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.

The area of triangle BMC is 28 yd² . The area of triangle AMD is 8 yd². The area of CMD is 20 yd².

Explanation:

It is given that the M is the midpoint of the side AB. The line MC is the median of the triangle ABC.

First we have to draw a figure according to the given information.

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 and area of triangle BMC and area of triangle AMC is equal.

Therefore the area of triangle BMC and triangle AMC is 28 yd².

Draw a perpendicular on AD from M as shown in the figure. i.e., MP.

Therefore the area of AMD is    part of the area of AMC.

Therefore the area of triangle AMD is 8 yd².

Therefore the area of triangle CMD is 20 yd².

b.my answer is b.

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