If the sides of a triangle are $6$, $10$ and $14$, then the triangle is:
A.Obtuse triangle
B.Acute triangle
C.Right triangle
D.Equilateral triangle
Answer
611.4k+ views
Hint: We will use the converse of the Pythagorean Theorem to check the type of triangle. If a, b and c are the sides of a triangle then:
If ${c^2} < {a^2} + {b^2}$, then the triangle is an acute triangle.
If ${c^2} = {a^2} + {b^2}$, then the triangle is a right triangle.
If ${c^2} > {a^2} + {b^2}$, then the triangle is an obtuse triangle.
So, we will check the relation between ${c^2}$ and ${a^2} + {b^2}$ for determining the type of the given triangle.
Complete step-by-step answer:
We are given the sides of a triangle as: $6$, $10$ and $14$.
We are required to determine the type of the triangle they form.
We know that the converse of Pythagorean Theorem states that: For a triangle with sides a, b and c,
(i)If ${c^2} < {a^2} + {b^2}$, then the triangle is an acute triangle.
(ii)If ${c^2} = {a^2} + {b^2}$, then the triangle is a right triangle.
(iii)If ${c^2} > {a^2} + {b^2}$, then the triangle is an obtuse triangle.
Let $a = 6$, $b = 10$ and $c = 14$.
We will check the relation between ${c^2}$ and ${a^2} + {b^2}$ for determining the type of the given triangle.
Here, ${c^2} = {\left( {14} \right)^2} = 196$ and ${a^2} + {b^2} = {\left( 6 \right)^2} + {\left( {10} \right)^2} = 36 + 100 = 136$.
We can see clearly that $196 > 136$ i.e., ${c^2} > {a^2} + {b^2}$. Hence, the given triangle is an obtuse triangle.
Therefore, option (A) is correct.
Note: In this question, we may get confused in the process of determining the type of triangle using this method. So, we can also use the method of calculating the largest angle of triangle by comparing the area of the triangle calculated by the Heron’s formula ($\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, c are the sides) with the formula: area of the triangle = $ = \dfrac{1}{2}ab\sin C$ and from here, we can get the value of C and determine the type of the triangle using the value of angle C.
If ${c^2} < {a^2} + {b^2}$, then the triangle is an acute triangle.
If ${c^2} = {a^2} + {b^2}$, then the triangle is a right triangle.
If ${c^2} > {a^2} + {b^2}$, then the triangle is an obtuse triangle.
So, we will check the relation between ${c^2}$ and ${a^2} + {b^2}$ for determining the type of the given triangle.
Complete step-by-step answer:
We are given the sides of a triangle as: $6$, $10$ and $14$.
We are required to determine the type of the triangle they form.
We know that the converse of Pythagorean Theorem states that: For a triangle with sides a, b and c,
(i)If ${c^2} < {a^2} + {b^2}$, then the triangle is an acute triangle.
(ii)If ${c^2} = {a^2} + {b^2}$, then the triangle is a right triangle.
(iii)If ${c^2} > {a^2} + {b^2}$, then the triangle is an obtuse triangle.
Let $a = 6$, $b = 10$ and $c = 14$.
We will check the relation between ${c^2}$ and ${a^2} + {b^2}$ for determining the type of the given triangle.
Here, ${c^2} = {\left( {14} \right)^2} = 196$ and ${a^2} + {b^2} = {\left( 6 \right)^2} + {\left( {10} \right)^2} = 36 + 100 = 136$.
We can see clearly that $196 > 136$ i.e., ${c^2} > {a^2} + {b^2}$. Hence, the given triangle is an obtuse triangle.
Therefore, option (A) is correct.
Note: In this question, we may get confused in the process of determining the type of triangle using this method. So, we can also use the method of calculating the largest angle of triangle by comparing the area of the triangle calculated by the Heron’s formula ($\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, c are the sides) with the formula: area of the triangle = $ = \dfrac{1}{2}ab\sin C$ and from here, we can get the value of C and determine the type of the triangle using the value of angle C.
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