Question

A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m (see Fig. 5.3). Find its length and breadth.

Answer 1

let us consider breadth of the rectangle path be x meter
then length=x+3
the base of the triangle=x meter
area of triangle=1/2(b)(h)
6x
area of rectangle=(l )(b)
x^2+3x
area of triangle+4= area of rectangle
6x+4=x^2+3x
x^2+3x-6x-4=0
x^2-3x-4=0
x=-1 x=4
since x must be positive number
so,x=4
x+3=7

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let the Length of park be x.

And the breadth be x - 3.

Area = x(x - 3) m²

Base of isosceles triangle = x - 3

Altitude = 12 m

Its Area = 1/2 × (12 × x - 3) m² = 6(x - 3) m²

According to the Question,

⇒ x(x - 3) = 6(x - 3) + 4

⇒ x² - 3x = 6x - 18 + 4

⇒ x² - 3x = 6x - 14

⇒ x² - 3x - 6x + 14 = 0

⇒ x² - 9x + 14 = 0

⇒ x² - 7x - 2x + 14 = 0

⇒ x(x - 7) - 2(x - 7) = 0

⇒ (x - 7) (x - 2) = 0

⇒ x = 7, 2 (Length can't be 2 because then breadth = - 1)

⇒ x = 7

Length = x = 7 m

Breadth = x - 3 = 7 - 3 = 4 m.