Question

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

Answer 1

SOLUTION :  Given :  Area of a ∆ = 48 cm² Let the altitude of a ∆ be x cm & base = x + 4 cm Area of triangle = 1/2×base×altitude Area or triangle =½ × (x +4) × x  48 = 1/2× x × (x+4)  48× 2=  x × (x+4) 96=  (x²+4x) x² + 4x- 96 = 0 x²+12x - 8x - 96 = 0  [By factorization] x(x+12) - 8(x+12) = 0 (x−8) (x+12) = 0 (x−8) = 0 or (x+12) = 0 x = 8 & x = -12 length cannot be negative so, x = 8. Altitude = x = 8 cm Base = (x + 4) = 8 + 4 = 12 cm  Hence, its base is 12 cm & altitude is 8 cm  HOPE THIS ANSWER WILL HELP YOU...

Answer 2

let the base of the triangle be x cm and altitude be y cm
Given base = 4 + y
Area of the triangle = 1/2 × base × altitude
48 = 1/2 × (4 + y) × y
⇒ 96 = 4y + y²
⇒ y²+ 4y – 96 = 0
⇒ y²+ 12y - 8 y – 96 = 0
⇒ y( y + 12) - 8 ( y + 12) = 0
⇒ (y - 8) ( y + 12) = 0
⇒ y = 8 or - 12
Since altitude cannot be negative ∴ y = 8
⇒ base = 4 + 8 = 12 cm
Hence, Base = 12 cm; Altitude = 8 cm