SAT Math 1.3 Geometry and Measurement

SAT Math 1.3 Geometry and Measurement. Find the length of CE.

Geometry	Define trigonometric ratios and solve problems involving right triangles Right Triangles and Trigonometry Similarity, right triangles, trigonometry, and dimensions
Language	English Language
Problem Solving and Data Analysis	Key features of graphs
Product Type	SAT Math
Right Triangles and Trigonometry	Pythagorean Theorem Right Triangles
SAT Math	Geometry and Measurement
Similarity, Right Triangles, and Trigonometry	Define trigonometric ratios and solve right triangle problems
Triangles	30°-60°-90° Triangles

00:22

We know right off the bat, for example, that triangle CDE is a 30-60-90 triangle.

00:37

…and we remember that the sides of a 30-60-90

00:40

triangle always have a ratio of x to x square root of 3 to 2x.

00:46

It’s pretty obvious that DE is the longer of the two legs…

00:50

…and even if we weren’t sure by eyeballing it, we know it has to be the side across from the 60…

00:55

…so we can use that information to solve for CE, the hypotenuse.

01:00

5 equals x square root of 3.

01:03

We want to get x alone…

01:06

…so we divide both sides by square root of 3…

01:09

…and x equals 5 over the square root of 3.

01:13

Our hypotenuse has to be 2x…so multiply that whole thing by 2 and we wind up with

01:18

10 over the square root of 3. Of course, we have this thing about leaving

01:23

square roots in the denominator… …so multiply both the top and bottom by

01:26

square root of 3 and we have 10 square root of 3 all over 3…

01:31

…which is answer B.

01:33

We didn’t even need triangle ABC. Who knows why they threw that in here.