Question

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.

Answer 1

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

Explanation:

It is given that the 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.

$\text{ Area of }\triangle BMC =\frac{1}{2}\times \text{ Area of }\triangle ABC}$

$\text{ Area of }\triangle BMC =\frac{1}{2}\times 56}$

$\text{ Area of }\triangle BMC =28$

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

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

$\frac{\text{ Area of }\triangle AMD}{\text{ Area of }\triangle AMC}= \frac{\frac{1}{2}\times AD\times ME}{\frac{1}{2}\times AC\times ME} =\frac{AD}{AC}= \frac{2}{7}$

Therefore the area of AMD is  $\frac{2}{7}th$  part of the area of AMC.

$\text{ Area of }\triangle AMD =\frac{2}{7}\times \text{ Area of }\triangle AMC}$

$\text{ Area of }\triangle AMD =\frac{2}{7}\times 28$

$\text{ Area of }\triangle AMD =8$

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.