Question

Is it possible to design a rectangular park of perimeter 80 metres and area 400m². If so, find its length and breadth.​

Answer 1

$\huge\underline\mathbb {SOLUTION:-}$

$\mathsf {Let\:length\:of\:park = x\:metres}$

$\mathsf {We\:have\:given\:area\:of\:rectangular\:park = 400\:m^2}$

$\therefore \mathsf {Breadth\:of\:park = \frac{400}{x}\:metres}$

• Area of rectangle = L × B

$\underline \mathsf {Here,}$

• L = Length
• B = Breadth

$\mathsf {Perimeter\:of\:rectangular\:park = 2(length + breadth) }$

$\implies \mathsf { 2\bigg(x + \frac{400}{x}\bigg)\:metres}$

$\mathsf {We\:have\:given\:perimeter\:of\:rectangle = 80\:metres}$

$\underline \mathsf \red {According\:to\:condition,}$

$\mathsf {2 \bigg(x + \frac{400}{x}\bigg) = 80}$

$\implies \mathsf {2\bigg(\frac{x^2 + 400}{x}\bigg) = 80}$

$\implies \mathsf {2x^2 + 800 = 80x }$

$\implies \mathsf {2x^2 - 80x + 800 = 0}$

$\implies \mathsf {x^2 - 40x + 400 = 0}$

$\underline \mathsf \blue {Comparing\:equation,}$

$\mathsf {x^2 - 40x + 400 = 0\:with\:general\:quadratic\:equation\:ax^2 + bx + c = 0,}$

$\underline \mathsf \red {We\:get\::-}$

• a = 1
• b = - 40
• c = 400

$\mathsf {Discrimination = b^2 - 4ac = (-40)^2 - 4(1)(400) = 1600 = 0}$

$\mathsf {Discrimination\:is\:equal\:to\:0.}$

$\therefore$ Two roots of equation are real and equal which means that it is possible to design a rectangular park of perimeters 80 meters and area 400m².

$\mathsf {Using\:quadratic\:formula\:x = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\:to\:solve\:equation,}$

$\mathsf {x = \frac{40 \pm \sqrt{0}}{2} = \frac{40}{2} = 20}$

$\mathsf {Both,the\:roots\:are\:equal\:to\:2.}$

$\therefore \mathsf {Length\:of\:rectangular\:parks = 20\:metres}$

$\mathsf \green {Breadth\:of\:rectangular\:park = \frac{400}{x} = \frac{400}{20} = 20\:metres}$

Answer 2

$\huge{\underline{\underline{\bf{\blue{Given}}}}}$

• Perimeter of rectangular park = 80m
• Area of rectangular park = 400m²

$\large\underline\bold\green{NoTE}$

• Area of rectangle = length×breadth
• perimeter of rectangle = 2(l+b)

$\huge{\underline{\underline{\bf{\blue{To\:find}}}}}$

• Length and breadth of rectangular park

$\huge{\underline{\underline{\bf{\blue{Solution}}}}}$

Let the length and breadth be l and b

Now,

According to question

Perimeter of rectangle = 80m

=> 2(l+b) = 80

=> l + b = 80/2

=> l + b = 40

=> l = (40 - b) --------(i)

Area of rectangle = 400m²

=> l × b = 400 -----(ii)

using (i)

=> (40-b) × b = 400

=> 40b - b²- 400 = 0

=> -b² + 40b - 400 = 0

splitting middle term

=> -b² +20b + 20b -400 = 0

=> -b(b-20)+20(b-20)=0

=> (-b+20)(b-20) = 0

Either

b = 20 or b = 20

Substitute the value of breadth in equation (ii)

=> l × b = 400

=> l × 20 = 400

=> l = 400/20 = 20m

$\large{\boxed{\red{Required\:length\:=\:20m}}}$

$\large{\boxed{\red{Required\:breadth\:=\:20m}}}$