To find the length of the longer leg [tex]\( I \)[/tex] in terms of the hypotenuse [tex]\( h \)[/tex], given that the hypotenuse is three times the shorter leg, we can use Pythagoras' theorem. Let's go step-by-step:
1. Let the shorter leg be denoted by [tex]\( a \)[/tex]. According to the problem, the hypotenuse [tex]\( h \)[/tex] is three times the shorter leg, so we have:
[tex]\[
h = 3a
\][/tex]
2. The relationship between the sides of a right triangle can be described using Pythagoras' theorem:
[tex]\[
h^2 = a^2 + I^2
\][/tex]
Substituting [tex]\( h = 3a \)[/tex] into this equation, we get:
[tex]\[
(3a)^2 = a^2 + I^2
\][/tex]
[tex]\[
9a^2 = a^2 + I^2
\][/tex]
3. Simplifying the equation:
[tex]\[
9a^2 - a^2 = I^2
\][/tex]
[tex]\[
8a^2 = I^2
\][/tex]
4. Taking the square root of both sides to solve for [tex]\( I \)[/tex]:
[tex]\[
I = \sqrt{8a^2}
\][/tex]
[tex]\[
I = \sqrt{8} \cdot a
\][/tex]
[tex]\[
I = 2\sqrt{2} \cdot a
\][/tex]
5. We already know that [tex]\( h = 3a \)[/tex], so we can express [tex]\( a \)[/tex] in terms of [tex]\( h \)[/tex]:
[tex]\[
a = \frac{h}{3}
\][/tex]
6. Substituting [tex]\( a = \frac{h}{3} \)[/tex] back into the equation for [tex]\( I \)[/tex]:
[tex]\[
I = 2\sqrt{2} \cdot \frac{h}{3}
\][/tex]
[tex]\[
I = \frac{2\sqrt{2}}{3} \cdot h
\][/tex]
Thus, the length of the longer leg [tex]\( I \)[/tex] in terms of [tex]\( h \)[/tex] is:
[tex]\[
I = \frac{2 \sqrt{2} h}{3}
\][/tex]
Therefore, the correct values for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] to plug into the given equation [tex]\( I = \frac{a \sqrt{b} h}{c} \)[/tex] are:
[tex]\[
a = 2, \quad b = 2, \quad c = 3
\][/tex]
So,
[tex]\[
I = \frac{2 \sqrt{2} h}{3}
\][/tex]
