Question

n △ABC points M and Q are points on the sides AB and BC respectively, so that AM=2MB and BQ=QC. Find area of quadrilateral MBQP if AQ ∩ CM =P and AABC=30 cm2.

Answer 1

Denote the area of quadilateral MBQP as A, the area of the triangle MBP as $A_1$ and the area of the triangle BPQ as $A_2.$

Note that:

1.

$A_{\triangle CMB}=\dfrac{1}{3}A_{\triangle ABC}=\dfrac{1}{3}\cdot 30=10\ cm^2.$

2.

$A_{\triangle CPQ}=A_2.$

3.

$A_{\triangle ABQ}=\dfrac{1}{2}A_{\triangle ABC}=\dfrac{1}{2}\cdot 30=15\ cm^2.$

4.

$A_{\triangle APM}=2A_1.$

Now

$A+A_{\triangle CPQ}=A_{\triangle CMB}\Rightarrow A+A_2=10\ cm^2.$

$A+A_{\triangle AMP}=A_{\triangle ABQ}\Rightarrow A+2A_1=15\ cm^2.$

You get the system of three equations:

$\left\{\begin{array}{l}A=A_1+A_2\\A+A_2=10\\A+2A_1=15\end{array}\right..$

Substitute the first equation into the last two:

$\left\{\begin{array}{l}A_1+A_2+A_2=10\\A_1+A_2+2A_1=15\end{array}\right.\Rightarrow \left\{\begin{array}{l}A_1+2A_2=10\\A_2+3A_1=15\end{array}\right..$

From the first equation $A_1=10-2A_2$ and then

$A_2+3(10-2A_2)=15,\\ \\A_2+30-6A_2=15,\\ \\-5A_2=-15,\\ \\A_2=3\ cm^2.$

Thus,

$A_1=10-2\cdot 3=10-6=4\ cm^2$

and

$A=4+3=7\ cm^2.$

Answer: 7 sq. cm

Answer 2

Answer:

In △ABC points M and Q are points on the sides

AB

 and

BC

 respectively, so that AM=2MB and BQ=QC. Find area of quadrilateral MBQP if

AQ

∩

CM

=P and AABC=30 cm2.

Answer : 7cm

Hope this Helps