
The area of a triangle whose sides are 4 cm, 13 cm and 15 cm, is
A.\[\sqrt {420} \] cm\[^2\]
B.24 cm\[^2\]
C.48 cm\[^2\]
D.56 cm\[^2\]
Answer
598.2k+ views
Hint: First, we will find the semi perimeter of the triangle, that is, \[s = \dfrac{{a + b + c}}{2}\]. Then use the formula of area of the triangle is \[Area = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \], where \[s\] is the semi perimeter of the triangle and \[a\], \[b\] and \[c\] are the sides of the triangle. Apply this formula of area of the triangle, and then use the given sides to find the required value.
Complete step-by-step answer:
We are given that the three sides of the triangle are 4 cm, 13 cm and 15 cm.
We know that the formula to calculate the semi perimeter of the circle is \[s = \dfrac{{a + b + c}}{2}\], where \[a\], \[b\], and \[c\] are the sides of the triangle.
We will find the values of \[a\], \[b\] and \[c\] are the sides of the triangle.
\[a = 4\]
\[b = 13\]
\[c = 15\]
We will now substitute these values in the above formula to find the semi perimeter of the given triangle.
\[
\Rightarrow s = \dfrac{{4 + 13 + 15}}{2} \\
\Rightarrow s = \dfrac{{32}}{2} \\
\Rightarrow s = 16{\text{ cm}} \\
\]
We know that the area of the triangle is calculated \[Area = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \], where \[s\] is the semi perimeter of the triangle and \[a\], \[b\] and \[c\] are the three sides of the triangle.
Substituting these values of the semi perimeter of the triangle and the sides of the triangle \[a\], \[b\] and \[c\] in the above formula of area of the triangle, we get
\[
Area = \sqrt {16\left( {16 - 4} \right)\left( {16 - 13} \right)\left( {16 - 15} \right)} \\
= \sqrt {16 \times 12 \times 3 \times 1} \\
= \sqrt {576} \\
= 24{\text{ c}}{{\text{m}}^2} \\
\]
Thus, the area of the triangle is 24 cm\[^2\].
Note: In solving this question, Heron’s formula should be used to compute the area of a triangle where sides of the triangle are given. While writing the answer to such types of questions, units need to be mentioned in the final value. We can also find the area of the triangle of using the formula, \[{\text{Area}} = \dfrac{1}{2} \times {\text{base}} \times {\text{height}}\], but this way could take a little longer as we have to find the height of the triangle first.
Complete step-by-step answer:
We are given that the three sides of the triangle are 4 cm, 13 cm and 15 cm.
We know that the formula to calculate the semi perimeter of the circle is \[s = \dfrac{{a + b + c}}{2}\], where \[a\], \[b\], and \[c\] are the sides of the triangle.
We will find the values of \[a\], \[b\] and \[c\] are the sides of the triangle.
\[a = 4\]
\[b = 13\]
\[c = 15\]
We will now substitute these values in the above formula to find the semi perimeter of the given triangle.
\[
\Rightarrow s = \dfrac{{4 + 13 + 15}}{2} \\
\Rightarrow s = \dfrac{{32}}{2} \\
\Rightarrow s = 16{\text{ cm}} \\
\]
We know that the area of the triangle is calculated \[Area = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \], where \[s\] is the semi perimeter of the triangle and \[a\], \[b\] and \[c\] are the three sides of the triangle.
Substituting these values of the semi perimeter of the triangle and the sides of the triangle \[a\], \[b\] and \[c\] in the above formula of area of the triangle, we get
\[
Area = \sqrt {16\left( {16 - 4} \right)\left( {16 - 13} \right)\left( {16 - 15} \right)} \\
= \sqrt {16 \times 12 \times 3 \times 1} \\
= \sqrt {576} \\
= 24{\text{ c}}{{\text{m}}^2} \\
\]
Thus, the area of the triangle is 24 cm\[^2\].
Note: In solving this question, Heron’s formula should be used to compute the area of a triangle where sides of the triangle are given. While writing the answer to such types of questions, units need to be mentioned in the final value. We can also find the area of the triangle of using the formula, \[{\text{Area}} = \dfrac{1}{2} \times {\text{base}} \times {\text{height}}\], but this way could take a little longer as we have to find the height of the triangle first.
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