Question

How do you find the remaining sides of a \[{{45}^{\circ }}-{{45}^{\circ }}-{{90}^{\circ }}\] triangle if the shortest sides are each \[\dfrac{1}{2}\]?

Answer 1

Hint: In order to find the solution to the given question that is to find the remaining sides of a \[{{45}^{\circ }}-{{45}^{\circ }}-{{90}^{\circ }}\] triangle if the shortest sides are each \[\dfrac{1}{2}\], we should apply the Pythagoras theorem for right-angled triangle which states that \[\text{Hypotenus}{{\text{e}}^{2}}~=\text{ Perpendicula}{{\text{r}}^{2}}~+\text{ Bas}{{\text{e}}^{2}}~\]where Perpendicular, Base and Hypotenuse are named as the sides of the triangle. In the question, it’s mentioned that the perpendicular side and base are equal to \[\dfrac{1}{2}\]. Here, the hypotenuse is the longest side, as it is opposite to the angle 90°, so we can find the hypotenuse by using Pythagoras theorem.

Complete step by step solution:
We have been given a \[{{45}^{\circ }}-{{45}^{\circ }}-{{90}^{\circ }}\] right triangle and the length of its shortest sides are each \[\dfrac{1}{2}\] according to the question, the angles of the triangle are \[{{45}^{\circ }},{{45}^{\circ }}\text{ }\!\!\And\!\!\text{ }{{90}^{\circ }}\] . Since two of the angles are the same and the third one is a right-angle. Therefore, it’s a right-angled triangle. The length of two sides perpendicular and Base is equal to \[\dfrac{1}{2}\] units.
Let us assume Hypotenuse to be equal to \[x\] units. Therefore, the triangle will look as follows:

Apply the Pythagoras theorem which states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides that is \[\text{Hypotenus}{{\text{e}}^{2}}~=\text{ Perpendicula}{{\text{r}}^{2}}~+\text{ Bas}{{\text{e}}^{2}}~\] “.
Now substitute the values from the figure in this formula we will get:
\[\Rightarrow {{x}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{1}{2} \right)}^{2}}\]
Simplifying it further we get:
\[\Rightarrow {{x}^{2}}=\left( \dfrac{1}{4} \right)+\left( \dfrac{1}{4} \right)\]
Taking LCM on the right-hand side, we get:
\[\Rightarrow {{x}^{2}}=\dfrac{1+1}{4}\]
\[\Rightarrow {{x}^{2}}=\dfrac{2}{4}\]
Take square root both the sides we get:
\[\Rightarrow x=\sqrt{\dfrac{2}{4}}\]
\[\Rightarrow x=\pm \dfrac{\sqrt{2}}{2}\]
We know that length of a side can never be negative which implies:
\[\Rightarrow x=\dfrac{\sqrt{2}}{2}\]
Therefore, the length of the remaining side of the \[{{45}^{\circ }}-{{45}^{\circ }}-{{90}^{\circ }}\] triangle is \[\dfrac{\sqrt{2}}{2}\] units.

Note: There’s alternative method to find the length of the hypotenuse:
We know that \[\sin \theta =\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}\], Therefore by substitute the values from the figure in this formula we will get:
\[\Rightarrow \sin {{45}^{\circ }}=\dfrac{\dfrac{1}{2}}{\text{x}}\]
Simplifying it further we get:
\[\Rightarrow \sin {{45}^{\circ }}=\dfrac{1}{\text{2x}}\]
We know that \[\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}\], using it in the above equation, we get:
\[\Rightarrow \dfrac{1}{\sqrt{2}}=\dfrac{1}{\text{2x}}\]
\[\Rightarrow x=\dfrac{\sqrt{2}}{2}\]
Therefore, the length of the remaining side (Hypotenuse) of the \[{{45}^{\circ }}-{{45}^{\circ }}-{{90}^{\circ }}\] triangle is \[\dfrac{\sqrt{2}}{2}\] units.