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  • Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares on their corresponding sides.      

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

 

 
 
 
 
 

Answers (1)

Given:    \rightarrow    \mathrm{\Delta ABC \sim\Delta PQR}

To prove:    \rightarrow    \mathrm{\frac{ar(ABC)}{ar(PQR)} =\left ( \frac{AB}{PQ} \right )^2 = \left ( \frac{BC}{QR} \right )^2 = \left ( \frac{CA}{RP} \right )^2}

Draw \mathrm{AM \perp BC} and \mathrm{PN \perp QR}

\mathrm{Ar(ABC) = \frac{1}{2}\times BC\times AM}\quad -(i)

\mathrm{Ar(PQR) = \frac{1}{2}\times QR\times PN}\quad -(ii)

Dividing eq (i) by (ii)

        \mathrm{\frac{ar(ABC)}{ar(PQR)} =\frac{\frac{1}{2}\times BC\times AM} {\frac{1}{2}\times QR\times PN}= \frac{BC\times AM}{QR\times PN}\qquad -(iii)}

Now in    \Delta\mathrm{ABM} and     \Delta\mathrm{PQN}

                \mathrm{\angle B = \angle Q\quad (\text{angles of similar } \Delta \text{ are equal})}

and        \mathrm{\angle M = \angle M\quad (\text{Both }90\degree)}

    \Rightarrow \Delta\mathrm{ABM}\sim\Delta\mathrm{PQN}\quad (\text{AA Similarity})

        \therefore \mathrm{\frac{AM}{PN} = \frac{AB}{PQ}\quad \text{(Corresponding sides of similar} \Delta\text{ are proportional)}\qquad -(iv)}

From (ii)

\begin{align*} \mathrm{\frac{ar(ABC)}{ar(PQR)}} & \mathrm{= \frac{BC}{QR}\times\frac{AM}{PN} }\\ & \mathrm{= \frac{BC}{QR}\times \frac{AB}{PQ}\quad [\text{From (iv)}]} \end{align*}

We have given,    \mathrm{\Delta ABC \sim \Delta PQR}

So,        \mathrm{\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} \qquad -(vi)}

\mathrm{\Rightarrow \frac{ar(ABC)}{ar(PQR)} = \frac{AB}{PQ}\times \frac{AB}{PQ}}

\mathrm{\Rightarrow \frac{ar(ABC)}{ar(PQR)} =\left ( \frac{AB}{PQ} \right )^2 }

Now, again using equation (vi), we get

        \mathrm{\frac{ar(ABC)}{ar(PQR)} =\left ( \frac{AB}{PQ} \right )^2 = \left ( \frac{BC}{QR} \right )^2 = \left ( \frac{CA}{RP} \right )^2}

Hence proved,

Posted by

Safeer PP

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