Mathematics — Standard (041) — CBSE Class 10 board question

Let altitude = x cm, then base = (x + 10) cm.

Area of right-angled triangle = ½ × base × altitude

$$\frac{1}{2} \times (x+10) \times x = 600$$

$$x(x+10) = 1200$$

$$x^2 + 10x - 1200 = 0$$

Factorising:

$$x^2 + 40x - 30x - 1200 = 0$$

$$x(x + 40) - 30(x + 40) = 0$$

$$(x - 30)(x + 40) = 0$$

So, $x = 30$ or $x = -40$

Since a dimension cannot be negative, $x = -40$ is rejected.

∴ Altitude = 30 cm, Base = 40 cm

Finding hypotenuse (by Pythagoras' theorem):

$$\text{Hypotenuse} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \text{ cm}$$

The three dimensions are: Altitude = 30 cm, Base = 40 cm, Hypotenuse = 50 cm.

Source: Chapter 4, Exercise 4.2

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• Let one unknown represent altitude; express base in terms of it using the given condition ("exceeds by 10 cm").
• Use area formula to form the quadratic equation, then factorise.
• Always reject the negative root with a reason (length cannot be negative).
• Don't forget the third dimension — the hypotenuse — found via Pythagoras. Examiners expect all three dimensions to be stated clearly at the end.