# Find The Area Of The Triangle Formed By Connecting Opposite Corners Of The Square And Centre Of The Circle

## Find the area of the triangle

From the figure, ABCD is a square with sides 6 cm in length. AP and BC are tangents of the circle. Connecting two corners of the square and centre of the circle to form a triangle, then find the area of the triangle

## Solution

Extent PO and DC then we get a rectangle APSD

Let’s assume the radius of the circle is r, then ST = 6 – 2r

Consider rectangle APSD

From the figure, Area of blue triangle = Area of rectangle APSD – Area of the red triangle – Area of the yellow triangle – Area of the green triangle

1. Area of rectangle APSD = AP × AD = (6 + r) × 6 = 36 + 6r cm²

2. Area of the red triangle = ½ × DA × DC = ½ × 6 × 6 = 18 cm²

3. Area of the yellow triangle = ½ × SC × SO = ½ × r × (6 – r) = 3r – r²/2 cm²

4. Area of the green triangle = ½ × PO × PA = ½ × r × (6 + r) = 3r + r²/2 cm²

We know, Area of blue triangle = Area of rectangle APSD – Area of the red triangle – Area of the yellow triangle – Area of the green triangle

⇒ Area of blue triangle = 36 + 6r – 18 – (3r – r²/2) – (3r + r²/2)

⇒ Area of blue triangle = 36 + 6r – 18 – 3r + r²/2 – 3r – r²/2

so ** Area of blue triangle = 18 cm²**