Given a square , let be a point within it. Join , , , (Fig. 6.33). What is the ratio of the areas of the red region ( and ) and the green region ( and )?

Understand the Question For any point inside a square with side length , connecting to all four vertices forms four triangles: , , , and . The sum of perpendicular distances from to two opposite sides always equals the side length of the square. Therefore, the sum of areas of any pair of opposite triangles is always equal to half the total area of the square, . Step 1 · Calculate the Area of the Red Region Let the side length of square be . Draw a line through parallel to and . Let the perpendicular distance from to be . Then the perpendicular distance from to is . Summing both areas for the total red area: Step 2 · Calculate the Area of the Green Region Draw a line through parallel to and . Let the perpendicular distance from to be . Then the perpendicular distance from to is . Summing both areas for the total green area: Step 3 · Find the Ratio of the Areas Comparing the two areas: Answer > Common Mistakes Assuming must be the center: The ratio holds for any point inside the square, not just when is at the exact center. Variable heights confusion: Forgetting that the sum of perpendiculars from an interior point to two opposite parallel sides of a square always equals the side length ().

Measuring Space: Perimeter and Area | Exercise 6.2

Question 10

Given a square $ABCD$ , let $P$ be a point within it. Join $PA$ , $PB$ , $PC$ , $PD$ (Fig. 6.33). What is the ratio of the areas of the red region ( $\Delta PAB$ and $\Delta PCD$ ) and the green region ( $\Delta PBC$ and $\Delta PDA$ )?

Solution

Understand the Question

For any point $P$ inside a square $ABCD$ with side length $a$ , connecting $P$ to all four vertices forms four triangles: $\Delta PAB$ , $\Delta PCD$ , $\Delta PBC$ , and $\Delta PDA$ .
The sum of perpendicular distances from $P$ to two opposite sides always equals the side length $a$ of the square.
Therefore, the sum of areas of any pair of opposite triangles is always equal to half the total area of the square, $\dfrac{1}{2}a^2$ .

Step 1 · Calculate the Area of the Red Region

Let the side length of square $ABCD$ be $a$ . Question diagram

Draw a line through $P$ parallel to $AB$ and $CD$ . Let the perpendicular distance from $P$ to $AB$ be $h$ . Then the perpendicular distance from $P$ to $CD$ is $a - h$ .

\begin{aligned} \text{Area}(\Delta PAB) &= \dfrac{1}{2} \times \text{base} \times \text{height} \\[0.6em] &= \dfrac{1}{2} \times AB \times h \\[0.6em] &= \dfrac{1}{2} \times a \times h \end{aligned}

\begin{aligned} \text{Area}(\Delta PCD) &= \dfrac{1}{2} \times \text{base} \times \text{height} \\[0.6em] &= \dfrac{1}{2} \times CD \times (a - h) \\[0.6em] &= \dfrac{1}{2} \times a \times (a - h) \end{aligned}

Summing both areas for the total red area:

\begin{aligned} \text{Total Red Area} &= \text{Area}(\Delta PAB) + \text{Area}(\Delta PCD) \\[0.6em] &= \dfrac{1}{2} ah + \dfrac{1}{2} a(a - h) \\[0.6em] &= \dfrac{1}{2} a [h + (a - h)] \\[0.6em] &= \dfrac{1}{2} a [h + a - h] \\[0.6em] &= \dfrac{1}{2} a [a] \\[0.6em] &= \dfrac{1}{2} a^2 \end{aligned}

Step 2 · Calculate the Area of the Green Region

Draw a line through $P$ parallel to $BC$ and $AD$ . Let the perpendicular distance from $P$ to $BC$ be $k$ . Then the perpendicular distance from $P$ to $AD$ is $a - k$ .

\begin{aligned} \text{Area}(\Delta PBC) &= \dfrac{1}{2} \times \text{base} \times \text{height} \\[0.6em] &= \dfrac{1}{2} \times BC \times k \\[0.6em] &= \dfrac{1}{2} \times a \times k \end{aligned}

\begin{aligned} \text{Area}(\Delta PDA) &= \dfrac{1}{2} \times \text{base} \times \text{height} \\[0.6em] &= \dfrac{1}{2} \times AD \times (a - k) \\[0.6em] &= \dfrac{1}{2} \times a \times (a - k) \end{aligned}

Summing both areas for the total green area:

\begin{aligned} \text{Total Green Area} &= \text{Area}(\Delta PBC) + \text{Area}(\Delta PDA) \\[0.6em] &= \dfrac{1}{2} ak + \dfrac{1}{2} a(a - k) \\[0.6em] &= \dfrac{1}{2} a [k + (a - k)] \\[0.6em] &= \dfrac{1}{2} a [k + a - k] \\[0.6em] &= \dfrac{1}{2} a [a] \\[0.6em] &= \dfrac{1}{2} a^2 \end{aligned}

Step 3 · Find the Ratio of the Areas

Comparing the two areas:

\begin{aligned} \text{Ratio} &= \text{Total Red Area} : \text{Total Green Area} \\[0.6em] &= \dfrac{1}{2} a^2 : \dfrac{1}{2} a^2 \\[0.6em] &= 1 : 1 \end{aligned}

Answer

$1 : 1$

Common Mistakes

Assuming $P$ must be the center: The ratio $1 : 1$ holds for any point $P$ inside the square, not just when $P$ is at the exact center.
Variable heights confusion: Forgetting that the sum of perpendiculars from an interior point to two opposite parallel sides of a square always equals the side length $a$ ( $h + (a - h) = a$ ).

Question 10

More questions in Exercise 6.2