4.2+Analyzing+Triangles+0910

A.R. 1/11/09 Block 8F Essential Question: How can Pythagoras help me? Big Idea: The dimensions of a right triangle can be determined with limited information. Notes: None


 * Problem 4.2 Analyzing Triangles**



Find the point halfway between vertices //B// and //C.// Label this point //P//. Point //P// is the //midpoint// of segment //BC//. Draw a segment from vertex //A// to point //P//. This divides triangle //ABC// into two triangles. Cut out triangle //ABC// and fold it along line //AP.//

A) Triangles ABP are both halves of the same equilateral triangle. They are symmetrical (the exact same), they both have a right angle and their other two angles are the same too.
 * A.** **How does triangle ABP compare with triangle ACP?**

B. **Find the measure of each angle in triangle ABP. Explain how you found each angle.** B) I know that angle P is 90° because it is a right angle. I also know that each triangle's angles add up to 180°, so 180º- 90º = 90º. So there are 90º left over. In an equilateral triangle, I know that all the angles are the same too, so when you cut the triangle in half you also cut the angle in half. So angle P is 90º, angle B is 60o, because 180÷3=60, and to find angle A all we have to do is, 90+60=150 and then 180-150= 30, so angle A is 30o.

C) Since the original triangle was an equilateral triangle, all the sides are the same. Since the triangle was cut in half that means that one of the sides stays the same, one is cut in half and the other one you have to find using the Pythagorean Theorem. The hypotenuse is 2; I know this because since the triangle was cut in two, the hypotenuse will stay the same, as it is the side length that did not get affected. Then, side AP is 1 because it used to be two until it the triangle was cut in half, and by doing so the line was cut in half. To find out B we have to use the Pythagorean theorem, but since we only know the length of the hypotenuse and one of the legs, we have to put in what we know and solve for what we don't know. So, 2. L1 = 1 3. H = 2 4. 12+X2 = 22 5. 2- 2+X2 = 4 -2 6. X2 = 2 7. X = √2 So the length of side AP is √2
 * C.** **Find the length of each side of triangle ABP. Explain how you found each length.**
 * 1.** L12+L22 = H2 [[image:problem4.1_advanced_math_8questionpic3.png align="right"]]

D. **Two line segments that meet at right angles are called perpendicular line segments. Find a pair of perpendicular line segments in the drawing above.** D) Segments AP and CP are perpendicular.

E) You can use the Pythagoras Theorem because it's a right triangle. It's a 30-60-90 triangle, which is half of an equilateral triangle.
 * E.** **What relationships do you observe among the side lengths of triangle ABP?**

F) Yes because they both are the same triangle put together to make an equilateral triangle.
 * F.** **Are these relationships true for triangle APC?**

**Follow Up**

1) **A right triangle with 60° angle is sometimes called a 30-60-90 triangle. This 30-60-90 triangle has a hypotenuse of length 6. What are the lengths of the other two sides?** 1) The side lengths are 3, 6, and √3. The way that I know these answers is because this was first an equilateral triangle, and then it was cut in half so the hypotenuse is still the same, and the base is divided by 2 so 6÷2 = 3. So that takes care of 6 and 3. Now to find the straight side you have to use the Pythagorean theorem So, 2. L1 = 3. H = 2 4. 12+X2 = 22 5. 2- 2+X2 = 4 -2 6. X2 = 2 7. X = √2 So, the last side length is √2
 * 1.** L12+L22 = H2 [[image:problem4.1_advanced_math_8questionpic4.png align="right"]]

2) **Square ABCD has side lengths of 1. On Lab sheet 4.2, draw a diagonal, dividing the square into two triangles.** a) **How do the 2 triangles compare?** They are both isosceles and are both right triangles b) **What are the angles?** The angles are, 90°, 60o, and 30o this is because all the angles must add up to 180° and the two given angles are, 30 and 60, so 30+60=90, and 180-90= 90, which tells us that the last triangle is 90o. Also the angle is a right angle, which are always 90o. c) **What is the length of the diagonal?** The length of the diagonal is the square root of 2 = 1.402. I know this because 1²+1²=√of 2. d) **If the side lengths of the squares were 5 then how would your answers to part a and b change?** The angles would stay the same, so they'd still be; C-90°, and B and D would be 45°. The lengths of the diagonal would change to 5²+5²= √50. This is because the side lengths are 5.