Question

Sides of triangles are given below. Determine which of them are right angled. In case of a right triangle, write the length of its hypotenuse.
(i) 7cm, 24cm, 25cm
(ii) 3cm, 8cm, 6cm
(iii) 50cm, 80cm, 100cm
(iv) 13cm, 12cm, 5cm

Answer 1

Hint: Here we use the concept that in any right angled triangle we can always apply Pythagoras theorem which gives us relation between squares of sides of the triangle. Use the given lengths to check if they satisfy Pythagoras theorem.
Pythagoras theorem states that in a right angled triangle, the sum of square of base and square of height is equal to square of the hypotenuse. In the right triangle the largest side opposite to the right angle is the hypotenuse.
If we have a right angled triangle, \[\vartriangle ABC\]with right angle, \[\angle B = {90^ \circ }\]

Then using the Pythagoras theorem we can write that \[A{C^2} = A{B^2} + B{C^2}\]

Complete step-by-step solution:
We have to check which part gives us a right triangle and then write its hypotenuse. We will add squares of two smaller sides and check if the sum of their squares equals the square of the largest side.
(i) 7cm, 24cm, 25cm
Here the largest side is 25cm.
We take other two sides of the triangle as 7cm and 24cm
Take square of the two sides and add them
\[ \Rightarrow {(7)^2} + {(24)^2} = 49 + 576\]
\[ \Rightarrow {(7)^2} + {(24)^2} = 625\]
Now we can write RHS of the equation as \[625 = {(25)^2}\]
\[ \Rightarrow {(7)^2} + {(24)^2} = {(25)^2}\]
The given sides suit the Pythagoras theorem as the sum of squares of two sides is equal to the square of the largest side.
\[\therefore \]The triangle formed by sides 7cm, 24cm, 25cm is a right triangle having hypotenuse 25cm.
(ii) 3cm, 8cm, 6cm
Here the largest side is 8cm.
We take other two sides of the triangle as 3cm and 6cm
Take square of the two sides and add them
\[ \Rightarrow {(3)^2} + {(6)^2} = 9 + 36\]
\[ \Rightarrow {(3)^2} + {(6)^2} = 45\]
Since RHS of the equation is 45 and we know \[{(8)^2} = 64\]
\[ \Rightarrow {(3)^2} + {(6)^2} \ne {(8)^2}\]
The given sides do not suit the Pythagoras theorem as the sum of squares of two sides is not equal to the square of the largest side.
\[\therefore \]The triangle formed by sides 3cm, 6cm, 8cm is not a right triangle.
(iii) 50cm, 80cm, 100cm
Here the largest side is 100cm.
We take other two sides of the triangle as 50cm and 80cm
Take square of the two sides and add them
\[ \Rightarrow {(50)^2} + {(80)^2} = 2500 + 6400\]
\[ \Rightarrow {(50)^2} + {(80)^2} = 8900\]
Since RHS of the equation is 8900 and we know \[{(100)^2} = 10000\]
\[ \Rightarrow {(50)^2} + {(80)^2} \ne {(100)^2}\]
The given sides do not suit the Pythagoras theorem as the sum of squares of two sides is not equal to the square of the largest side.
\[\therefore \]The triangle formed by sides 50cm, 80cm, 100cm is not a right triangle.
(iv) 13cm, 12cm, 5cm
Here the largest side is 13cm.
We take other two sides of the triangle as 5cm and 12cm
Take square of the two sides and add them
\[ \Rightarrow {(5)^2} + {(12)^2} = 25 + 144\]
\[ \Rightarrow {(5)^2} + {(12)^2} = 169\]
Now we can write RHS of the equation as \[169 = {(13)^2}\]
\[ \Rightarrow {(5)^2} + {(12)^2} = {(13)^2}\]
These suit the Pythagoras theorem as the sum of squares of two sides is equal to the square of the largest side.
\[\therefore \]The triangle formed by sides 5cm, 12cm, 13cm is a right triangle having hypotenuse 13cm.

Note: Students are likely to make mistake while applying Pythagoras theorem as many students write any side on any side of the equation, keep in mind the hypotenuse (largest side) is on one side of the equation and base and perpendicular are on the other side of the equation.