Find the area of the triangle formed by the mid-points of the sides of $△$ ABC, where A(3,2), B(-5,6) and C(8,3).

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Solution

Area of $△$ ABC = $\frac{1}{2} ∣ ∣ ∣ \begin{matrix} 3 - (- 5) & 2 - 6 - 5 - 8 & 6 - 3 \end{matrix} ∣ ∣ ∣$

$= \frac{1}{2} ∣ ∣ ∣ \begin{matrix} 8 & - 4 - 13 & 3 \end{matrix} ∣ ∣ ∣ = \frac{1}{2} | - 28 | = 14$ sq. units

Hence, area of triangle formed by the mid-points of the sides of the $t r i a n g l e A B C$ = $dfrac14(14)=3.5$ sq. units

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Find the area of the triangle formed by the mid-points of the sides of △ ABC, where A(3,2), B(-5,6) and C(8,3).

Area of △ ABC = 12∣∣∣3−(−5)2−6−5−86−3∣∣∣=12∣∣∣8−4−133∣∣∣=12|−28|=14 sq. unitsHence, area of triangle formed by the mid-points of the sides of the triangleABC = dfrac14(14)=3.5 sq. units

Find the area of the triangle formed by the mid-points of the sides of $△$ ABC, where A(3,2), B(-5,6) and C(8,3).