Measuring Space: Perimeter and Area | Exercise 6.2

Question 8

If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.)

Solution

Understand the Question

Let $ABCD$ be any 4-gon (quadrilateral) with $P, Q, R, S$ as the mid-points of sides $AB, BC, CD, DA$ respectively.
A line segment joining the midpoints of two sides of a triangle forms a corner triangle whose area is $\dfrac{1}{4}$ of that triangle's area.
By drawing diagonals $AC$ and $BD$ , we can find the sum of the areas of the four corner triangles $(\triangle PBQ, \triangle SDR, \triangle APS, \triangle CQR)$ .
Subtracting the sum of these four corner areas from the total area of $ABCD$ gives the area of the inner parallelogram $PQRS$ , which equals $\dfrac{1}{2} \times \text{Area}(ABCD)$ .

Step 1 · Prove that PQRS is a Parallelogram

Let $ABCD$ be a 4-gon with $P, Q, R, S$ as the midpoints of sides $AB, BC, CD, DA$ respectively. Diagram 1

Join diagonal $AC$ .

In $\triangle ABC$ , $P$ and $Q$ are the midpoints of $AB$ and $BC$ . By the Midpoint Theorem: $PQ \parallel AC \quad \text{and} \quad PQ = \dfrac{1}{2}AC$

In $\triangle ADC$ , $S$ and $R$ are the midpoints of $AD$ and $CD$ . By the Midpoint Theorem: $SR \parallel AC \quad \text{and} \quad SR = \dfrac{1}{2}AC$

Therefore: $PQ \parallel SR \quad \text{and} \quad PQ = SR$

Since one pair of opposite sides is equal and parallel, $PQRS$ is a parallelogram.

Step 2 · Calculate the Area of Parallelogram PQRS

Diagram 2

The segment joining the midpoints of two sides of a triangle divides its area such that the smaller triangle has an area equal to $\dfrac{1}{4}$ of the original triangle.

Considering diagonal $AC$ : $\text{Area}(\triangle PBQ) = \dfrac{1}{4} \text{Area}(\triangle ABC)$ $\text{Area}(\triangle SDR) = \dfrac{1}{4} \text{Area}(\triangle ADC)$

Adding these two areas:

\begin{aligned} \text{Area}(\triangle PBQ) + \text{Area}(\triangle SDR) &= \dfrac{1}{4} \text{Area}(\triangle ABC) + \dfrac{1}{4} \text{Area}(\triangle ADC) \\[0.6em] &= \dfrac{1}{4} \left(\text{Area}(\triangle ABC) + \text{Area}(\triangle ADC)\right) \\[0.6em] &= \dfrac{1}{4} \text{Area}(ABCD) \end{aligned}

Similarly, considering diagonal $BD$ : $\text{Area}(\triangle APS) = \dfrac{1}{4} \text{Area}(\triangle ABD)$ $\text{Area}(\triangle CQR) = \dfrac{1}{4} \text{Area}(\triangle BCD)$

Adding these two areas:

\begin{aligned} \text{Area}(\triangle APS) + \text{Area}(\triangle CQR) &= \dfrac{1}{4} \text{Area}(\triangle ABD) + \dfrac{1}{4} \text{Area}(\triangle BCD) \\[0.6em] &= \dfrac{1}{4} \left(\text{Area}(\triangle ABD) + \text{Area}(\triangle BCD)\right) \\[0.6em] &= \dfrac{1}{4} \text{Area}(ABCD) \end{aligned}

Sum of all four corner triangles:

\begin{aligned} \text{Total corner area} &= \text{Area}(\triangle PBQ) + \text{Area}(\triangle SDR) + \text{Area}(\triangle APS) + \text{Area}(\triangle CQR) \\[0.6em] &= \dfrac{1}{4} \text{Area}(ABCD) + \dfrac{1}{4} \text{Area}(ABCD) \\[0.6em] &= \dfrac{1}{2} \text{Area}(ABCD) \end{aligned}

Subtracting the corner areas from the total area:

\begin{aligned} \text{Area}(PQRS) &= \text{Area}(ABCD) - \text{Total corner area} \\[0.6em] &= \text{Area}(ABCD) - \dfrac{1}{2} \text{Area}(ABCD) \\[0.6em] &= \dfrac{1}{2} \text{Area}(ABCD) \end{aligned}

Answer

$\text{Area}(\text{parallelogram } PQRS) = \dfrac{1}{2} \times \text{Area}(\text{4-gon } ABCD)$

Common Mistakes

Area Ratio Misunderstanding: Assuming that joining the midpoints gives triangles with $\dfrac{1}{2}$ the area of the original triangle instead of $\left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$ .
Special Case Assumption: Incorrectly assuming this property only holds for regular shapes (like rectangles or squares); this property (Varignon's Theorem) holds for any general 4-gon.

Question 8

More questions in Exercise 6.2