triangles.
$\therefore \dfrac{ar\left( \Delta AMT \right)}{ar\left( \Delta AHE \right)}=\dfrac{\dfrac{1}{2}\times MA\times TM\times \sin \left( \angle AMT \right)}{\dfrac{1}{2}\times HA\times EH\times \sin \left( \angle AHE \right)}=\dfrac{MA}{HA}\times \dfrac{TM}{EH}\times \dfrac{\sin \left( \angle AMT \right)}{\sin \left( \angle AHE \right)}$
Now if we substitute the values of $\dfrac{MA}{HA}\text{ and }\dfrac{TM}{EH}$ from equation (i), we get
$\therefore \dfrac{ar\left( \Delta AMT \right)}{ar\left( \Delta AHE \right)}={{k}^{2}}\times \dfrac{\sin \left( \angle AMT \right)}{\sin \left( \angle AHE \right)}$
Also, from equation (ii), we know $\angle AMT=\angle AHE$ , so their sine would also be the same.
$\therefore \dfrac{ar\left( \Delta AMT \right)}{ar\left( \Delta AHE \right)}={{k}^{2}}......(iii)$
Also, we know that area of a triangle is half of the product of the base and the height.
$\therefore \dfrac{ar\left( \Delta AMT \right)}{ar\left( \Delta AHE \right)}=\dfrac{\dfrac{1}{2}\times MA\times {{h}_{1}}}{\dfrac{1}{2}\times HA\times {{h}_{2}}}=\dfrac{MA}{HA}\times \dfrac{{{h}_{1}}}{{{h}_{2}}}$
Now if we substitute the values of $\dfrac{MA}{HA}\text{ and }\dfrac{ar\left( \Delta AMT \right)}{ar\left( \Delta AHE \right)}$ from equation (i) and equation (iii), we get
$\therefore {{k}^{2}}=k\dfrac{{{h}_{1}}}{{{h}_{2}}}$
$\Rightarrow k=\dfrac{{{h}_{1}}}{{{h}_{2}}}$
$\therefore \dfrac{ar\left( \Delta AMT \right)}{ar\left( \Delta AHE \right)}={{k}^{2}}......(iii)$$\Rightarrow {{k}^{2}}={{\left( \dfrac{{{h}_{1}}}{{{h}_{2}}} \right)}^{2}}..........(iv)$
So, if we compare equation (iii) and (iv), we get
$\therefore \dfrac{ar\left( \Delta AMT \right)}{ar\left( \Delta AHE \right)}={{\left( \dfrac{{{h}_{1}}}{{{h}_{2}}} \right)}^{2}}$
Therefore, we have proved that the areas of similar triangles have the same ratio as the squares of the corresponding altitudes.

Note: While using the relation of similarity, generally the students mix up the sides while it is clearly mentioned that the relation is valid only for the corresponding sides of similar triangles. If you want, you can solve the question using the heron’s formula of area of a triangle as well, but that would be a bit lengthier.