Question

In △ABC, D is the midpoint of segment BC. M is the midpoint of
segment AD. If area of the △CMD is 17, then find the area of △ABC.

Answer 1

$\large\underline{\sf{Solution-}}$

Given that, In △ ABC

• D is the midpoint of segment BC.

• M is the midpoint of segment AD.

• Area of the △CMD is 17 square units.

Now, In △ ADC

• M is the midpoint of segment AD.

We know, median of a triangle divides it in to two triangles of equal area.

$\rm\implies \:ar( \triangle \: ADC) = 2 \: ar( \triangle \: CMD)$

$\rm\implies \:ar( \triangle \: ADC) = 2 \times 17 = 34 \: sq. \: units$

Now, In △ ABC

• D is the midpoint of segment BC.

We know, median of a triangle divides it in to two triangles of equal area.

$\rm\implies \:ar( \triangle \: ABC) \: = \: 2 \: ar( \triangle \: ADC)$

$\rm\implies \:ar( \triangle \: ABC) \: = \: 2 \times 34$

$\rm\implies \:ar( \triangle \: ABC) \: = \: 68 \: sq. \: units$

$\rule{190pt}{2pt}$

Short Cut Trick :-

In △ ABC, D is the midpoint of segment BC. M is the midpoint of segment AD, then the area of △ABC = 4 area of △ CMD.

$\rule{190pt}{2pt}$

Additional information :-

1. Diagonal of a parallelogram divides it in to two triangles of equal area.

2. Parallelograms on the same base and between same parallels are equal in area.

3. Triangles on the same base and between same parallels are equal in area.

4. If parallelogram and triangle are on the same base and between same parallels, then area of triangle is equal to half the area of parallelogram.