Question

in the given figure,P is a point on side BC of triangle ABC such that BP:PC=1:2 and Q is a point on AP such that PQ :QA=2:3 .show that area of triangle AQC :area of triangle ABC=2:5 in the given figure,P is a point on side BC of triangle ABC such that BP:PC=1:2 and Q is a point on AP such that PQ :QA=2:3 .show that area of triangle AQC :area of triangle ABC=2:5

Answer 1

Answer:

The area of triangle AQC :area of triangle ABC=2:5

Step-by-step explanation:

Given information: BP:PC=1:2 and PQ :QA=2:3.

Let the area of triangle ABC be x.

If a point divides a side in m:n, then the line joining the opposite vertex and the point divides the area of triangle in m:n.

BP:PC=1:2,

$A(\triangle ABP):A(\triangle APC)=1:2$

$A(\triangle ABP)=\frac{1}{2}x$

$A(\triangle APC)=\frac{2x}{3}$

PQ :QA=2:3,

$A(\triangle PQC):A(\triangle AQC)=2:3$

$A(\triangle AQC)=\frac{2}{5}\times A(\triangle APC)=\frac{3}{5}\times\frac{2x}{3}=\frac{2x}{5}$

$\frac{A(\triangle AQC)}{A(\triangle ABC)}=\frac{\frac{2x}{5}}{x}= \frac{2}{5}$

Hence proved that area of triangle AQC :area of triangle ABC=2:5.

Answer 2

Answer:

This is the answer of the question......