Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD. What is the ratio of the areas Class 9

Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD. What is the ratio of the areas Class 9

Question 1.

Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD. What is the ratio of the areas of the red region (∆PAB and ∆PCD) and the green region (∆PBC and ∆PDA)? Class 9

Solution:

Let side of square ABCD be a units

Let h₁ and h₂ be the lengths of perpendiculars drawn from P to sides AB and CD respectively.

Using, Area of $\frac{1}{2}$ × base × height

∴ A(∆PAB) = $\frac{1}{2}$ × a × h₁ and

A(∆PCD) = $\frac{1}{2}$ × a × h₂

Adding above areas, we get

A(∆PAB) + A(∆PCD)

$\begin{aligned} & =\frac{1}{2} a h_1+\frac{1}{2} a h_2 \\ & =\frac{1}{2} \times a \times\left(h_1+h_2\right) \\ & =\frac{1}{2} \times a \times a \\ & \quad\left[\because h_1+h_2=a\right]\end{aligned}$

= $\frac{1}{2}$ × a ²

Area of red region = $\frac{1}{2}$ × Area of Square .. .(i)

Similarly, we can show that

Area of Green region (∆PBC and ∆PDA)

= $\frac{1}{2}$ × Area of Square ... (ii)

From (i) and (ii)

Area of red region = Area of green region

$\therefore \quad \frac{\text { Area of red region }}{\text { Area of Green region }}=\frac{1}{1}=1: 1$