Question

is it possible to design a rectangular park of perimeter 320m and area 4800m²....if so find it's length and breadth​

Answer 1

Answer:

$\huge\underline\bold\red{Answer!!}$

yes,it is possible to design a rectangular park of perimeter 320m and area 4800m²

$if \: the \: length \: be \: l \:,breadth \: be \: b$

$area = lb = 4800 = 40 \times 120m {}^{2}$

$perimeter = 2(l + b)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2(40 + 120)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 320m$

$length = 120m$

$breadth = 40m$

Hope it helps you ❤️ ✨

Answer 2

Your Answer:

Given:-

$\tt \boxed{ \tt \star Perimeter = 320m} \\\\ \tt \boxed{ \star\tt Area = 4800m^2}$

Solution:-

$\tt l\times b= 4800 \rightarrow \rightarrow \rightarrow (1) \\\\ \tt So, l = 4800/b$

And

$\tt l + b = 160 \\\\ \tt Replacing \ \ values \\\\ \tt \Rightarrow \dfrac{4800}{b} + b = 160 \\\\ \tt \Rightarrow \dfrac{4800+b^2}{b} = 160 \\\\ \tt \Rightarrow b^2 - 160b + 4800 = 0$

So, finding discriminant

$\tt D=b^2 -4ac \\\\ \tt \Rightarrow D =( -160)^2 - 4(1)(4800) \\\\ \tt \Rightarrow D = 25600 - 19200 \\\\ \tt \Rightarrow D=6400 \\\\\ \tt \Rightarrow \sqrt{D} = 80$

So D is in positive. So this case is possible

$\tt Now \ finding \ Values \\\\ \tt \Rightarrow \dfrac{-b\pm D}{2a} \\\\ \tt \Rightarrow \dfrac{-(-160)\pm 80}{2} \\\\ \tt \Rightarrow \dfrac{160\pm 80}{2} \\\\ \tt b= \dfrac{240}{2} \ \ and \ \ \dfrac{80}{2} \\\\ \tt b = 120 \ \ and \ \ 40$

So, In first Case where b = 120

l = 40

And in Second case where b = 40

l = 120

So, 120 and 40 are the length and breadth of Rectangle