Area of Parallelogram

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Problem: Given midpoints of three sides of a parallelogram. Find the area of the parallelogram.

Observation:Consider a parallelogram $ABCD$ and three points $P, \, Q$ and $R$ as midpoints of the sides $AB, \, BC$ and $CD$ respectively.

Let’s connect the points to each other with straight lines.Now, $PR$ is parallel to both $BC$ and $AD$ as it divides the sides $AB$ and $CD$ proportionally. So, $PR$ divides $ABCD$ into two equal parallelograms $PBCR$ and $PRDA$ .

\therefore ABCD = 2 \times PBCR

Let’s draw a perpendicular from $BC$ on point $E$ such that $QE$ be the height of the triangle $PQR$ and parallelogram $PBCR$ .

We know,

\text{Area Parallelogram} = Base \times Height \\ \text{And Area of Triangle} = \frac{1}{2} \times Base \times Height

So,

\text{Area of } PBCR = PR \times QE \\ \text{Area of } \bigtriangleup PQR = \frac{1}{2} \times PR \times QE

Hence,

PBCR = 2 \times \bigtriangleup PQR

Therefore,

ABCD = 2 \times PBCR = 4 \times \bigtriangleup PQR

Solution Idea: As the coordinates of the vertices of the triangle is given, we can use shoelace theorem to find the area.

\begin{array}{c} \text{Area of } ABCD = 4 \times \frac{1}{2} \times \begin{vmatrix} x_1 \quad x_2 \quad x_3 \quad x_1 \\ y_1 \quad y_2 \quad y_3 \quad y_1 \end{vmatrix} \\ = 2 \times (x_1 y_2 + x_2 y_3 + x_3 y_1 - x_2 y_1 - x_3 y_2 - x_1 y_3) \end{array}

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