rectangle is 4 times as long as it is wide. a second rectangle is 5 centimeters longer and2 centimeters wider?

Answer 1

Let the width of the first rectangle be x.
Therefore the length of the rectangle is 4x.
Therefore is area is:
A = LW
A = 4x^2

The second rectangle has a length of 4x+5
The second rectangle has a width of x+2
Therefore its area is:
A = LW
A = (4x+5)(x+2)
A = 4x^2 + 13x + 10

These expressions can be compared to deduce various things. Eg: if the second rectangle is twice as big as the first - we can deduce their actual size:

2(4x^2) = 4x^2 + 13x + 10
4x^2 - 13x - 10 = 0

By the quadratic formula:
x = 3.8923 (ignoring the negative)

Therefore the original rectangle has dimensions of 3.9cm by 15.6cm.

Answer 2

I love these! They really make you think.
Hmmm....let's try to fill in the 'punch-line'!
Trying to find the question; now that's fun!

Given that rectangle A has length 4 times greater than its width, and rectangle B is 5 cm longer and 2 cm wider than rectangle A,
what is the difference in area?
Express algebraically.

Answer 3

What you talkin' about, Willis?

Given the question, there are an infinite number of possible lengths and widths for rectangle A, and therefore an infinite number of lengths and widths for rectangle b.

For A:
l = 4w

For B:
length of B = length of A +5
width of b = 4(length of A) + 2

If it wants you to express the dimensions of B in terms of one variable, that's it. But if it wants something numeric, it's impossible to answer without more information.

Answer 4

What are you trying to answer? Area? Perimeter?

Answer 5

please tell us what is the question