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Hint: The ratios of the corresponding sides of two similar triangles are the same. Use this theorem and find the area of triangles in terms of their side and altitude and then compare areas of both triangles.

Suppose we have two similar triangles of sides ${a_1},{b_1},{c_1}$and ${a_2},{b_2},{c_2}$respectively. Further let ${h_1}$and ${h_2}$be the altitudes of the triangles drawn from the opposite vertices on the sides of length ${a_1}$and ${a_2}$of respective triangles.

We know that if two triangles are similar then the ratios of their corresponding sides are the same. Therefore, using this property we have:

$ \Rightarrow \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} = \dfrac{{{h_1}}}{{{h_2}}}$

Now, if have to calculate the areas of triangles we can apply formula as:

$ \Rightarrow $Area $ = \dfrac{1}{2} \times $Base $ \times $Height.

Thus, the area of the first will be:

$

\Rightarrow {A_1} = \dfrac{1}{2} \times {a_1} \times {h_1}, \\

\Rightarrow {A_1} = \dfrac{1}{2}{a_1}{h_{1.}} \\

$

Similarly the area of the second circle will be:

$

\Rightarrow {A_2} = \dfrac{1}{2} \times {a_2} \times {h_2}, \\

\Rightarrow {A_2} = \dfrac{1}{2}{a_2}{h_2}. \\

$

The ratio of their areas is:

$

\Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\dfrac{1}{2}{a_1}{h_{1.}}}}{{\dfrac{1}{2}{a_2}{h_2}}}, \\

\Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \left( {\dfrac{{{a_1}}}{{{a_2}}}} \right).\left( {\dfrac{{{h_1}}}{{{h_2}}}} \right) \\

$

And since we have already determined$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} = \dfrac{{{h_1}}}{{{h_2}}}$. Using this result, we’ll get:

$ \Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{{a_1}^2}}{{{a_2}^2}} = \dfrac{{{b_1}^2}}{{{b_2}^2}} = \dfrac{{{c_1}^2}}{{{c_2}^2}} = \dfrac{{{h_1}^2}}{{{h_2}^2}}$

Therefore, the ratio of the areas of similar triangles is equal to the ratio of the squares of their corresponding sides. Hence, this is the required proof.

Note: If two triangles are similar then the ratio of their corresponding sides are same along with the ratio of their corresponding altitudes, their circumradius and their inradius. The measurement of their corresponding angles is also the same.

Suppose we have two similar triangles of sides ${a_1},{b_1},{c_1}$and ${a_2},{b_2},{c_2}$respectively. Further let ${h_1}$and ${h_2}$be the altitudes of the triangles drawn from the opposite vertices on the sides of length ${a_1}$and ${a_2}$of respective triangles.

We know that if two triangles are similar then the ratios of their corresponding sides are the same. Therefore, using this property we have:

$ \Rightarrow \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} = \dfrac{{{h_1}}}{{{h_2}}}$

Now, if have to calculate the areas of triangles we can apply formula as:

$ \Rightarrow $Area $ = \dfrac{1}{2} \times $Base $ \times $Height.

Thus, the area of the first will be:

$

\Rightarrow {A_1} = \dfrac{1}{2} \times {a_1} \times {h_1}, \\

\Rightarrow {A_1} = \dfrac{1}{2}{a_1}{h_{1.}} \\

$

Similarly the area of the second circle will be:

$

\Rightarrow {A_2} = \dfrac{1}{2} \times {a_2} \times {h_2}, \\

\Rightarrow {A_2} = \dfrac{1}{2}{a_2}{h_2}. \\

$

The ratio of their areas is:

$

\Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\dfrac{1}{2}{a_1}{h_{1.}}}}{{\dfrac{1}{2}{a_2}{h_2}}}, \\

\Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \left( {\dfrac{{{a_1}}}{{{a_2}}}} \right).\left( {\dfrac{{{h_1}}}{{{h_2}}}} \right) \\

$

And since we have already determined$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} = \dfrac{{{h_1}}}{{{h_2}}}$. Using this result, we’ll get:

$ \Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{{a_1}^2}}{{{a_2}^2}} = \dfrac{{{b_1}^2}}{{{b_2}^2}} = \dfrac{{{c_1}^2}}{{{c_2}^2}} = \dfrac{{{h_1}^2}}{{{h_2}^2}}$

Therefore, the ratio of the areas of similar triangles is equal to the ratio of the squares of their corresponding sides. Hence, this is the required proof.

Note: If two triangles are similar then the ratio of their corresponding sides are same along with the ratio of their corresponding altitudes, their circumradius and their inradius. The measurement of their corresponding angles is also the same.

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