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Fifth Grade Math Triangle Challenge

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Here’s an especially tricky problem from 5th grade geometry. Everyone knows that the area of a triangle is 1/2 (b * h). But with this particular triangle, what qualifies as “the height” is difficult to see. At least it was for me the first time I looked at it.

I don’t know–maybe you’ll look at this problem and say “Duh, Jenny.” But for me, the scalene triangle was strange looking.

When I first looked at this I saw that it would be easy to solve with the Pythagorean Theorem. But Houghton Mifflin Math Expressions hadn’t covered that yet. So there was another even easier way to solve this problem that wasn’t jumping out to my me or my son.

Can you figure out what it is?

From Houghton Mifflin Math Expressions Grade 5 Chapter Two

From Houghton Mifflin Math Expressions Grade 5 Chapter Two

Figuring out the perimeter of the red triangle is easy. That’s 17 + 9 + 10 = 36 cm. But what about the area?

First, I’ll show the way that ends up being the most complicated: using the Pythagorean Theorem.

Using a squared + b squared = c squared, find out the area of the yellow triangle.

Using a squared + b squared = c squared, find out the area of the yellow triangle.

Now that you know b = 6, this lets you figure out that the length of the rectangle is 15 cm. That lets you figure out the area of the whole rectangle.

Now that you know b = 6, this lets you figure out that the length of the rectangle is 15 cm. That lets you figure out the area of the whole rectangle.

Using the Pythagorean Theorem to find the area of the yellow triangle.

Use the Pythagorean Theorem to find the area of the yellow triangle.

Then you figure out the area of the green triangle, and subtract green and yellow from the area of the rectangle, finally finding your answer.

Then you figure out the area of the green triangle, and subtract green and yellow from the area of the rectangle, finally find your answer.

This is a perfect example of how being algorithm dependent can screw up your number sense. I was so sure the Pythagorean Theorem was the way to go, I initially missed seeing the easier solution.

Another look at the original problem.

Another look at the original problem.

Okay, so everyone knows that the area of a triangle is 1/2 b * h. But with this particular triangle, that's tricky to see.

Triangles can always become parallelograms, which can be easier to deal with.

Now it's super easy to see that 8 cm = the height of the triangle, right?

Here’s the “Duh!” moment. Now it’s super easy to see that 8 cm = the height of the parallelogram which means it also = the height of the triangle.

1/2 the area of the parallelogram is the area of your triangle.

1/2 the area of the parallelogram is the area of your triangle.

Now after all of that, let’s look at the original problem and try a third method to solve this problem, using the formula 1/2 (b*h). This is arguably the easiest method.

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1/2 (9 * 8) = 36 sq cm.

Okay, so why didn’t I use the formula to begin with? When my son first looked at this, why didn’t I say “Dude, plug in the formula 1/2 (b * h),”?

Because that’s not what good math teachers do. Math is more than memorizing and applying formulas. Math is about experimenting, visualizing, internalizing and sometimes struggling until you reach a higher level of understanding.

This is an example of a problem that is simple yet confusing. Those are the best types! I’ve gone through college level calculus and I still looked at the picture and couldn’t viscerally understand why 8 cm was the height of the triangle. Neither could anyone in my family. (My husband, btw, is a lot smarter in math than me!)

So we played with this problem. We turned it inside out. Now, it makes sense. Along the way, we got to do a lot of cool math.


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