# Solution to the Circle Inside a Triangle Puzzle +-- {.image} [[CircleInsideaTriangle.png:pic]] > What’s the area of the circle? =-- ## Solution by [[Area of a Triangle]] +-- {.image} [[CircleInsideaTriangleLabelled.png:pic]] =-- Decomposing the triangle as above shows that its area can be written as: $$ \frac{1}{2} \times 15 \times h + \frac{1}{2} \times 14 \times h + \frac{1}{2} \times 15 \times h = \frac{15 + 14 + 13}{2} h = 21 h $$ 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 $\frac{15 + 14 + 13}{2} = 21$. $$ A = \sqrt{ 21 ( 21 - 15) (21 - 14) ( 21 - 13)} = \sqrt{ 21 \times 6 \times 7 \times 8} = 84 $$ To use [[Pythagoras' theorem]], mark a point $D$ directly below $C$ on the edge $A B$ so that $A D C$ forms a [[right-angled triangle]]. +-- {.image} [[CircleInsideaTrianglePythagoras.png:pic]] =-- Applying [[Pythagoras' theorem]] to the two [[right-angled triangles]] $A D C$ and $C D B$ yields: $$ \begin{aligned} 13^2 &= y^2 + x^2 \\ 14^2 &= y^2 + (15 - x^2) = y^2 + 15^2 - 30 x + x^2 = 15^2 + 13^2 - 30 x \end{aligned} $$ So $x = \frac{15^2 + 13^2 - 14^2}{30} = \frac{33}{5}$ and $y = \frac{56}{5}$. The area of the triangle is therefore $$ A = \frac{1}{2} \times 15 \times \frac{56}{5} = 84 $$