Question

 Is it possible to design a rectangular park of perimeter 80 and area 400 sq.m.? If so find its length and breadth.​

Answer 1

$\huge \sf \red{Question \to}$

 Is it possible to design a rectangular park of perimeter 80 and area 400 sq.m.? If so find its length and breadth

$\huge \sf \purple{Solution \to}$

Let the length and breadth of the park be L and B.

Perimeter of the rectangular park $\tt = 2 (L + B) = 80$

So, $\tt L + B = 40$

Or, $\tt B = 40 – L$

Area of the rectangular park $\tt = L × B = L(40 – L) = 40L – L^2 = 400$

$\implies\tt L^2 – 40L + 400 = 0$

Which is a quadratic equation

Comparing the equation with $\large\tt ax^2 + bx + c = 0$, we get

$\sf\red{a = 1}$

$\sf\blue{b = -40}$

$\sf\pink{c = 400}$

Since, $\large\sf Discriminant = b^2– 4ac$

$\implies\tt (-40)^2 – 4 × 400$

$\implies\tt 1600 – 1600$

$\implies\tt 0$

Thus, $\sf b^2 – 4ac = 0$

Therefore, this equation has equal real roots. Hence, the situation is possible.

Root of the equation,

$\tt \: L = \frac{ - b}{2a}$

$\tt L = \frac{40}{2(1)} = \frac{40}{2} = 20$

Therefore, length of rectangular park,

$\tt L =$ $\sf\red{ 20 m}$

And breadth of the park,

$\tt B = 40 – L = 40 – 20 =$ $\sf\red{20 m}$

$\huge\ \green{\underline {\boxed { \mathfrak {⋐Thanks⋑}}}}$

Answer 2

$\bold{\huge{\fbox{\color{blue}{ANSWER}}}}$

Given ---

Area of rectangle = 400m²

Perimeter = 80m

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Perimeter of rectangle =

$2(l + b) = 80m$

Area of rectangle =

$l \times b = 400m {}^{2}$

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$2(l + b) = 80$

$l + b$

$= 82 \\ 2$

$= 40$

$l = 40 - b$

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$(40 - b)(b) = 400$

$40b - b {}^{2} = 40$

$b {}^{2} - 40b + 400 = 0$

by splitting middle term"

____________________

$b {}^{2} - 20b - 20b + 400 = 0$

$b(b - 20) \: \: - 20(b - 20) = 0$

$(b - 20) \:( b - 20) = 0$

$(b - 20) {}^{2} = 0$

$b = 20$

$breadth = 20m$

$length = 40 - b$

$= 40 - 20$

$= 20m$

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hope it helps you

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