Question

11. The length of a rectangular room is twice its breadth.
If the perimeter of the room is 180 cm, find the sides
of the room.​

Answer 1

We are given,
l = 2b. eqn1
We know, perimeter of a rectangle = 2 ( l + b ) = 180 cm. eqn 2


Putting the value of l from equation 1 in equation 2,
2( 2b + b ) = 180
3b = 90
b = 30 cm
We know, l = 2b, therefore l = 60 cm.
Therefore the length and breadth of the room are 60 cm and 30 cm respectively :)

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

• The length of a rectangular room is twice its breadth.  If the perimeter of the room is 180 cm, Find the sides  of the room.​

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♣ ᴀɴꜱᴡᴇʀ :

• Breadth of the rectangular room = 30 cm

• Length of the rectangular room =  60 cm

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 60 cm }\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 30 cm }\put(-0.5,-0.4){\bf C}\put(-0.5,3.2){\bf A}\put(5.3,-0.4){\bf D}\put(5.3,3.2){\bf B}\end{picture}$

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

Let Breadth of the rectangular room be x

Then :

Length of the rectangular room = 2x

Perimeter of Rectangle = 2 × (Length + Breadth)

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

Perimeter of Rectangle = 2 × (x + 2x)

180 cm = 2 × (x + 2x)

180 cm = 2 × (3x)

180 cm = 6x

Dividing both sides by 6 :

(180 cm)/6 = (6x)/6

30 cm = x

x = 30 cm

Now we can easily findout values of Length and Breadth from the value of x

We have :

Breadth of the rectangular room = x

Length of the rectangular room = 2x

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

Substituting the values :

Breadth of the rectangular room = 30 cm

Length of the rectangular room = 2 × 30 cm = 60 cm

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 60 cm }\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 30 cm }\put(-0.5,-0.4){\bf C}\put(-0.5,3.2){\bf A}\put(5.3,-0.4){\bf D}\put(5.3,3.2){\bf B}\end{picture}$