Question

How do you use Heron's formula to find the area of a triangle with sides of lengths 5,6 and 7?

Answer 1

Hint: To find the area of a triangle, we will be using the Heron’s formula which is given as $\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$ , where a, b and c are the sides of the triangle and s is the semi-perimeter and is given as $s=\dfrac{a+b+c}{2}$ .We will consider $a=5,b=6,c=7$ and substitute these values to get s. Then we will substitute the value onto the Heron’s formula to get the area of the given triangle.

Complete step-by-step solution:
We need to find the area of a triangle with sides of lengths 5,6 and 7 using Heron's formula. Let us see Heron's formula. Heron's formula is given as:
$\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}...(i)$ , where a, b and c are the sides of the triangle and s is the semi-perimeter and is given as
$s=\dfrac{a+b+c}{2}$
Let us consider $a=5,b=6,c=7$
First let us find the semi-perimeter.
$s=\dfrac{5+6+7}{2}=\dfrac{18}{2}=9$ units.
Now, we have to substitute the value of s in (i). We will get
$\sqrt{9\left( 9-5 \right)\left( 9-6 \right)\left( 9-7 \right)}$
Let us solve this by simplifying the values inside the brackets.
$\Rightarrow \sqrt{9\times 4\times 3\times 2}=\sqrt{216}$
Let us simplify $\sqrt{216}$ .
$\begin{align}
  & 2\text{ }\left| \!{\underline {\,
  216 \,}} \right. \\
 & 2\text{ }\left| \!{\underline {\,
  106 \,}} \right. \\
 & 53\left| \!{\underline {\,
  53 \,}} \right. \\
 & \text{ 1} \\
\end{align}$
We can write it as $2\sqrt{53}$ .

Hence, the area of the given triangle is $2\sqrt{53}$ square units.

Note: Students may make mistake when writing Heron’s formula by writing as \[\sqrt{s\left( s+a \right)\left( s+b \right)\left( s+c \right)}\] . They may also make mistakes when finding the semi-perimeter by writing the formula as $s=a+b+c$ . Students must try to simplify the answer as much as possible. Never forget to write the units even though it is not specified in the question.