The length of a rectangle is 11ft less than three times the width, and the area of the rectangle is 42ft^2. Fine the dimensions of the rectangle.

Answer 1

To solve this problem, it might be helpful to draw the rectangle and label what we are given. The question says that the length of the rectangle is 11ft less than three times the width. We can convert this into a mathematical equation which states L=3W-11 where W is the width of the rectangle. From here, we also define the area of the rectangle to be A=42ft².

We know that to find the area of a rectangle we must multiply the length by the width. This gives us A=L×W. If we substitute 3W-11 in for L we find A=(3W-11)×W. This can be expanded out by using the distributive property to obtain A=3W²-11W or 0=3W²-11W-A. If we plug in the known value for A we get 0=3W²-11W-42 which is a quadratic equation for W. If we make use of the quadratic formula we can calculate that W is either 6 or -7/3. Now, it doesn't make sense for the width of the rectangle to be negative, so we'll use W=6ft. Plugging this into L=3W-11 we see that L=7ft.

We now have the length and the width of the rectangle. All we have to do is plug them into the area formula and make sure they give us a result which agrees with the given information. Doing this yields A=L×W=7ft×6ft=42ft² which is in agreement with the information given in the problem.

Answer 2

L = length, and W = width

From the information given, we can write 2 equations:

L = 3W - 11

L*W = 42

Solve this pair of equations using one of the standard methods.....eg it looks like you could solve the second equation for either L or W and then substitute the resulting expression in the first equation.

Answer 3

w(3w - 11) = 42

3w² - 11w - 42 = 0

(3w + 7)(w - 6) = 0

3w + 7 = 0

3w = -7

w = -7/3

We don't measure using negative numbers, so this is out.

w - 6 = 0

w = 6 ft width

3(6) - 11 = 7 ft length