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

Step 1: Understanding the Concept:
We use the area formula for a triangle and then apply the Pythagoras theorem to find the third side.
Step 2: Key Formula or Approach:
Area = \( \frac{1}{2} \times \text{Base} \times \text{Altitude} \).
Pythagoras Theorem: \( \text{Base}^2 + \text{Altitude}^2 = \text{Hypotenuse}^2 \).
Step 3: Detailed Explanation:
Let the altitude be \( x \) cm.
Then the base is \( (x + 10) \) cm.
Area \( = \frac{1}{2} \times (x + 10) \times x = 600 \).
\[ x(x + 10) = 1200 \implies x^2 + 10x - 1200 = 0 \]
Factoring the quadratic:
\[ x^2 + 40x - 30x - 1200 = 0 \]
\[ x(x + 40) - 30(x + 40) = 0 \]
\[ (x - 30)(x + 40) = 0 \]
Thus, \( x = 30 \) (ignoring \( x = -40 \) as length is positive).
Altitude = 30 cm.
Base = \( 30 + 10 = 40 \text{ cm} \).
To find the hypotenuse (\( h \)):
\[ h^2 = 30^2 + 40^2 = 900 + 1600 = 2500 \]
\[ h = \sqrt{2500} = 50 \text{ cm} \).
Step 4: Final Answer:
The dimensions of the triangle are 30 cm, 40 cm, and 50 cm.