Question

The area of a right angled triangle is 165 m². Determine its base and altitude if the latter exceeds the former by 7 m.

Answer 1

SOLUTION :  

Let the base of a ∆  be x m, then the altitude be (x + 7) m.

Area of a ∆ = 1/2 x Base x Altitude

1/2 x  (x + 7) = 165

x (x + 7) = 165 × 2

x² + 7x = 330

x² + 7x - 330 = 0

x² + 22x – 15x – 330 = 0

[By middle term splitting]

x(x + 22) – 15(x + 22) = 0

(x + 22)(x – 15) = 0

(x + 22)  or (x – 15) = 0

x = - 22 or x = 15

Since, side can’t be negative, so x ≠ - 22 , Therefore x = 15 m

Base = x = 15 m

Altitude = x + 7 = 15 + 7 = 22 m

Hence, the base of a ∆  is 15 m and the altitude is 22 m.

HOPE THIS ANSWER WILL HELP YOU.. ..

Answer 2

Question :

The area of a right angled triangle is 165 m². Determine its base and altitude if the latter exceeds the former by 7 m.

Solution :

Let the base = x m

Then altitude = x+7 m

Area of right angled triangle =

½ × Base × Height = 165

165= ½ × x × (x+7)

165×2 = x²+7x

330= x²+7x

x²+7x-330 = 0

x²+22x-15x-330 = 0

x(x+22)-15(x+22)

(x+22)(x-15)

Then x = -22 and x = 15

The base can't be in negatives so ,

x = 15 m is the base

Height = x+7 = 15+7 = 22 m