Certainly! Let's solve the problem step by step.
We are given that the hypotenuse of a right triangle is three times as long as the shorter leg. Let's designate:
- [tex]\( h \)[/tex] as the length of the hypotenuse,
- [tex]\( a \)[/tex] as the length of the shorter leg,
- [tex]\( l \)[/tex] as the length of the longer leg.
From the problem, we know:
[tex]\[ h = 3a \][/tex]
In a right triangle, the Pythagorean theorem states:
[tex]\[ a^2 + l^2 = h^2 \][/tex]
Substitute [tex]\( h \)[/tex] with [tex]\( 3a \)[/tex]:
[tex]\[ a^2 + l^2 = (3a)^2 \][/tex]
Simplify the equation:
[tex]\[ a^2 + l^2 = 9a^2 \][/tex]
Rearrange to solve for [tex]\( l^2 \)[/tex]:
[tex]\[ l^2 = 9a^2 - a^2 \][/tex]
[tex]\[ l^2 = 8a^2 \][/tex]
Take the square root of both sides to solve for [tex]\( l \)[/tex]:
[tex]\[ l = \sqrt{8a^2} \][/tex]
[tex]\[ l = \sqrt{8} \cdot \sqrt{a^2} \][/tex]
[tex]\[ l = 2\sqrt{2} \cdot a \][/tex]
Since [tex]\( a = \frac{h}{3} \)[/tex], substitute [tex]\( a \)[/tex] back into the equation:
[tex]\[ l = 2\sqrt{2} \cdot \frac{h}{3} \][/tex]
[tex]\[ l = \frac{2\sqrt{2}h}{3} \][/tex]
Therefore, the length of the longer leg [tex]\( l \)[/tex] in terms of the hypotenuse [tex]\( h \)[/tex] is:
[tex]\[ l = \frac{2\sqrt{2}h}{3} \][/tex]
So, in the desired form:
[tex]\[ I = \frac{2 \cdot \sqrt{2h}}{3} \][/tex]
This establishes the correct values for [tex]\( a, b, \)[/tex] and [tex]\( c \)[/tex]:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = 2 \)[/tex]
- [tex]\( c = 3 \)[/tex]
Therefore:
[tex]\[ I=\frac{2 \sqrt{2h}}{3} \][/tex]
