This is an algebra question, so the student should be comfortable represent unknown values with variables.Let w = width and h = height of the rectangle.To get area, you multiply. So in this case w x h = 144.To get perimeter you add all the sides: w + h + w + h = 48. If you
simplify this, you get:
2w + 2h = 48.
If you divide both sides of this by 2, you get
w + h = 24.So we are looking for two numbers that when multiplied give 144 and when added give 24. At this point, you can decide whether the student should guess and then check the guess by putting the answers into the two equations, or proceed analytically. For this question, it is easy now to guess and check, but if you want to teach your child more advanced methods, read on.Now solve for one of the variables (it doesn't matter which):
w + h = 48
(w+h) - h = 24 - h
w = 24 - hWe set that aside for a moment and similarly solve for w using the
area equation.
w x h = 144
w = 144 / hSo we have two different expressions that both equal w. You can refer to logic and say that if thing 1 is the same as thing 2 and thing 3 is the same as thing 2, then thing 1 must be the same as thing 3.So
24 - h = 144 / hNow make the left hand side 0 by subtracting (24 - h) from both sides:
24 - h - (24 - h) = 144/h - (24 - h)
0 = 144/h - (24 - h)
and let's get rid of that fraction by multiplying both sides by h:
0h = h ( 144/h - 24 + h)
0 = 144 - 24h + h^2
0 = h^2 - 24h + 144This is called a quadratic expression. The easiest way to solve for the actual height is to factor it. Note that at this point we are effectively using guess-and-check again: we are trying to find two numbers that multiplied together give 144 and added together give -24.0 = (h - 12) (h - 12)What this means is that h is some value that makes the expression above true, which is only the value h = 12.Now that we have h, we can use either of the original equations to solve for w.
w = 144/h
w = 144/12 = 12.It's a long question with many steps for a grade six student, but an advanced learner will learn a lot from seeing the analysis (and then practicing the technique with other numbers, like area 56 and perimeter 30, or area 150 and perimeter 70).Stephen La Rocque.>
and another that doesn't use algebra:
Anne,there is an algebraic way in which to solve this but I'd say at the grade 6 level the teacher expects a guess and check approach. Since the area is the product of the length of the sides it might be useful to factor 144 and see what the perimeter might be, i.e. is it 48? For example, 144 is 6 x 24 but if that was the shape of the rectangle the perimeter would be 6+6+24+24 = 60, not 48. Similarly 144 = 8 x 18 but this leads to a perimeter of 8+8+18+18 = 52, not 48. What else can you try?
