An altitude of the triangle is five third the lenght of its corresponding base. If the altitude is increased by 4cm and base be decreased by 2cm, the area of the triangle remains the same. Find the base and altitude of the triangle

Question

An altitude of the triangle is five third the lenght of its
corresponding base If the altitude is increased by 4cm and base be
decreased by 2cm, the area of the triangle remains the same Find
the base and altitude of the triangle

Ankush Jain · Accepted Answer

Let the base of the triangle be x cm Case 1: $Altitude of triangle = \frac{5}{3} \times x = \frac{5}{3} x cm So, area of Δ = \frac{1}{2} \times base \times altitude = \frac{1}{2} \times x \times \frac{5}{3} x = \frac{5}{6} x^{2} {cm}^{2}$ Case 2: When altitude is increased by 4 cm and base is decreased by 2 cm, then also area remains unchanged, i e $\frac{5}{6} x^{2} = \frac{1}{2} \times (\frac{5}{3} x + 4) \times (x - 2) \Rightarrow \frac{5}{6} x^{2} = \frac{1}{2} (\frac{5 x + 12}{3}) (x - 2) \Rightarrow \frac{5}{6} x^{2} = \frac{1}{6} (5 x^{2} - 10 x + 12 x - 24) \Rightarrow 5 x^{2} = 5 x^{2} + 2 x - 24 \Rightarrow 2 x = 24 \Rightarrow x = 12 cm$ So, base of the triangle = 12 cm and Altitude of the triangle = $\frac{5}{3} \times 12$ = 20 cm