The base of a triangle is 4 cm longer than its altitude. If the area o

1.	The base of a triangle is 4 cm longer than its altitude. If the area of the triangle is 48 sq.cm, then find its base and altitude.
Answer» Let the altitude of the triangle h = x cm Then its base ‘b’ = x + 4. Area = 1/2 × base × height = 1/2 (x + 4)(x) = \(\frac{x^2+4x}{2}\) By problem \(\frac{x^2+4x}{2}\) = 48 ⇒ x² + 4x = 2 × 48 ⇒ x² + 4x – 96 = 0 ⇒ x² + 12x – 8x – 96 = 0 ⇒ x(x + 12) – 8(x + 12) = 0 ⇒ (x + 12)(x – 8) = 0 ⇒ x + 12 = 0 (or) x – 8 = 0 ⇒ x = -12 (or) x = 8 But x can’t be negative. ∴ x = 8 and x + 4 = 8 + 4 = 12 Hence altitude = 8 cm and base = 12 cm.

The base of a triangle is 4 cm longer than its altitude. If the area of the triangle is 48 sq.cm, then find its base and altitude.

Answer»

Let the altitude of the triangle h = x cm

Then its base ‘b’ = x + 4.

Area = 1/2 × base × height

= 1/2 (x + 4)(x)

= \(\frac{x^2+4x}{2}\)

By problem \(\frac{x^2+4x}{2}\) = 48

⇒ x² + 4x = 2 × 48

⇒ x² + 4x – 96 = 0

⇒ x² + 12x – 8x – 96 = 0

⇒ x(x + 12) – 8(x + 12) = 0

⇒ (x + 12)(x – 8) = 0

⇒ x + 12 = 0 (or) x – 8 = 0

⇒ x = -12 (or) x = 8

But x can’t be negative.

∴ x = 8 and x + 4 = 8 + 4 = 12

Hence altitude = 8 cm and base = 12 cm.