Sure, let's solve this problem step-by-step.
### Step 1: Understanding the Properties of a [tex]$45^{\circ}-45^{\circ}-90^{\circ}$[/tex] Triangle
A [tex]$45^{\circ}-45^{\circ}-90^{\circ}$[/tex] triangle is a special type of right triangle. In such a triangle:
- The two legs are congruent (i.e., they have the same length).
- The hypotenuse (the side opposite the right angle) is [tex]$\sqrt{2}$[/tex] times longer than each leg.
### Step 2: Given Information
The hypotenuse of the triangle measures [tex]$10 \sqrt{5}$[/tex] inches.
### Step 3: Setting Up the Relationship Between the Legs and the Hypotenuse
Let's denote the length of each leg as [tex]\( a \)[/tex]. According to the properties of a [tex]$45^{\circ}-45^{\circ}-90^{\circ}$[/tex] triangle, the relationship between the hypotenuse and a leg is given by:
[tex]\[ \text{Hypotenuse} = a \sqrt{2} \][/tex]
### Step 4: Substituting the Known Value
We know the hypotenuse measures [tex]$10 \sqrt{5}$[/tex] inches:
[tex]\[ 10 \sqrt{5} = a \sqrt{2} \][/tex]
### Step 5: Solving for [tex]\( a \)[/tex]
To solve for [tex]\( a \)[/tex], we need to isolate [tex]\( a \)[/tex] from the equation:
[tex]\[ a = \frac{10 \sqrt{5}}{\sqrt{2}} \][/tex]
### Step 6: Simplifying the Expression
Divide the numerator and the denominator inside the fraction:
[tex]\[ a = \frac{10 \sqrt{5}}{\sqrt{2}} = 10 \left(\frac{\sqrt{5}}{\sqrt{2}}\right) \][/tex]
[tex]\[ \frac{\sqrt{5}}{\sqrt{2}} = \sqrt{\frac{5}{2}} \][/tex]
So,
[tex]\[ a = 10 \sqrt{\frac{5}{2}} \][/tex]
### Step 7: Numerical Value of the Leg Length
Now, compute the numerical value of [tex]\( a \)[/tex]:
[tex]\[ \sqrt{\frac{5}{2}} \approx 1.58113883008 \][/tex]
Therefore,
[tex]\[ a = 10 \times 1.58113883008 = 15.811388300841896 \][/tex]
### Step 8: Results
We have the measurements for both the hypotenuse and the legs of the triangle:
- The hypotenuse is [tex]\( 22.360679774997898 \)[/tex] inches.
- Each leg is [tex]\( 15.811388300841896 \)[/tex] inches.
These results give the final dimensions of the [tex]$45^{\circ}-45^{\circ}-90^{\circ}$[/tex] triangle.
