How to Solve a Right Triangle Given Its Perimeter and Altitude

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

• 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: \[ a^2 + b^2 = H^2 \]
• 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 \(h = 48\): \[ ab = 48H \]
• From the perimeter equation: \[ a + b = 240 - H \]
• Square both sides: \[ (a + b)^2 = (240 - H)^2 \] \[ a^2 + b^2 + 2ab = 240^2 + H^2 - 480H \]
• Subtract \(a^2 + b^2 = H^2\): \[ 2ab = 240^2 - 480H \]
• Substitute \(ab = 48H\): \[ 2(48H) = 240^2 - 480H \] \[ 96H = 57600 - 480H \]
• Solve for \(H\): \[ 576H = 57600 \] \[ H = 100 \text{ cm} \]
• Substitute \(H = 100\): \[ a + b = 140 \] \[ ab = 4800 \]
• From \(ab = 4800\), write: \[ b = \frac{4800}{a} \]
• Substitute into \(a + b = 140\): \[ a + \frac{4800}{a} = 140 \]
• Multiply by \(a\): \[ a^2 + 4800 = 140a \] \[ a^2 - 140a + 4800 = 0 \]
• Solve the quadratic equation: \[ a = 80 \text{ cm} \quad \text{or} \quad a = 60 \text{ cm} \]
• Using \(b = 4800/a\): \[ b = 60 \text{ cm} \quad \text{or} \quad b = 80 \text{ cm} \]

Final Answer

Assuming \(a < b\), the triangle dimensions are:

This is a classic 60–80–100 right triangle, which is a scaled version of the 3–4–5 triangle.

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