 ###### Brian McCall

Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school

Brian was a geometry teacher through the Teach for America program and started the geometry program at his school

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# Perimeter - Problem 1

Brian McCall ###### Brian McCall

Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school

Brian was a geometry teacher through the Teach for America program and started the geometry program at his school

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Remember that the perimeter of a polygon is the sum of the lengths of the sides of that polygon. For rectangles, we know that the perimeter is P = 2b + 2h, where b is the length of the base, and h is the length of the height. This is because rectangles are made up of two pairs of congruent segments.

If given a perimeter of a rectangle that is 44 cm, the length of the height is 12 cm, and the length of the base is unknown x, you can use the formula for the perimeter to solve for the length of the base. Doing so, you get:

44 = 2x + 2(12)

44 = 2x + 24

Then, subtracting 24 from both sides:

20 = 2x

Dividing both sides by 2:

10 = x

So, the length of the base is 10 cm.

In this problem you are given the perimeter of this quadrilateral and you’re being asked to find this missing variable x. Well before we start, what is perimeter? Perimeter is the sum of the lengths of all the sides of a polygon and for rectangles we can use the short-cut that perimeter is equal to 2 times the base plus 2 times the height. So we can go back to our problem here, to our rectangle and we can say that our perimeter is equal to 2 times b, where b is you base plus 2 times h, where h is your height. And we have to identify what are b and h. Well the base is this side right here which is congruent to x, so we’re going to substitute in x for b. Your height, h is going to be the other side, so we’re going to have 12 and units here are centimeters. And as we’ve already indicated we know our perimeter is 44cm.

So if we substitute in for b, h and p into your equation, then we can solve for x. So instead of writing p I’m going to write 44. We’d have 2 time b, but we said is the same thing as x; so we’d have 2x plus 2 times our height and our height we said was 12cm. So now we have one equation, one variable x that we can solve.

So 44 is equal to 2x pus 2 times 12 is 24. So I’m going to subtract 24 from both sides. So this is just regular old algebra, I’m not doing anything you hopefully haven’t seen before. 44 minus 24 is 20 and that equals 2x. And last we’re going to divide by 2 and you find out that x must be 10. So up here where it says x equals? we’re going to make sure to include our units and we find out that x is 10. Where the key to this problem was remembering your perimeter formula, identifying your variables, substituting and solving.