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.
2 answers:
Denote the area of quadilateral MBQP as A, the area of the triangle MBP as
and the area of the triangle BPQ as ![A_2.](https://tex.z-dn.net/?f=A_2.)
Note that:
1.
![A_{\triangle CMB}=\dfrac{1}{3}A_{\triangle ABC}=\dfrac{1}{3}\cdot 30=10\ cm^2.](https://tex.z-dn.net/?f=A_%7B%5Ctriangle%20CMB%7D%3D%5Cdfrac%7B1%7D%7B3%7DA_%7B%5Ctriangle%20ABC%7D%3D%5Cdfrac%7B1%7D%7B3%7D%5Ccdot%2030%3D10%5C%20cm%5E2.)
2.
![A_{\triangle CPQ}=A_2.](https://tex.z-dn.net/?f=A_%7B%5Ctriangle%20CPQ%7D%3DA_2.)
3.
![A_{\triangle ABQ}=\dfrac{1}{2}A_{\triangle ABC}=\dfrac{1}{2}\cdot 30=15\ cm^2.](https://tex.z-dn.net/?f=A_%7B%5Ctriangle%20ABQ%7D%3D%5Cdfrac%7B1%7D%7B2%7DA_%7B%5Ctriangle%20ABC%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%2030%3D15%5C%20cm%5E2.)
4.
![A_{\triangle APM}=2A_1.](https://tex.z-dn.net/?f=A_%7B%5Ctriangle%20APM%7D%3D2A_1.)
Now
![A+A_{\triangle CPQ}=A_{\triangle CMB}\Rightarrow A+A_2=10\ cm^2.](https://tex.z-dn.net/?f=A%2BA_%7B%5Ctriangle%20CPQ%7D%3DA_%7B%5Ctriangle%20CMB%7D%5CRightarrow%20A%2BA_2%3D10%5C%20cm%5E2.)
![A+A_{\triangle AMP}=A_{\triangle ABQ}\Rightarrow A+2A_1=15\ cm^2.](https://tex.z-dn.net/?f=A%2BA_%7B%5Ctriangle%20AMP%7D%3DA_%7B%5Ctriangle%20ABQ%7D%5CRightarrow%20A%2B2A_1%3D15%5C%20cm%5E2.)
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..](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7DA%3DA_1%2BA_2%5C%5CA%2BA_2%3D10%5C%5CA%2B2A_1%3D15%5Cend%7Barray%7D%5Cright..)
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..](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7DA_1%2BA_2%2BA_2%3D10%5C%5CA_1%2BA_2%2B2A_1%3D15%5Cend%7Barray%7D%5Cright.%5CRightarrow%20%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7DA_1%2B2A_2%3D10%5C%5CA_2%2B3A_1%3D15%5Cend%7Barray%7D%5Cright..)
From the first equation
and then
![A_2+3(10-2A_2)=15,\\ \\A_2+30-6A_2=15,\\ \\-5A_2=-15,\\ \\A_2=3\ cm^2.](https://tex.z-dn.net/?f=A_2%2B3%2810-2A_2%29%3D15%2C%5C%5C%20%5C%5CA_2%2B30-6A_2%3D15%2C%5C%5C%20%5C%5C-5A_2%3D-15%2C%5C%5C%20%5C%5CA_2%3D3%5C%20cm%5E2.)
Thus,
![A_1=10-2\cdot 3=10-6=4\ cm^2](https://tex.z-dn.net/?f=A_1%3D10-2%5Ccdot%203%3D10-6%3D4%5C%20cm%5E2)
and
![A=4+3=7\ cm^2.](https://tex.z-dn.net/?f=A%3D4%2B3%3D7%5C%20cm%5E2.)
Answer: 7 sq. cm
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
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