course Mth 152 ?wz????????e????assignment #023
......!!!!!!!!...................................
17:37:15 **** query 9.4.6 ABC, DEF transversed by EOB at rt angles; OB = EO; show triangles ABO and DOF congruent.
......!!!!!!!!...................................
RESPONSE --> ?
.................................................
......!!!!!!!!...................................
17:37:18 SAS: Angle AOB and Angle FOE are equal because they are vertical angles, so we have 2 sides and the included angle of triangle AOB equal, respectively, to 2 sides and the included angle of triangle FOE. Thus, the Side-Angle-Side property holds that triangle AOB is congruent to triangle FOE.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:37:25 **** Explain the argument you used to show that the triangles were congruent.
......!!!!!!!!...................................
RESPONSE --> ?
.................................................
......!!!!!!!!...................................
17:38:57 **** query 9.4.18 ACB and QPR similar triangles, C and P rt angles, A=42 deg **** List the measures of the three angles of each triangle and explain how you obtained each.
......!!!!!!!!...................................
RESPONSE --> B = 48, P = 90, Q = 42, R = 48
.................................................
......!!!!!!!!...................................
17:39:05 It is given that Angle A = 42 deg. and Angle C = 90 deg. Since all three angles must add up to equal 180 then Angle B = 48 deg. In the second triangle, Angle P must equal 90 deg. since it is a right angle. To find Angle R, 90(48) = 90R sp 4320 = 90R and 48 = R Angle R = 48 deg. To find Angle Q, 90/90 = Q/42 Q = 42 Angle Q = 42 deg.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:41:33 **** query 9.4.24 similar triangles, corresp sides a, b, 75; 10, 20, 25 **** What are the lengths of sides a and b and how did you obtain each?
......!!!!!!!!...................................
RESPONSE --> b/20 = 75/25 b = 60 a/10 = 75/25 a = 30
.................................................
......!!!!!!!!...................................
17:41:36 To find a, 75 (10) = 25a 750 = 25a a= 30 To find b, 75/25 = b/20 1500/25 = 25b/25 so b = 60. a = 30, b = 60 and c = 75. These values are triple the values of the similar triangle.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:43:13 **** query 9.4.42 rt triangle a = 7, c = 25, find b **** What is the length of side b and how did you obtain it?
......!!!!!!!!...................................
RESPONSE --> 7^2 + b^2 = 25^2 b = 24
.................................................
......!!!!!!!!...................................
17:43:46 By the Pythagorean Theorem a^2 + b^2 = c^2. So we have 49 + b^2 = 625 Subtract 49 from both sides to get b^2 = 576. Take the square root of both sides to get b = 24.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:44:20 **** What does the Pythagorean Theorem say about the triangle as given and how did you use this Theorem to find the length of b?
......!!!!!!!!...................................
RESPONSE --> a^2 + b^2 = c^2
.................................................
......!!!!!!!!...................................
17:44:24 Student Response: It says the sum of the squares of the lengths of the legs is equal to the square of the hypotenuse. I showed that this is true in the previous problem. I squared the legs and they equaled the hyppotenuse squared.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:44:45 **** query 9.4.60 m, (m^2 +- 1) / 2 gives Pythagorean Triple **** What Pythagorean Triple is given by m = 5?
......!!!!!!!!...................................
RESPONSE --> 5, 12, 13
.................................................
......!!!!!!!!...................................
17:44:54 ** If m = 5 then (m^2 + 1) / 2 = (5^2 + 1 ) / 2 = 26 / 2 = 13 (m^2 - 1) / 2 = (5^2 - 1 ) / 2 = 24 / 2 = 12 So the Pythagorean triple is 5, 12, 13. We can verify this: 5^2 + 12^2 should equal 13^2. 5^2 + 12^2 = 25 + 144 = 169. 13^2 = 169. The two expressions are equal so this is indeed a Pythagorean triple. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:46:06 **** How did you verify that your result is indeed a Pythagorean Triple?
......!!!!!!!!...................................
RESPONSE --> Test it using: a^2 + b^2 = c^2
.................................................
......!!!!!!!!...................................
17:46:11 Student Answer: The numbers checked out when substituted into the Pythagorean Theorem.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:46:33 **** query 9.4.75 10 ft bamboo broken, upper end touches ground 3 ft from stem. **** How high is the break, and how
......!!!!!!!!...................................
RESPONSE --> 4.55 ft
.................................................
......!!!!!!!!...................................
17:46:37 did you obtain your result?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:46:44 ** If the break is at height x then the hypotenuse, consisting of the broken part, is at height 10 - x. The triangle formed by the vertical side, the break and the ground therefore has legs x and 3 and hypotenuse 10-x. So we have x^2 + 3^2 = (10-x)^2. Squaring the 3 and the right-hand side: x^2 + 9 = 100 - 20 x + x^2. Subtracting x^2 from both sides 9 = 100 - 20 x so that -20 x = -91 and x = 4.55. The break occurs at height 4.55 ft and the broken part has length 10 - 4.55 = 5.45, or 5.45 feet. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:47:07 **** How did the Pythagorean Theorem allow you to solve this problem?
......!!!!!!!!...................................
RESPONSE --> a^2 + b^2 = c^2
.................................................
......!!!!!!!!...................................
17:47:11 I substituted the numbers into the Pythagorean Theorem.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:47:28 **** query 9.4.84 isosceles triangle perimeter 128 alt 48 **** What is the area of the triangle and how did you find
......!!!!!!!!...................................
RESPONSE --> ?
.................................................
......!!!!!!!!...................................
17:47:31 it?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:48:51 ** This problem is algebraically demanding. Your text might have a slicker way to do this, but the following works: If the equal sides are x then the base is 128 - 2 x. The altitude forms a right triangle with half the base and one of the equal sides. The sides of this right triangle are therefore 48, 1/2 (128 - 2x) = 64 - x, and x. The right angle is formed between base and altitude so x is the hypotenuse. We therefore have 48^2 + (64 - x)^2 = x^2 so that 48^2 + (64 - x) ( 64 - x) = x^2 or 48^2 + 64 ( 64-x) - x(64 - x) = x^2 or 48^2 + 64^2 - 64 x - 64 x + x^2 = x^2 or 48^2 + 64^2 - 128 x + x^2 = x^2. Subtracting x^2 from both sides we get 48^2 + 64^2 - 128 x = 0. Adding 128 x to both sides we get 48^2 + 64^2 = 128 x. Multiplying both sides by 1/128 get have (48^2 + 64^2) / 128 = x. Evaluating this expression we end up with x = 50. The base of the triangle is therefore 128 - 2x = 128 - 2 * 50 = 128 - 100 = 28. So its area is 1/2 b h = 1/2 * 28 * 48 = 672. ** DRV
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:49:15 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> PROOFS! Eghh
.................................................
......!!!!!!!!...................................
17:49:20 I had some trouble with the last 2 problems.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................