Question

PQ is parallel to BC and AP:PB=1:2 in triangle ABC.find ar(apq)/ar(abc).

Answer 1

Answer:

$\frac{ar(\triangle APQ)}{ar(\triangle ABC)}=1:9$

Step-by-step explanation:

Given PQ//BC

AP:PB = 1:2,

AP: AB = 1:3,

∆APQ ~ ∆ABC

$\frac{ar(\triangle APQ)}{ar(\triangle ABC)}\\=\left(\frac{AP}{AB}\right)^{2}$

/* The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides */

$=\left(\frac{1}{3}\right)^{2}\\=\frac{1}{9}$

Therefore,

$\frac{ar(\triangle APQ)}{ar(\triangle ABC)}=1:9$

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Answer 2

Answer:

AP:PB

1:2

so,

AP:AB will be 1:3

now,

∆APQ/∆ABC=(1/3)2

=1/9

hope it will help you

thank you ☺️