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:

- Every Diagonal of a Parallelogram Divides it into Two Triangles of Equal Area
- Area of a Triangle is Half that of a Parallelogram on the Same Base and between the Same Parallels
- Triangles on the Same Base and between the Same Parallels are Equal in Area

### 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.