Find the area of the triangle formed by joining the midpoints of the triangles with vertices (2,1),(4,3) and (2.5).

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Solution

Given vertices of triangle are
$(2, 1)$ $(4, 3)$ & $(2, 5)$
then
area of triangle is
$Δ = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 & 1 4 & 3 2 & 5 2 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$
$Δ \Rightarrow \frac{1}{2} | [4 + 6 + 10 - (6 + 20 + 2)] |$
$Δ = \frac{1}{2} | [20 - 20 - 8] | = 4$ sq. units.

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Find the area of the triangle formed by joining the midpoints of the triangles with vertices (2,1),(4,3) and (2.5).

Given vertices of triangle are(2,1) (4,3) & (2,5)thenarea of triangle isΔ=12∣∣ ∣ ∣ ∣∣21432521∣∣ ∣ ∣ ∣∣Δ⇒12|[4+6+10−(6+20+2)]|Δ=12|[20−20−8]|=4 sq. units.