Question

The area of a right-angled triangle is \( 600 \text{ cm}^2 \). If the base of the triangle exceeds the altitude by 10 cm, find all the three dimensions of the triangle.

Hint:

Notice that the dimensions (30, 40, 50) are just 10 times the standard Pythagorean triplet (3, 4, 5).

Answer 1

Given:
Area of a right-angled triangle = 600 cm²
Base exceeds altitude (height) by 10 cm.

Let the altitude = \( h \) cm.
Then the base = \( h + 10 \) cm.

Step 1: Use the area formula of a right-angled triangle
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] Substitute the values: \[ \frac{1}{2} \cdot h \cdot (h + 10) = 600 \] Multiply both sides by 2: \[ h(h + 10) = 1200 \] Expand: \[ h^2 + 10h - 1200 = 0 \]

Step 2: Solve the quadratic
\[ h^2 + 10h - 1200 = 0 \] Factorise: \[ h^2 + 40h - 30h - 1200 = 0 \] \[ (h + 40)(h - 30) = 0 \] So, \[ h = 30\ \text{or}\ h = -40 \] Negative value is rejected.

Thus, \[ \text{Altitude} = h = 30\ \text{cm} \] \[ \text{Base} = h + 10 = 40\ \text{cm} \]

Step 3: Find the hypotenuse
Use Pythagoras theorem: \[ \text{Hypotenuse} = \sqrt{30^2 + 40^2} \] \[ = \sqrt{900 + 1600} \] \[ = \sqrt{2500} = 50\ \text{cm} \]

Final Answer:
The three sides of the right-angled triangle are:
Altitude = 30 cm
Base = 40 cm
Hypotenuse = 50 cm