Question

The length of a rectangle is four times its width. If the perimeter of the rectangle is 80 yd, find its length and width.​

Answer 1

>>> Length = 32 yd

>>> Width = 8 yd

Explanation:

Given ,

• Length of rectangle is 4 times its Width

• Perimeter of rectangle is 80 yd

We have to find ,

• Length and Width of the rectangle

Solution :

For finding Length and Width of rectangle, We need to assume the width and then form an equation according to given condition and after that we have to form an equation with the help of its perimeter. And at Last we have to solve the equation to get the value of Width and Length.

• Let Width of rectangle be "x yd"

According to condition , Length of rectangle is 4 times its Width which means we have to multiply 4 with x , So :

• Length of rectangle is "4x yd"

Now ,

`\: \: \: \: \: \: \: \: \boxed{\sf{Perimeter_((Rectangle)) = 2( l + w )}}`

Where ,

• l refers to length of rectangle

• w refers to width of rectangle

`\longmapsto \: \: \sf{2(4x + x) = 80}`

Step 1 : Dividing both sides with 2 :

`\longmapsto \: \: \sf{ \frac{ \cancel{2}(4x + x) }{ \cancel{2}}= \cancel{( 80)/(2)}}`

We get ,

`\longmapsto \: \: \sf{(4x + x) = 40}`

Step 2 : Adding 4x and x :

`\longmapsto \: \: \sf{5x = 40}`

Step 3 : Dividing both sides with 5 :

`\longmapsto \: \: \sf{ \frac{ \cancel{5}x}{ \cancel{5} }= \cancel{( 40)/(5)}}`

We get ,

`\longmapsto \: \: \underline{ \boxed{\sf{ \bold{x = 8 \: yd}}} } \: \: \: \bigstar`

• Therefore, Width of rectangle is "8 yd"

We know that Length of rectangle is 4 times its breadth that means :

`\longmapsto \: \: \sf{Length = 4x}`

`\longmapsto \: \: \sf{Length = 4 * 8}`

`\longmapsto \: \: \underline{ \boxed{ \sf{ \bold{Length = 32 \: yd}}}} \: \: \: \bigstar`

• Therefore , Length of rectangle is "32 yd"

Verification :

• 2 ( l + w ) = 80

• 2 ( 32 + 8 ) = 80

• 2 ( 40 ) = 80

• 80 = 80

• LHS = RHS

• Hence , Verified

Therefore, we can say that our answer is correct :)

Answer 2

The length is 8. the width is 32.