Question

Find the area of a triangle having perimeter 32 cm, if one of its sides is equal to 11 cm, and the difference of the other two sides is 5 cm.

Answer 1

Hint:
Here, we need to find the area of the triangle. We will assume the other two sides are of length to be \[x\] cm and \[y\] cm respectively. We will use the given information to form an equation in terms of \[x\] and \[y\]. Then, using the formula for the perimeter of a triangle, we will form another equation in terms of \[x\] and \[y\]. We will solve these two equations to find the lengths of the sides of the triangle. Finally, we will use Heron’s formula to find the area of the triangle using the lengths of its sides.
Formula Used: We will use the formula of the perimeter of a triangle, \[\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \], where \[s\] is the semi-perimeter of the triangle, and \[a\], \[b\], and \[c\] are the lengths of the sides of the triangle.

Complete step by step solution:
Let the other two sides be of length \[x\] cm and \[y\] cm respectively.
The difference in the other two sides is 5 cm.
Therefore, we get
\[x - y = 5\]
Rewriting the equation, we get
\[ \Rightarrow x = 5 + y\]
Now, it is given that the perimeter of the triangle is 32 cm.
The perimeter of a triangle is the sum of the lengths of all the three sides.
The lengths of the three sides of the given triangle are 11 cm, \[x\] cm, and \[y\] cm.
Therefore, we get
\[11 + x + y = 32\]
Substituting \[x = 5 + y\] in the equation, we get
\[ \Rightarrow 11 + 5 + y + y = 32\]
Adding the like terms, we get
\[ \Rightarrow 16 + 2y = 32\]
Subtracting 16 from both sides, we get
\[\begin{array}{l} \Rightarrow 16 + 2y - 16 = 32 - 16\\ \Rightarrow 2y = 16\end{array}\]
Dividing both sides by 2, we get the value of \[y\] as
\[ \Rightarrow y = \dfrac{{16}}{2} = 8\]
Therefore, one of the other two sides is of length 8 cm.
Substitute \[y = 8\] in the equation \[x = 5 + y\], we get
\[x = 5 + 8 = 13\]
Therefore, we get the sides of the triangle as 8 cm, 11 cm, and 13 cm.
Now, we will find the area of the triangle.
We will use Heron’s formula to find the area of the triangle using the length of its sides.
Heron’s formula states that the area of a triangle is given by \[\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \], where \[s\] is the semi-perimeter of the triangle, and \[a\], \[b\], and \[c\] are the lengths of the sides of the triangle.
The semi-perimeter of a triangle is half of its perimeter.
Therefore, we get
\[s = \dfrac{{32}}{2} = 16\] cm
Now, substituting \[s = 16\] cm, \[a = 8\] cm, \[b = 11\] cm, and \[c = 13\] cm in Heron’s formula, we get
Area of the triangle \[ = \sqrt {16\left( {16 - 8} \right)\left( {16 - 11} \right)\left( {16 - 13} \right)} {\text{ c}}{{\text{m}}^2}\]
Subtracting the terms in the parentheses, we get
\[ \Rightarrow \]Area of the triangle \[ = \sqrt {16\left( 8 \right)\left( 5 \right)\left( 3 \right)} {\text{ c}}{{\text{m}}^2}\]
Rewriting the expression, we get
\[ \Rightarrow \]Area of the triangle \[ = \sqrt {{4^2} \times 4 \times 2 \times 5 \times 3} {\text{ c}}{{\text{m}}^2} = \sqrt {{4^2} \times {2^2} \times 2 \times 5 \times 3} {\text{ c}}{{\text{m}}^2}\]
Simplifying the expression, we get
\[ \Rightarrow \]Area of the triangle \[ = \left( 4 \right)\left( 2 \right)\sqrt {2 \times 5 \times 3} {\text{ c}}{{\text{m}}^2} = 8\sqrt {30} {\text{ c}}{{\text{m}}^2}\]

Therefore, we get the area of the triangle as \[8\sqrt {30} {\text{ c}}{{\text{m}}^2}\].

Note:
We have used the Heron’s formula instead of the formula \[\dfrac{1}{2} \times {\text{Base}} \times {\text{Altitude}}\] to find the area of the triangle because we found the lengths of the sides of the triangle. We can also use Pythagoras’s theorem to find the altitude of the triangle, and then use the formula \[\dfrac{1}{2} \times {\text{Base}} \times {\text{Altitude}}\] to find the area of the triangle.