The length of diagonal of a square is 9√2 cm. the square is reshaped to form a triangle. what is the area (in cm2) of largest incircle that can be formed in that triangle?
Answers
The area (in cm²) of largest incircle that can be formed in that triangle is 12П cm².
Given, diagonal of the square = 9√2 cm.
Let us consider the side of square to be 'x' cm.
As, the sides of square subtend an angle of 90° , so from Pythagoras theorem,
x² + x² = (9√2)²
⇒ 2x² = 81×2
⇒ x = ±√81 = ±9 cm
As side cannot be in negative, so side of square = 9 cm.
∴ Perimeter of square = 4 × Side = 4 × 9 = 36cm
Since the square is reshaped to a triangle, the perimeter of equilateral triangle = Perimeter of square = 36 cm.
As all the sides of equilateral triangle are equal, so
Each side = 36/3 = 12 cm [Triangle has 3 sides]
We know, for an equilateral triangle of side 'a', its incircle would have a radius of:
r = a/(2√3)
Here, a = 12 cm,
So, inradius r = 12/(2√3) = 2√3 cm.
So, area of the circle is Пr² = П (2√3)² = 12П cm²
Area is 12П cm².