Help with daughter's math problem

You might need to cut some paper.

Cut a square and explain that a square has as its area the length of one of its sides squared (times itself).

Fold the square in half along one side, and ask what the area of the folded square is (half of the square) and that this is product of half width and the width.

Then ask what is the area of both figures added together. Then explain one half is 50% and a whole is 100%. So when you add then together this is the same as 150% (or 50% longer).

Then (this is the tough part), ask if this 150% larger area is equal to 294 in^2, what is the length.

Then show in this relation length is 1.5 time the width and the area is the length (1.5w) times the width (w) and this is 294in^2. So when you crunch the numbers 1.5w*w=294 or using the rules of arithmetic w*w=294/1.5. So the width is 14 inches and the length is 50% more then the width (14 inches and 7 inches).

Rap

Couldn't you use x's and y's and just explain why you have those equations? She'll need to know how to do that eventually.

So x * (1.5)x = 294? or x^2 = 194 so x = 14 (width) and length = 1.5 * 14 = 21 inches

Nicely explained above.

Eventually yes, but there is a strong relation between elementary algebra and geometry--and geometry is well--more tactile.

There is a teaching tool, known as algebra blocks, that in the hands of a skilled math educator, can demonstrate tactile algebraic manipulations as complex as solutions and derivations of the quadratic equation.

In the old days (before plastic) they would use cut paper to do the same thing.

Rap

Re: Help with daughter's math problem

L = length, w = width
L = 1.5W
Area = LW = (1.5W)W = 1.5 W^2 = 294
w^2 = 294/1.5 = 196

W = 14
L = 1.5W = 21

Thank You, great explanations!