In this problem, we determine the sides and hypotenuse of a right triangle when its perimeter and the altitude to the hypotenuse are known.
Problem
The perimeter of a right triangle is 240 cm. The altitude drawn perpendicular to the hypotenuse has length 48 cm. Find the two legs and the hypotenuse of the triangle.
Step-by-Step Solution
Step 1: Setting up Variables & Basic Equations
Let the legs be \(a\) and \(b\), the hypotenuse be \(H\), and the altitude be \(h = 48\).
Perimeter equation:
\[ a + b + H = 240 \]Pythagorean theorem:
\[ a^2 + b^2 = H^2 \]Step 2: Area Relations
Area using the legs:
\[ A = \frac{1}{2}ab \]Area using altitude and hypotenuse:
\[ A = \frac{1}{2}hH \]Since both expressions equal the same area:
\[ ab = hH \]Substitute \(h = 48\):
\[ ab = 48H \]Step 3: Squaring the Perimeter Equation
From the perimeter equation:
\[ a + b = 240 - H \]Square both sides:
\[ (a + b)^2 = (240 - H)^2 \] \[ a^2 + b^2 + 2ab = 240^2 + H^2 - 480H \]Step 4: Solving for the Hypotenuse \(H\)
Substitute \(a^2 + b^2 = H^2\):
\[ H^2 + 2ab = 240^2 + H^2 - 480H \] \[ 2ab = 240^2 - 480H \]Substitute \(ab = 48H\):
\[ 2(48H) = 57600 - 480H \] \[ 96H = 57600 - 480H \]Solve for \(H\):
\[ 576H = 57600 \] \[ H = 100 \text{ cm} \]Step 5: Finding the Legs \(a\) and \(b\)
Substitute \(H = 100\) into the sum and product equations:
\[ a + b = 140 \] \[ ab = 4800 \]Express \(b\) as \(b = \frac{4800}{a}\) and substitute into the sum equation:
\[ a + \frac{4800}{a} = 140 \]Multiply by \(a\) to form a quadratic equation:
\[ a^2 - 140a + 4800 = 0 \]Solve the quadratic equation:
\[ a = 80 \text{ cm} \quad \text{or} \quad a = 60 \text{ cm} \]Using \(b = \frac{4800}{a}\), we get:
\[ b = 60 \text{ cm} \quad \text{or} \quad b = 80 \text{ cm} \]Final Answer
Assuming \(a < b\), the triangle dimensions are:
\[ a = 60 \text{ cm}, \quad b = 80 \text{ cm}, \quad H = 100 \text{ cm} \]This is a classic 60–80–100 right triangle, which is a scaled version of the 3–4–5 triangle.