Scalene area
Can you find the area of this scalene triangle?
Age
11 to 14
Challenge level
Problem
The diagram shows a scalene triangle with side lengths 13 cm, 21 cm and 20 cm.
Image
From a point O inside the triangle, line segments can be drawn at right angles to each of the sides.
The lengths of these line segments are 9 cm, 3 cm and 3.6 cm, as shown in the diagram.
What is the area of the triangle?
This problem is adapted from the World Mathematics Championships
Student Solutions
Drawing lines from O to each vertex
Drawing lines from O to each vertex splits the triangle into three smaller triangles, as shown below.
For each smaller triangle, the 'base' and 'height' are shown (the 'bases' are the sides of the original triangle, which are not horizontal). So, using Area $=\dfrac{1}{2}$ base$\times$height, the areas of the smaller triangles can be found:
Red triangle area $\dfrac{1}{2}\times13\times9$
Yellow triangle area $\dfrac{1}{2}\times20\times3.6$
Green triangle area $\dfrac{1}{2}\times21\times3$
So the total area of the triangle is
$\dfrac{13\times9+20\times3.6+21\times3}{2}=\dfrac{117+72+63}{2}=126$cm$^2$
Using trigonometry
The cosine rule can be used to find one of the angles in the triangle, and then the formula $\frac{1}{2}ab\sin{C}$ can be used to find the area.
For example, the angle between the 13 cm side and the 20 cm side can be found using the cosine rule:$$\begin{align}&21^2=13^2+20^2-2\times 13\times20\times\cos{\theta}\\\Rightarrow&441=569-520\cos{\theta}\\\Rightarrow&\cos{\theta}=\frac{569-441}{520}=\frac{128}{5200}=\frac{16}{65}\\\Rightarrow&\theta=75.75^{\text{o}}\end{align}$$
So the area of the triangle is $\frac{1}{2}\times13\times20\times\sin{75.75^{\text{o}}}=126$ cm$^2$.
Using Heron's Forumla
Heron's formula for the area of a triangle with sides $a$, $b$ and $c$ is $\sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is half of the perimeter of the triangle.
For this triangle, $a$, $b$ and $c$ are $13$, $20$ and $21$, and the perimeter is $54$, so $s=27$.
So the area is $$\begin{split}\sqrt{27\times14\times7\times6}&=\sqrt{3^3\times2\times7\times2\times3}\\&=\sqrt{2\times2\times3^2\times3^2\times7\times7}\\&=\sqrt{(2\times3^2\times7)^2}\\&=2\times3^2\times7\\&=126\end{split}$$
So the area is $126$ cm$^2$.
Drawing lines from O to each vertex splits the triangle into three smaller triangles, as shown below.
Image
For each smaller triangle, the 'base' and 'height' are shown (the 'bases' are the sides of the original triangle, which are not horizontal). So, using Area $=\dfrac{1}{2}$ base$\times$height, the areas of the smaller triangles can be found:
Red triangle area $\dfrac{1}{2}\times13\times9$
Yellow triangle area $\dfrac{1}{2}\times20\times3.6$
Green triangle area $\dfrac{1}{2}\times21\times3$
So the total area of the triangle is
$\dfrac{13\times9+20\times3.6+21\times3}{2}=\dfrac{117+72+63}{2}=126$cm$^2$
Using trigonometry
The cosine rule can be used to find one of the angles in the triangle, and then the formula $\frac{1}{2}ab\sin{C}$ can be used to find the area.
For example, the angle between the 13 cm side and the 20 cm side can be found using the cosine rule:$$\begin{align}&21^2=13^2+20^2-2\times 13\times20\times\cos{\theta}\\\Rightarrow&441=569-520\cos{\theta}\\\Rightarrow&\cos{\theta}=\frac{569-441}{520}=\frac{128}{5200}=\frac{16}{65}\\\Rightarrow&\theta=75.75^{\text{o}}\end{align}$$
So the area of the triangle is $\frac{1}{2}\times13\times20\times\sin{75.75^{\text{o}}}=126$ cm$^2$.
Using Heron's Forumla
Heron's formula for the area of a triangle with sides $a$, $b$ and $c$ is $\sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is half of the perimeter of the triangle.
For this triangle, $a$, $b$ and $c$ are $13$, $20$ and $21$, and the perimeter is $54$, so $s=27$.
So the area is $$\begin{split}\sqrt{27\times14\times7\times6}&=\sqrt{3^3\times2\times7\times2\times3}\\&=\sqrt{2\times2\times3^2\times3^2\times7\times7}\\&=\sqrt{(2\times3^2\times7)^2}\\&=2\times3^2\times7\\&=126\end{split}$$
So the area is $126$ cm$^2$.