Question

in triangle ABC .a PQ meets AB in p and AC in Q . if AP =1 cm , PB is 3cm ,AQ is 1.5 cm ,QC is 4.5 cm

prove that area of triangle APQ is one sixteenth of the triangle ABC.

if will be satisfied then answer will be mark as brainliest

*****20 points***********

Answer 1

Answer:

Step-by-step explanation:15-28x=36

15-36=28x

21=28x

21÷28=x

3/4=x

Answer 2

To prove that area of the triangle APQ is one-sixteenth of the triangle ABC.

Step-by-step explanation:

Given:
$AP=1cm$

$PB=3cm$

$AQ=1.5cm$

$QC=4.5cm$

To find:

To prove that area of the triangle APQ is one-sixteenth of the triangle ABC.

Solution:

In Δ$APQ$ and Δ$ABC$

∠$A=$∠$A$

$\frac{AP}{AB}=\frac{AQ}{AC}$

By area of similar triangle theorem

$\frac{area(\triangle APQ)}{area (\triangle ABC)} =\frac{1^{2} }{4^{2} }$

$\frac{area(\triangle APQ)}{area (\triangle ABC)} =\frac{1^{2} }{16^{2} }\times area(\triangle ABC)$

Hence, we proved that area of the triangle APQ is one-sixteenth of the triangle ABC.

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