Find the dimensions of a rectangle with a perimeter of 54 meters, if its length is 3 meters less than twice its width.
Answers
Answer:
Let L = length
Let W = width
L = 2W - 3, since the length is 3 meters less than twice the width
The formula for perimeter is P = W + L + W + L = 2W + 2L
So 2W + 2L = 54
Substitute the value for L from the 1st equation in to the 2nd equation to get
2W + 2(2W - 3) = 54
2W + 4W - 6 = 54
6W = 60
W = 10
So L = 2W - 3 = 2(10) - 3 = 20 - 3 = 17
The width is 10 meters and the length is 17 meters.
Given data:-
- Perimeter of rectangle is 54 meter.
- Length of rectangle is 3 meter less than twice its width.
Solution:-
Let, L & W be the length & width of rectangle respectively.
{According to question & given}
Perimeter of rectangle = 54 meter .....( 1 )
L = 2W - 3 meter .....( 2 )
Now, we use formula of perimeter of rectangle :
—› Perimeter of rectangle = 2 [ L + W ]
{ from ( 1 ) & ( 2 )}
—› 54 = 2 [ 2W - 3 + W ]
—› 54 = 2 [ 3W - 3 ]
—› 54 = 6W - 6 i.e.
—› 6W = 54 + 6
—› 6W = 60
—› W = 60/6
—› W = 10 meter {Width = 10 meter}
Put value of W in eq. ( 2 )
—› L = 2W - 3
—› L = 2 ( 10 ) - 3
—› L = 20 - 3
—› L = 17 meter {Length = 17 meter}