Question

In an isosceles right- angled triangle one of the congruent sides is of length 10cm. Find the length of the hypotenuse
(A) $10\sqrt 2 cm$
(B) $5cm$
(C) $\sqrt 2 cm$
(D) $3\sqrt 2 $

Answer 1

Hint: An isosceles right angled triangle is almost similar to a right angled triangle but there is a difference that separates it from right angles triangle. In an isosceles triangle there are three sides and two sides are similar in length and corresponding angles are also similar (congruent).
Also, we know that the sum of all angles of any triangle is ${180^ \circ }$. Now using properties of the isosceles triangle, we know two angles are equal, we can find the angle of the given triangle and then using suitable trigonometric ratios we can find the hypotenuse.

Complete step-by-step answer:
Now we are given an isosceles right- angled triangle with one of the congruent sides is of length 10cm.
Let ABC be an isosceles triangle right angled at B.
So, we can make the diagram of a triangle using the given information.

Now we know that a triangle is isosceles so its two sides will have equal length.
Therefore,
$ \Rightarrow $$AB = BC = 10cm$ ------(1)
Since, triangle ABC is right-angled at B so
$ \Rightarrow $$\angle B = {90^ \circ }$ --------(2)
 Now we know that as triangle is isosceles so its two sides will have equal length and the corresponding angles are also equal, so,
$ \Rightarrow $$\angle A = \angle C$ --------(3)
Now we know that the sum of all angles of any triangle is equal to ${180^ \circ }$.
So, also for a given triangle ABC we have the sum of all of its angles equal to ${180^ \circ }$.
$ \Rightarrow $$\angle A + \angle B + \angle C = {180^ \circ }$
Now substituting value of $\angle B$ from (2), we get,
$
 \Rightarrow \angle A + {90^ \circ } + \angle C = {180^ \circ } \\
  \Rightarrow \angle A + \angle C = {180^ \circ } - {90^ \circ } = {90^ \circ } \\
 $
Now using (3) we get
\[
 \Rightarrow \angle A + \angle A = {90^ \circ } \\
  \Rightarrow 2\angle A = {90^ \circ } \\
 \]
Now further solving this, we get,
\[\angle A = {45^ \circ }\] -------(4)
Now again from (3), we can write that
\[\angle C = \angle A = {45^ \circ }\]
Now, \[\angle C = {45^ \circ }\] -------(5)
Therefore, all the three angles are $90^\circ ,45^\circ and45^\circ .$
Given, that the length of one congruent side is $10cm$
$ \Rightarrow $$AB = BC = 10cm$
Now using trigonometric ratio $\sin \theta $ which involves hypotenuse, we know that
$ \Rightarrow $$\sin \theta = \dfrac{{{\text{prependicular}}}}{{hypo.}}$
Now using the above formula of $\sin \theta $, where $\theta = \angle C$, we get
$ \Rightarrow $ $\sin \angle C = \dfrac{{{\text{prependicular}}}}{{hypo.}} = \dfrac{{AB}}{{AC}}$
Now substituting values of AB from (1) and $\angle C$ from (5), we get
$ \Rightarrow $$\sin {45^ \circ } = \dfrac{{10}}{{AC}}$
Now we know that $\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$, we get
$ \Rightarrow $$\dfrac{1}{{\sqrt 2 }} = \dfrac{{10}}{{AC}}$
Now we get
$ \Rightarrow $$AC = 10\sqrt 2 cm$
Therefore, the length of hypotenuse is $10\sqrt 2 cm$.

Hence option (A) is the correct answer.

Note: Alternative method to solve this question is-
We will solve the question similarly up to (2), equation.
And then we know the length of two sides of the isosceles triangle right angled at B.
Now we know Pythagoras theorem, which says that
${\left( h \right)^2} = {\left( p \right)^2} + {\left( b \right)^2}$, where $h,p,b$ are hypotenuse, perpendicular and base of the right-angled triangle.
Now using the Pythagoras theorem for triangle ABC, where $h = AC,p = AB,b = BC$, we get
$
 \Rightarrow {\left( {AC} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2} \\
 \Rightarrow AC = \sqrt {{{\left( {AB} \right)}^2} + {{\left( {BC} \right)}^2}} \\
 $
Now substituting values of AB and BC from (1), we get
$
  \Rightarrow AC = \sqrt {{{\left( {10} \right)}^2} + {{\left( {10} \right)}^2}} \\
 \Rightarrow AC = \sqrt {100 + 100} \\
 \Rightarrow AC = \sqrt {200} \\
 \Rightarrow AC = 10\sqrt 2 cm \\
 $
Hence, we can solve this question using this method also.