Question

a rectangular park is to be designed whose breadth is 3 m les than its length.its area is to be 4 square metres more than area of a park that has already been made in shape of isosceles triangle with base as breadth of rectangular parkland of alititude 12m.find its length and breadth.

Answer 1

length is x
breadth is x-3
area of rectangular park is 2(l+b)
2(x+x-3)=4x-6
base of isosceles triangle is x-3
height is 12
area of isosceles triangle is bh/2
area is 6x-18
on equating
6x-18 =4x-6+4
x=8
y=5
length is 8
breadth is 5

Answer 2

Let the length of the rectangular park = x m

Then, it's length = (x+3)m

$\therefore$ Area of the rectangular park =

$\sf{x(x + 3) = ( {x}^{3} + 3x) {m}^{2} }$

Now, base of the isosceles triangle = x m

$\therefore$ It's area = $\sf{\frac{1}{2}×x×12=6x{m}^{2}}$

A/C,

Area of rectangular park = Area of the triangular park + $\sf{4{{m}^{2}}$

$\longrightarrow$$\sf{ {x}^{2} + 3x = 6x + 4}$

$\longrightarrow$$\sf{ {x}^{2} - 3x - 4 = 0}$

$\longrightarrow$$\sf{{x}^{2} - 4x + x - x = 0}$

$\longrightarrow$$\sf{(x + 1)(x - 4) = 0}$

$\longrightarrow$$\sf{x = - 1 \: or \: x = 4}$

As the breadth of the park cannot be negative, x ≠ -1, x = 4

Hence, the breadth of the park = 4m

And it's length = 4 + 3 = 7m

Verification

Area of rectangular park = 4m × 7m = $\sf{28{m}^{2}}$

Area of triangular park = $\sf{ \frac{1}{2} \times 4 \times 12 = 24 {m}^{2} = (28 - 4) {m}^{2}}$