
How do you determine if the length 9, 40, 41 form a right triangle?
Answer
543.9k+ views
Hint: In this question, we are given three lengths of the sides of a triangle. We need to check if the triangle is a right angled triangle. For this we will check if the sides satisfy the Pythagoras theorem. The hypotenuse will be the greatest length and rest of them will be perpendicular and base. Pythagoras theorem in a right angled triangle is given by ${{\left( \text{hypotenuse} \right)}^{2}}={{\left( \text{base} \right)}^{2}}+{{\left( \text{perpendicular} \right)}^{2}}$.
Complete step by step answer:
Here we are given the length of the three sides of the triangle as 9, 40, 41. We need to check if they form a right angled triangle. As we know that, every right angled triangle follows the Pythagoras theorem. So, let us try to check if these sides satisfy the Pythagoras theorem. A right angled triangle has three sides, hypotenuse, base and perpendicular. As hypotenuse is always greater than the other two sides. So let us suppose the hypotenuse as 41 because it is larger than 9 and 40.
As both base and perpendicular are on one side so we can suppose any one of them as base and other as the perpendicular.
Here let us suppose the base as 9 and the perpendicular as 40.
So we have hypotenuse = 41, perpendicular = 40 and base = 9.
The Pythagoras theorem is a right angled triangle is given by ${{\left( \text{hypotenuse} \right)}^{2}}={{\left( \text{base} \right)}^{2}}+{{\left( \text{perpendicular} \right)}^{2}}$.
So let us put all the values and check ${{\left( \text{41} \right)}^{2}}={{\left( \text{9} \right)}^{2}}+{{\left( \text{40} \right)}^{2}}$.
Let us calculate ${{\left( \text{41} \right)}^{2}}$ we have,
\[\begin{align}
& 0041 \\
& \times 041 \\
& \overline{\begin{align}
& 0041 \\
& 164\times \\
& \overline{1681} \\
\end{align}} \\
\end{align}\]
We know that ${{\left( \text{40} \right)}^{2}}=40\times 40\text{ and }{{\left( \text{9} \right)}^{2}}=9\times 9$. So putting the values we get $1681=81+1600\Rightarrow 1681=1681$.
Left side is equal to the right side. Hence the three sides 9, 40, 41 satisfy the Pythagoras theorem. Therefore, these three sides form the sides of a right angled triangle.
Note:
Students should note that they can switch the length of base and perpendicular but hypotenuse is always greater than both base and perpendicular in a right angled triangle. Take care of the calculation of squares.
Complete step by step answer:
Here we are given the length of the three sides of the triangle as 9, 40, 41. We need to check if they form a right angled triangle. As we know that, every right angled triangle follows the Pythagoras theorem. So, let us try to check if these sides satisfy the Pythagoras theorem. A right angled triangle has three sides, hypotenuse, base and perpendicular. As hypotenuse is always greater than the other two sides. So let us suppose the hypotenuse as 41 because it is larger than 9 and 40.
As both base and perpendicular are on one side so we can suppose any one of them as base and other as the perpendicular.
Here let us suppose the base as 9 and the perpendicular as 40.
So we have hypotenuse = 41, perpendicular = 40 and base = 9.
The Pythagoras theorem is a right angled triangle is given by ${{\left( \text{hypotenuse} \right)}^{2}}={{\left( \text{base} \right)}^{2}}+{{\left( \text{perpendicular} \right)}^{2}}$.
So let us put all the values and check ${{\left( \text{41} \right)}^{2}}={{\left( \text{9} \right)}^{2}}+{{\left( \text{40} \right)}^{2}}$.
Let us calculate ${{\left( \text{41} \right)}^{2}}$ we have,
\[\begin{align}
& 0041 \\
& \times 041 \\
& \overline{\begin{align}
& 0041 \\
& 164\times \\
& \overline{1681} \\
\end{align}} \\
\end{align}\]
We know that ${{\left( \text{40} \right)}^{2}}=40\times 40\text{ and }{{\left( \text{9} \right)}^{2}}=9\times 9$. So putting the values we get $1681=81+1600\Rightarrow 1681=1681$.
Left side is equal to the right side. Hence the three sides 9, 40, 41 satisfy the Pythagoras theorem. Therefore, these three sides form the sides of a right angled triangle.
Note:
Students should note that they can switch the length of base and perpendicular but hypotenuse is always greater than both base and perpendicular in a right angled triangle. Take care of the calculation of squares.
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