Question

prove that the ratio of two similar triangles is equal to the ratio of square of their corresponding sides

Answer 1

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Answer 2

Ratio of two similar triangles is equal to the ratio of square of their corresponding sides is proved

Solution:

To prove:  Ratio of two similar triangles is equal to the ratio of square of their corresponding sides

Consider two similar triangles ABC and PQR

The figure is attached below

We have to prove:

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

Let us first calculate the area of triangles

We know that,

$\text { area of triangle }=\frac{1}{2} \times \text { base } \times \text { height }$

$\begin{array}{l}{\text { Area of triangle } \mathrm{ABC}=\frac{1}{2} \times B C \times A M} \\\\ {\text { Area of triangle } \mathrm{PQR}=\frac{1}{2} \times Q R \times P N}\end{array}$

$\frac{area(ABC)}{area(PQR)} ==\frac{\frac{1}{2} \times B C \times A M}{\frac{1}{2} \times Q R \times P N}=\frac{B C \times A M}{Q R \times P N}$  ---- eqn 1

If we compare triangle ABM and PQN, we observe

Angle B = angle Q     { since ,triangle ABC and PQN are similar}

Angle M = angle N      { both are 90 degree}

So, triangle ABM and PQN are similar by Angle angle criterion

$\frac{A M}{P N}=\frac{A B}{P Q}$  ----- eqn 2

Also, triangle ABC and PQR are similar

$\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P}$  ---- eqn 3

Therefore,

$\frac{area(A B C)}{area(P Q R)}=\frac{A B}{P Q} \times \frac{A M}{P N} \quad\{\text { from }(1) \text { and }(3)\}$

$\frac{area(A B C)}{area(P Q R)}=\frac{A B}{P Q} \times \frac{A B}{P Q}=\left(\frac{A B}{P Q}\right)^{2}$

From equation(3), we get

$\frac{area(A B C)}{area(P Q R)}=\left(\frac{A B}{P Q}\right)^{2}=\left(\frac{B C}{Q R}\right)^{2}=\left(\frac{C A}{R P}\right)^{2}$

Thus proved

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