Question

Points M, N, and P are respectively the midpoints of sides AC , BC , and AB of △ABC. Prove that the area of △MNP is on fourth of the area of △ABC.

Answer 1

Answer:

The area of △MNP is one fourth of the area of △ABC.

Step-by-step explanation:

It is given that the points M, N, and P are the midpoints of sides AC, BC and AB respectively. It means AC, BC and AB are median of the triangle ABC.

Median divides the area of a triangle in two equal parts.

Since the points M, N, and P are the midpoints of sides AC, BC and AB respectively, therefore MN, NP and MP are midsegments of the triangle.

Midsegments are the line segment which are connecting the midpoints of tro sides and parallel to third side. According to midpoint theorem the length of midsegment is half of length of third side.

Since MN, NP and MP are midsegments of the triangle, therefore the length of these sides are half of AB, AC and BC respectively. In triangle ABC and MNP corresponding side are proportional.

$\triangle ABC \sim \triangle NMP$

$MP\parallel BC$

$MP=\frac{BC}{2}$

By the property of similar triangles,

$\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{PM^2}{BC^2}$

$\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{(\frac{BC}{2})^2}{BC^2}$

$\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{1}{4}$

Hence proved.