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.
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.
[tex]\text{Area of }\triangle BMC=\frac{1}{2}\times \text{Area of }\triangle ABC[/tex]
[tex]\text{Area of }\triangle BMC=\frac{1}{2}\times 56[/tex]
[tex]\text{Area of }\triangle BMC=\frac{1}{2}\times =28[/tex]
Therefore the area of BMC and AMC is 28 yd square.
Draw a perpendicular on AD from M as shown in the figure.
[tex]\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}[/tex]
[tex]\frac{\text{Area of }\triangle AMD}{\text{Area of }\triangle AMC}=\frac{AD}{AC} =\frac{2}{7}[/tex]
Therefore the area of AMD is [tex]\frac{2}{7}[/tex]th part of the area of AMC.
[tex]\text{Area of }\triangle AMD=\frac{2}{7}\times \text{Area of }\triangle AMC[/tex]
[tex]\text{Area of }\triangle AMD=\frac{2}{7}\times 28=8[/tex]
Therefore the area of AMD is 8 yd square.
[tex]\text{Area of }\triangle CMD=\text{Area of }\triangle ABC-\text{Area of }\triangle AMD-\text{Area of }\triangle BMC[/tex]
[tex]\text{Area of }\triangle CMD=56-8-28=20[/tex]
Therefore the area of CMD is 20 yd square.
