Question

Is it possible to design a rectangular park of perimeter 80m & area 400m^2? If so, find its length & breadth?

Answer 1

l  = x

2(l + b) = 80

b = 40 - x

area = lb = 400

x  x (40 - x) = 400

40x - x² - 400

x² - 40x + 400 = 0

(x - 20)² ie. (x-20) (x - 20)

x = 20

Length = 20

Breadth = 40 - 20 = 20

Answer 2

$\bf\huge\boxed{\boxed{\bf\huge\:Hello\:Mate}}}$

$\bf\huge Let\: length\: be\: x\: metres\: and\: breadth\: be\: y\: metres$

$\bf\huge Perimeter = 80m$

$\bf\huge => 2(x + y) = 80$

$\bf\huge => x + y = 40 (Eqn1)$

$\bf\huge Area = 400 m$

$\bf\huge => xy = 400$

$\bf\huge => x(40 - x) = 400 (From Eqn 1)$

$\bf\huge =>40x - x^2 = 400$

$\bf\huge => x^2 - 40x + 400 = 0$

$\bf\huge Hence\: , a = 1 , \:b = -40 \:and\: c = 400$

$\bf\huge => D = b^2 - 4ac$

$\bf\huge => D = (-40)^2 - 4\times 1 \times 400$

$\bf\huge => D = 1600 - 1600 = 0$

$\bf\huge Equation\: has\: equal\: roots$

$\bf\huge Length \:and\: Breadth$

$\bf\huge => x^2 - 40x + 400 = 0$

$\bf\huge => (x - 20)^2 = 0$

$\bf\huge => x = 20 , 20$

$\bf\huge Length = 20m\: and\: Breadth = 20m$

$\bf\huge Hence\:Design\: is\: Possible$

$\bf\huge\boxed{\boxed{\:Regards=\:Yash\:Raj}}}$