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SAT Math 3.4 Geometry and Measurement 213 Views


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Description:

SAT Math 3.4 Geometry and Measurement

Language:
English Language

Transcript

00:02

Here’s your shmoop du jour, brought to you by missing values.

00:06

Charles Manson could teach a master class on the subject.

00:10

What is the value of x in this figure?

00:12

And here are the potential answers...

00:15

Okay, so what we have here is a tale of two right triangles.

00:19

Let’s start with the bigger of the two.

00:21

The hypotenuse is 11…one leg is the dotted vertical line, or the height,

00:26

and the other leg is 6 plus x… whatever “x” is.

00:29

The smaller triangle has a hypotenuse of 7, the same height, and plain ol’ x as its base.

00:35

So, of course, because we’re dealing with triangles, we can use the

00:38

Pythagorean Theorem to solve for the missing leg, x

00:41

For the bigger triangle, (x + 6) squared + h squared = 11 squared.

00:47

For the smaller one, x squared + h squared = 7 squared.

00:51

We can simplify the second one a bit....

00:53

So now we have x squared + h squared = 49.

00:57

By subtracting x squared from both sides, we now have h squared = 49 - x squared…

01:02

…and we can now take the second part – the 49 minus x squared…

01:06

…and use it to replace the h squared in our first equation…

01:10

…giving us x squared + 12x + 36 + (49 - x squared) = 121.

01:23

We can add together our 36 and 49, like this…

01:26

…and our two x squareds cancel out…

01:27

…giving us 12x + 85 = 121.

01:30

We’ll knock 85 off both sides to get 12x = 36…

01:34

…and finally divide both sides by 12 to get x = 3.

01:38

Choice B.

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