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In △ABC, point M is the midpoint of AB , point D∈ AC so that AD:DC=2:5. If AABC=56 yd2, find ABMC, AAMD, and ACMD.

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Answer: The area of triangle BMC is 28 yd square, the area of AMD is 8 yd square and the area of CMD is 20 yd square.

Step-by-step explanation:

Given information: M is the midpoint of the side AB. The line MC is the median of the triangle ABC.

A median divides the area of triangle in two equal parts, therefore the area of triangle BMC is half of the area of triangle ABC.


\text{Area of }\triangle BMC=(1)/(2)* \text{Area of }\triangle ABC


\text{Area of }\triangle BMC=(1)/(2)* 56


\text{Area of }\triangle BMC=(1)/(2)* =28

Therefore the area of BMC and AMC is 28 yd square.

Draw a perpendicular on AD from M as shown in the figure.


\frac{\text{Area of }\triangle AMD}{\text{Area of }\triangle AMC}= ((1)/(2)* AD* ME)/((1)/(2)* AC* ME)


\frac{\text{Area of }\triangle AMD}{\text{Area of }\triangle AMC}=(AD)/(AC) =(2)/(7)

Therefore the area of AMD is
(2)/(7)th part of the area of AMC.


\text{Area of }\triangle AMD=(2)/(7)* \text{Area of }\triangle AMC


\text{Area of }\triangle AMD=(2)/(7)* 28=8

Therefore the area of AMD is 8 yd square.


\text{Area of }\triangle CMD=\text{Area of }\triangle ABC-\text{Area of }\triangle AMD-\text{Area of }\triangle BMC


\text{Area of }\triangle CMD=56-8-28=20

Therefore the area of CMD is 20 yd square.

In △ABC, point M is the midpoint of AB , point D∈ AC so that AD:DC=2:5. If AABC=56 yd-example-1
User Ahmad Abdelghany
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