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the legnth of a rectangle is three times its width. if the perimeter of the rectangle is 64 in, find the length and width

Sagot :

Step-by-step explanation:

According to the question,

Let the width of rectangle be x and length of rectangle be 3x

Perimeter of Rectangle :- 2(L+B) = 64 in

Putting the values we get ,

2(3x+x) = 64 in

8x = 64 in

x = 8 in

Putting the value of x ,

Width :- 8 Inch

Length :- 24 inch

Given:-

[tex]\rightarrow[/tex] Length(l) of the rectangle is three times it's width(w) = 3w.

[tex]\rightarrow[/tex] Width(w) of the rectangle = w.

[tex]\rightarrow[/tex] Perimeter of the rectangle = 64in.

To Find:-

[tex]\rightarrow[/tex]Length and width of the rectangle.

Solution:-

[tex]\rightarrow[/tex] Perimeter of rectangle = [tex]\sf{2(l+w)}[/tex] putting the value of perimeter, l and w from the above given)

[tex]\rightarrow[/tex] 64 = [tex]\sf{2(3w+w)}[/tex]

[tex]\rightarrow[/tex] 64 = [tex]\sf{2(4w)}[/tex]

[tex]\rightarrow[/tex] 64 = [tex]\sf{8w}[/tex]

[tex]\rightarrow[/tex] [tex]\sf{\frac{64}{8}}[/tex]= [tex]\sf{w}[/tex]

[tex]\rightarrow[/tex] [tex]\sf{8}[/tex]= [tex]\sf{w}[/tex]

Therefore, width of the rectangle = 8in.

And Length = 3(8)in. = 24in.

To check whether the answer is correct or not, we can put the value of length and width in the formula = [tex]\sf{2(l+w)}[/tex]

[tex]\rightarrow[/tex] [tex]\sf{=\: 2(24+8)}[/tex]

[tex]\rightarrow[/tex] [tex]\sf{=\: 2(32)}[/tex]

[tex]\rightarrow[/tex] [tex]\sf{=\: 64in.}[/tex]

Since, the perimeter of the rectangle is same as given in the question, therefore the value of length and width are correct.

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