Area of the Triangle formed by Joining the Middle Points of the Sides of a Triangle is Equal to One-fourth Area of the given Triangle

In this article we have to show that the Area of the Triangle formed by Joining the Middle Points of the Sides of a Triangle is Equal to One-fourth Area of the given Triangle.

Area of the Triangle formed by Joining the Middle Points of the Sides of a Triangle is Equal to One-fourth Area of the given Triangle – Theorem

Theorem:
Prove that Area of the Triangle formed by Joining the Middle Points of the Sides of a Triangle is Equal to One-fourth Area of the given Triangle?
Proof:
Given:
X, Y, and Z are the middle points of sides QR, RP and PQ respectively of the triangle PQR.
To prove that:
ar(∆XYZ) = 1/4 × ar(∆PQR)
Proof:
ZY = ∥QX [ Z, Y are the midpoints of PQ and PR respectively. So, using the Midpoint Theorem we get it]…. equation 1
QXYZ is a parallelogram.[equation 1 implies it]…… equation 2
ar(∆XYZ) = ar(∆QZX).[XZ is a diagonal of the parallelogram QXYZ]…. equation 3
ar(∆XYZ) = ar(∆RXY), and ar(∆XYZ) = ar(∆PZY).[ Similarly as equation 3 ]…. equation 4
3 × ar(∆XYZ) = ar(∆QZX) + ar(∆RXY) = ar(∆PZY). [Adding from equation 3 and 4 ]…. equation 5
4 × ar(∆XYZ) = ar(∆XYZ) + ar(∆QZX) + ar(∆RXY) = ar(∆PZY). [ Adding ar(∆XYZ) on both side of equality in equations ]…. equation 6
4 × ar(∆XYZ) = ar(∆PQR), i.e.,
ar(∆XYZ) = 1/4 × ar(∆PQR) [By addition axiom for area]….. equation 7
Hence proved.

Also, See:

FAQs on Area of the Triangle formed by Joining the Middle Points of the Sides of a Triangle is Equal to One-fourth Area of the given Triangle

1. What is the perimeter of the triangle formed joining the midpoints of the sides?

From the midpoint theorem, the line joining the midpoints of two sides of a triangle is half the third side. Thus the perimeter of the triangle formed by joining the midpoints of the sides of a triangle is half of the original triangle.

2. What is the relation between the area of triangles standing on the same base and between the same parallels?

The triangle and a parallelogram are having the same base and are between the same parallels. Therefore the area of the triangle is equal to half the area of the parallelogram.

3. Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

Given:
△ABC and △ABD are two triangles on the same base AB such that area(△ABC) = area(△ABD).
Join CD
To prove:
CD ∥ AB
Construction:
Draw the altitudes of △ABC and △ABD through C and D and meeting base at E and F respectively.
Proof:
Since CE⊥AB and DF⊥AB(by construction)
Therefore lines perpendicular to the same line are parallel to each other.
Therefore CE ∥ DF….. equation 1
Now, area(△ABC)= ½×AB×CE
area(△ABD) = 1/2 ×AB×DF
Since area(△ABC)=area(△ABD)(Given)
Therefore CF=DF….. equation 2
Now, In CDFE, CF∥DF, and CE∥DF
Since one pair of opposite sides are equal and parallel
Hence CDFE is a parallelogram.
So, CD ∥ EF (opposite sides of the parallelogram are parallel)
Therefore CD∥AB
Hence, proved.

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