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Find the area of the triangle formed by joining the midpoints

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Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are $A (2, 1), B (4, 3)$ and $C (2, 5)$ .

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Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are

A (2, 1), B (4, 3)

and

C (2, 5)

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Text solutionVerified

Given: A triangle whose vertices are

A (2, 1), B (4, 3)

and

C (2, 5)

Let D, E and F are the midpoints of the sides CB, CA and AB respectively of

Δ A BC

, as shown in the figure.

Find vertices of D, E and F:

Midpoint formula:

(x, y) = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

Vertices of D:

= (\frac{4 + 2}{2}, \frac{3 + 5}{2})

= (\frac{6}{2}, \frac{8}{2}) = (3, 4)

Vertices of E:

= (\frac{2 + 2}{2}, \frac{5 + 1}{2})

= (\frac{4}{2}, \frac{6}{2}) = (2, 3)

Vertices of F:

= (\frac{2 + 4}{2}, \frac{1 + 3}{2})

= (\frac{6}{2}, \frac{4}{2}) = (3, 2)

Area of triangle DEF:

We know that:

Area of

Δ A BC = \frac{1}{2} [x_{1} (y_{2} - y_{2}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]

Area of triangle DEF

=

= \frac{1}{2} [3 (3 - 2) + 2 (2 - 4) + 3 (4 - 3)]

= \frac{1}{2} [3 \times 1 + 2 \times (- 2) + 3 \times 0]

= \frac{1}{2} \times 2 = 1

sq. units.

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Question Text	Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are $A (2, 1), B (4, 3)$ and $C (2, 5)$ .
Answer Type	Text solution:1
Upvotes	150

Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are A(2,1),B(4,3) and C(2,5).

Text solutionVerified

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Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are $A (2, 1), B (4, 3)$ and $C (2, 5)$ .