The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Step-by-Step Solution
Key Concept: Use Pythagoras theorem for a right‑angled triangle and express one side in terms of the other using the given relation. This leads to a quadratic equation whose positive root gives the required lengths.
1. Let the base of the right triangle be $b$ cm.
2. Since the altitude is 7 cm less than the base, altitude $a = b-7$ cm.
3. For a right triangle, Pythagoras theorem gives
$$a^{2}+b^{2}=\text{hypotenuse}^{2}=13^{2}=169.$$
4. Substitute $a = b-7$:
$$(b-7)^{2}+b^{2}=169.$$
5. Expand and simplify:
\[b^{2}-14b+49+b^{2}=169\]
\[2b^{2}-14b-120=0\]
Divide by 2:
\[b^{2}-7b-60=0\]
6. Solve the quadratic equation $b^{2}-7b-60=0$ using factorisation or the quadratic formula.
\[b = \frac{7\pm\sqrt{7^{2}+4\times60}}{2}=\frac{7\pm\sqrt{289}}{2}=\frac{7\pm17}{2}\]
The two roots are $b=12$ cm and $b=-5$ cm. Since a length cannot be negative, take $b=12$ cm.
7. Find the altitude:
$$a = b-7 = 12-7 = 5\text{ cm}.$$
8. Hence the two sides (other than the hypotenuse) are:
- Base = $12\text{ cm}$
- Altitude = $5\text{ cm}$.
9. Verify: $5^{2}+12^{2}=25+144=169=13^{2}$, which satisfies Pythagoras theorem.
Correct Answer: Base = 12 cm, Altitude = 5 cm