26 inches. Technically, since this is a perimeter problem, you need to add each length to itself and then the others. If this is confusing, you essentially need to do 5+5+8+8=26 because you have two of the fives and two eights. Perimeter is much easier than area, I promise you that
the idea of showing that multiplying by 1 is true because as long as A and B follow the same ratio pattern (ie. 1n:2n is the same as 2n:4n) the statement will always be true
We assume w is the number of <em>inches</em> of width (as opposed to <em>feet</em> or some other measure). The length is 7 inches more than the width, so is w+7. The area is the product of these dimensions, and the problem statement tells us that area is greater than 375 in².
... w(w+7) > 375 . . . . . the inequality that can be used to find dimensions
Hello from MrBillDoesMath!
Answer:
23
Discussion:
In ΔABC side AC is 11, but as the triangle is isosceles the other leg also has length 11. Let the third, unknown side, be called x. Then
11 + 11 + x = Perimeter = 45.
11 + 11 + x = 45 => as 11 + 11 = 22
22 + x = 45 => subtract 22 from both sides
x = 45 -22 = 23
Thank you,
MrB
