Question

The altitude of a right-angled triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Hint:

Always check if the sides form a Pythagorean triplet. Here, $5,12,13$ is a well-known set, which quickly confirms the solution.

Answer 1

Solution (Right triangle):
 

Step 1: Define variables
Let base $=x$ cm. Then altitude $=x-7$ cm. Hypotenuse $=13$ cm.
 

Step 2: Apply Pythagoras theorem
\[ x^2+(x-7)^2=13^2. \] \[ x^2+x^2-14x+49=169. \] \[ 2x^2-14x-120=0 \ \Rightarrow \ x^2-7x-60=0. \]

Step 3: Solve quadratic
\[ x=\frac{7\pm\sqrt{(-7)^2-4(1)(-60)}}{2}=\frac{7\pm\sqrt{49+240}}{2}=\frac{7\pm17}{2}. \] So $x=\frac{24}{2}=12$ or $x=\frac{-10}{2}=-5$ (reject negative).
 

Step 4: Find altitude
Base $=12$ cm, Altitude $=12-7=5$ cm.
\[ \boxed{\text{Base }=12\ \text{cm}, \ \text{Altitude }=5\ \text{cm}} \]