Find the ratio of the areas of two similar triangles ABC and PQR shown in the figure where AM and PN are the medians of the $△$ ABC and $△$ PQR respectively.

AB²/PQ²

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AM²/PN²

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Both a and b

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PQ²/QR²

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Solution

The correct option is C
Both a and b

We are given two triangles ABC and PQR such that $△$ ABC ~ $△$ PQR

So, $\frac{A B}{P Q} = \frac{B C}{Q R} = \frac{A C}{P R}$

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

$∠$ B = $∠$ Q

$△$ ABM ~ $△$ PQN {SAS similarity}

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

So , $\frac{a r e a o f △ A B C}{a r e a o f △ P Q R}$ = $\frac{A B^{2}}{P Q^{2}} = \frac{A M^{2}}{P N^{2}}$

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If $△$ ABC and $△$ PQR are two similar triangles shown in the figure. AM and PN are the medians on $△$ ABC and $△$ PQR respectively. The ratio of areas of $△$ ABC and $△$ PQR is 9:25. If AM = PO = 5 cm. Find the value of 3ON.

$△$ ABC and $△$ PQR are two similar triangles as shown in the figure such that $\frac{A r e a o f A B C}{A r e a o f P Q R}$ = $\frac{9}{25}$ . AM and PN are the medians on $△$ ABC and $△$ PQR respectively. If AM = PO = 5 cm, find the value of 3ON.

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