Notes
circle inside a triangle solution

Solution to the Circle Inside a Triangle Puzzle

Circle Inside a Triangle

What’s the area of the circle?

Solution by Area of a Triangle

Circle inside a triangle labelled

Decomposing the triangle as above shows that its area can be written as:

12×15×h+12×14×h+12×15×h=15+14+132h=21h \frac{1}{2} \times 15 \times h + \frac{1}{2} \times 14 \times h + \frac{1}{2} \times 15 \times h = \frac{15 + 14 + 13}{2} h = 21 h

The area can be established either using Heron's formula or Pythagoras' theorem.

Using Heron's formula, the area is written in terms of the semi-perimeter which is 15+14+132=21\frac{15 + 14 + 13}{2} = 21.

A=21(2115)(2114)(2113)=21×6×7×8=84 A = \sqrt{ 21 ( 21 - 15) (21 - 14) ( 21 - 13)} = \sqrt{ 21 \times 6 \times 7 \times 8} = 84

To use Pythagoras' theorem, mark a point DD directly below CC on the edge ABA B so that ADCA D C forms a right-angled triangle.

Circle inside a triangle Pythagoras

Applying Pythagoras' theorem to the two right-angled triangles ADCA D C and CDBC D B yields:

13 2 =y 2+x 2 14 2 =y 2+(15x 2)=y 2+15 230x+x 2=15 2+13 230x \begin{aligned} 13^2 &= y^2 + x^2 \\ 14^2 &= y^2 + (15 - x^2) = y^2 + 15^2 - 30 x + x^2 = 15^2 + 13^2 - 30 x \end{aligned}

So x=15 2+13 214 230=335x = \frac{15^2 + 13^2 - 14^2}{30} = \frac{33}{5} and y=565y = \frac{56}{5}. The area of the triangle is therefore

A=12×15×565=84 A = \frac{1}{2} \times 15 \times \frac{56}{5} = 84