Find the side if a square whose area is equal to the area of a triangle whose vertices are $A (3, 2), B (11, 8)$ and $C (8, 12)$ .

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Solution

$Δ A B C = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 3 & 2 & 1 11 & 8 & 1 8 & 12 & 1 \end{matrix} ∣ ∣ ∣ ∣$

$R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$

$Δ A B C = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 3 & 2 & 1 8 & 6 & 0 5 & 10 & 0 \end{matrix} ∣ ∣ ∣ ∣ = \frac{1}{2} ∣ ∣ ∣ \begin{matrix} 8 & 6 5 & 10 \end{matrix} ∣ ∣ ∣ = 25$

Hence side of the square is $= \sqrt{25} = 5$

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