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ACT Math 5.2 Plane Geometry 251 Views


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ACT Math: Plane Geometry Drill 5, Problem 2. What is the area of the shaded sections?

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Transcript

00:02

Here's your shmoop du jour, brought to you by Equilateral Triangles. It's always the

00:07

same thing with them. This figure is made up of two equilateral

00:11

triangles, one rotated 60° about the center of the other, each with side lengths of 6.

00:18

Find the area of the shaded sections.

00:21

And here are your potential answers...

00:25

At first, we may only see 6 itty-bitty equilateral triangles... but on second glance, we see

00:32

that there are, in fact, 2 big honkin' equilateral triangles here.

00:37

So we know the side length of the biggies...

00:39

and no matter which way we look at it, each side of the big triangles is cleanly cut into

00:43

thirds. Since the big sides all measure 6 units, each

00:47

third measures 2 units. These thirds make up the sides of the smaller triangles, which

00:52

are equilateral

00:53

We have the base -- 2 -- but we still need to find the height.

00:57

To do that, we split the small triangle in half and use the Pythagorean theorem... a

01:01

squared plus b squared equals c squared... to solve for the height. It looks something

01:02

like this: We add 1 squared plus x squared to equal 2

01:05

squared, and by solving for x, we get the square root of three.

01:09

The area of a triangle is 1/2 times the base times the height 2 times the

01:14

square root of three times one half... we get square root of three as the area of the

01:20

small triangle. Since there

01:21

are six small triangles, we have to multiply the square root of three by six and that gives us 6 root three.

01:28

The correct answer is D.

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