Draw a line segment AB = 3.5 cm. Locate a point P such that PA = 2.5 cm and PB = 3 cm. Produce PA upto point A' such that PA' = 3 times PA and produce PB upto B' such that PB' = 3 times PB. Join A and B. Find the ratio $\frac{A^{'} B^{'}}{A B}$

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Solution

Given, $△ P A B$ is extended to form $△ P A^{'} B^{'}$ such that $P A^{'} = 3 P A$
and $P B^{'} = 3 P B$
Since, the triangle is extended, $△ P A B \sim △ P A^{'} B^{'}$

Hence, $\frac{P A^{'}}{P A} = \frac{A^{'} B^{'}}{A B}$

$\frac{A^{'} B^{'}}{A B} = \frac{3 P A}{P A}$

$\frac{A^{'} B^{'}}{A B} = 3$

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A B = 3.5

P A = 2.5

P B = 3

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P A^{'} = 3 P A

P B

B^{'}

P B^{'} = 3 P B

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\frac{A^{'} B^{'}}{A B} .

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= 10

∠

= 60^{o}

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