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

Step-by-step 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 and area of triangle BMC and area of triangle AMC is equal.

`\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 =28`

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

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) =(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`

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

Therefore the area of triangle AMD is 8 yd².

`\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 triangle CMD is 20 yd².