Question

20. The dimension of a plot is 20 cm x 15 cm. Find the length of the wire which is required to fence
the plot?​

Answer 1

Answer:

$\blue\bigstar$ The length of the wire which is required to fence the plot = 70cm.

SOLUTION

Length of the plot = 20cm.

Breadth of the plot = 15cm.

Here,the plot is of rectangular shape.

Fencing is done along the sides of the plot.

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 20 cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 15 cm}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

$\boxed{\sf Length \: of \: wire \: required\: to \: fence \: the \: plot = Perimeter \: of \: the\: rectangular \: plot}$

Perimeter of a rectangle = 2(l+b) , Where l is the length and b is the breadth of the rectangle.

Here,

l = 20cm.

b = 15cm.

⇒ Perimeter of the rectangular plot = 2(20+15)

⇒ Perimeter of the rectangular plot = 2 × 35

⇒  Perimeter of the rectangular plot = 70cm.

So, the length of the wire which is required to fence the plot = 70cm.

Answer 2

Given:

• The dimensions of a plot are 20cm × 15cm

To Find:

• the length of the wire which is required to fence the plot?

Diagram:

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 20 m}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 15 m}\end{picture}$

Solution:

➸ Here, we have been provided with the dimensions of a plot and said that the plot needs to be fenced and that we should find the length of the wire required to fence it respectively.

~ So, we have to find its perimeter to know the length of wire required to fence it as we fend plots throughout their boundary (perimeter)

$\star \: \blue{ \underline{ \boxed{ \pink{ \mathfrak{ perimeter _{(rectangle)} =2(lenght + breadth}}}}}$

✪ Now, let's substitute the dimensions accordingly ツ

$\longrightarrow \tt \: perimeter = 2(l + b) \: \: \: \: \: \\ \\ \\ \longrightarrow \tt \: perimeter = 2(20 + 15) \\ \\ \\ \longrightarrow \tt \: perimeter = 2(35) \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \tt \: perimeter = { \boxed{ \tt{70m} \bigstar}} \: \: \: \: \: \: \:$

Hence:

• The required length of wire is 50 meters

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