The length of a rectangle is 5cm more than twice the width. The perimeter of the rectangle is 34 cm. Find the dimensions of the rectangle. So confused on how to do this problem please help me thanks.
Perimeter = L + L + W + W
or P = 2L + 2W
L = 5+ 2w
P = 2(5+2W) + 2W
I replaces the L in the original formula with the 5 + 2w because that equals L according to the problem.
34 = 10 + 4W + 2W
Just solve for W.
And then you can find L.
Thanks
You are welcome.
the length of marshalls rectangular poster is 2 times its width .if the perimeter is 24 inches what is the area of the poster?
To solve this problem, we need to set up an equation based on the information given and then solve for the dimensions of the rectangle.
Let's start by assigning variables to the dimensions of the rectangle. Let's say the width of the rectangle is represented by 'w' (in cm), and the length of the rectangle is represented by 'l' (in cm).
Based on the given information, we can create two equations:
1) The length of the rectangle is 5cm more than twice the width:
l = 2w + 5
2) The perimeter of the rectangle is 34cm:
Perimeter = 2(length + width)
Plugging in the values, we get:
34 = 2(l + w)
Now, we can substitute the value of 'l' from the first equation into the second equation:
34 = 2((2w + 5) + w)
Simplifying further:
34 = 2(3w + 5)
34 = 6w + 10
24 = 6w
4 = w
Now, we can substitute this value of 'w' back into the first equation to find 'l':
l = 2w + 5
l = 2(4) + 5
l = 8 + 5
l = 13
Therefore, the width of the rectangle is 4cm and the length of the rectangle is 13cm.