# Is a triangle with sides of $3,\,4,\,6$ a right triangle?

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Hint: We know that a triangle is a right triangle if and only if the sum of squares of the first two sides is equal to the square of the third side. In short, we have to check if Pythagoras theorem holds here or not.

In the above question, we have to check if the sides are of a right triangle. To check this we have to check if the sum of the squares of the two smaller sides equals the length of the square of the longest side we have to check Pythagoras theorem.
Therefore, we have to check if ${3^2} + {4^2}$ is equal to ${6^2}$ or not.
Let $A = {3^2} + {4^2}$
$\Rightarrow A = 9 + 16$
$\Rightarrow A = 25$
Now, let $B = {6^2}$
$\Rightarrow B = 36$
Therefore,
${3^2} + {4^2} \ne {6^2}$
Therefore, the triangle with sides of $3,\,4,\,6$ is not a right triangle

Note: The idea is to use the Pythagoras Theorem to check if a right-angled triangle is possible or not. Calculate the length of the three sides of the triangle by joining the given coordinates. Let the sides be A, B, and C. The given triangle is right-angled if and only if ${A^2} = {B^2} + {C^2}$ where A is the longest side and the other two are the shorter ones.