Question

The lengthand the breadth of arectengular field are in the ratio 5:3. if its perimeter is 128m, find the dimensions of the field​

Answer 1

Answer:

Length of the rectangular field is 40 m

Breadth of the rectangular field is 24 m

Step-by-step explanation:

the length and the breadth of a rectangular field are in the ratio 5:3

Then, Let, length of the rectangular field is 5x m and breadth of the rectangular field is 3x m

Perimeter of the rectangular field is 128 m

Then, 2(5x + 3x) = 128

=> 8x = 64

=> x = 8

Hence, The length of the rectangular field is (5×8) = 40 m

and the breadth of the rectangular field is (3×8) = 24 m

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

The lengthand the breadth of a rectengular field are in the ratio 5:3.

Let assume that

Length of a rectangular field = 5x m

Breadth of a rectangular field = 3x m

Further, given that

Perimeter of rectangular field = 128 m

We know,

Perimeter of a rectangle is given by

$\boxed{\rm{  \:Perimeter_{(rectangle)} \: = \: 2(Length + Breadth) \: \: }}$

So, on substituting the values, we get

$\rm \: 2(5x + 3x) = 128$

$\rm \: 2(8x) = 128$

$\rm \: 16x = 128$

$\rm \: x = \dfrac{128}{16}$

$\rm\implies \:x = 8$

Hence, Dimensions of rectangular field are

Length of a rectangular field = 5x = 5 × 8 = 40 m

Breadth of a rectangular field = 3x = 3 × 8 = 24 m

$\rule{190pt}{2pt}$

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$