Question 10

We will calculate the areas of the red and green regions separately and then find their ratio.

Step 1 — Set up the square and point P

Let the side length of the square ABCD be a. Let P be any point inside the square.

Step 2 — Calculate the area of the red region

The red region consists of two triangles. These triangles are $\Delta$PAB and $\Delta$PCD. Let's draw a line through P parallel to AB and CD. This line is perpendicular to AB and CD. Let the perpendicular distance from P to AB be h. Then the perpendicular distance from P to CD will be a - h. The base of $\Delta$PAB is AB, which is a. The base of $\Delta$PCD is CD, which is a. Area of $\Delta$PAB is: $= \frac{1}{2} \times \text{base} \times \text{height}$ $= \frac{1}{2} \times \text{AB} \times h$ $= \frac{1}{2} \times a \times h$ Area of $\Delta$PCD is: $= \frac{1}{2} \times \text{base} \times \text{height}$ $= \frac{1}{2} \times \text{CD} \times (a - h)$ $= \frac{1}{2} \times a \times (a - h)$ Now, let's find the total red area. Total red area is Area($\Delta$PAB) + Area($\Delta$PCD). $= \frac{1}{2} ah + \frac{1}{2} a(a - h)$ $= \frac{1}{2} a [h + (a - h)]$ $= \frac{1}{2} a [h + a - h]$ $= \frac{1}{2} a [a]$

$\boxed{\frac{1}{2} a^2}$

Step 3 — Calculate the area of the green region

The green region also consists of two triangles. These triangles are $\Delta$PBC and $\Delta$PDA. Let's draw a line through P parallel to BC and AD. This line is perpendicular to BC and AD. Let the perpendicular distance from P to BC be k. Then the perpendicular distance from P to AD will be a - k. The base of $\Delta$PBC is BC, which is a. The base of $\Delta$PDA is AD, which is a. Area of $\Delta$PBC is: $= \frac{1}{2} \times \text{base} \times \text{height}$ $= \frac{1}{2} \times \text{BC} \times k$ $= \frac{1}{2} \times a \times k$ Area of $\Delta$PDA is: $= \frac{1}{2} \times \text{base} \times \text{height}$ $= \frac{1}{2} \times \text{AD} \times (a - k)$ $= \frac{1}{2} \times a \times (a - k)$ Now, let's find the total green area. Total green area is Area($\Delta$PBC) + Area($\Delta$PDA). $= \frac{1}{2} ak + \frac{1}{2} a(a - k)$ $= \frac{1}{2} a [k + (a - k)]$ $= \frac{1}{2} a [k + a - k]$ $= \frac{1}{2} a [a]$

$\boxed{\frac{1}{2} a^2}$

Step 4 — Find the ratio of the areas

We found the total red area. We also found the total green area. The total red area is $\frac{1}{2} a^2$. The total green area is $\frac{1}{2} a^2$. Let's find the ratio of the red region to the green region. Ratio = Total red area : Total green area $= \frac{1}{2} a^2 : \frac{1}{2} a^2$ $= 1 : 1$ The ratio is 1:1.

Answer

The ratio of the areas of the red region and the green region is 1 : 1.