Question

The difference between the semi perimeter and the sides of a triangle ABC are 8cm, 7cm, and 5cm respectively. Find the area of the triangle?

Answer 1

Hint: In this question we first assume its semi perimeter. Then all three sides will be computed as the difference between these and the semi perimeter. Further apply formula for semi perimeter to compute the semi perimeter value. Further apply Heron's formula to compute the area of the triangle.

Complete step-by-step answer:
Let us suppose the semi perimeter of the given triangle ABC is “s”. Its three sides are of the lengths “a”, “b” and “c” , as shown in the diagram.

AS given that difference between semi perimeter and sides are 8 cm , 7 cm and 5 cm.
Thus, in the triangle ABC , we have
$s-a = 8$
$s-b = 7$
$s- c = 5$
Thus three sides will be as follows,
$a = s-8$
$b = s-7$
and $c = s-5$
Since formula for semi perimeter is,
$s = \dfrac{{a + b + c}}{2}$
Now, substitute the values in above formula, we get
$s = \dfrac{{(s - 8) + (s - 7) + (s - 5)}}{2}$
Now do simplification, to get
$  s = \dfrac{{(s - 8) + (s - 7) + (s - 5)}}{2} $
$   \Rightarrow 2s = 3s - 20 $
$   \Rightarrow s = 20 $
Thus, the semi perimeter is 20 cm.
Now, with the help of this semi perimeter of the triangle and the given differences, we can find out the area of this triangle. This will be done by Heron’s formula:
Heroin formula is,
${\text{Area of triangle = }}\sqrt {{\text{s}}\left( {{\text{s - a}}} \right)\left( {{\text{s - b}}} \right)\left( {{\text{s - c}}} \right)} $
Substitute the values in above formula, we get
${\text{Area of triangle}} = \sqrt {20 \times 8 \times 7 \times 5} $
$   \Rightarrow {\text{Area of triangle}} = \sqrt {2 \times 2 \times 5 \times 2 \times 2 \times 2 \times 7 \times 5} $
$   \Rightarrow {\text{Area of triangle}} = 20\sqrt {14} c{m^2} $
$\therefore $ The value of the semi perimeter will be 20 cm and area of the triangle given is $20\sqrt {14} c{m^2}$.

Note: This type of question is solved in a very limited method. The above method is a very good method to solve such questions in which there will be less chances of doubt and errors. Here we have used the area formula for triangles based on Heron’s method.