course Mth 152 ÆaèìÕæ¸ß©äÝB©ÇŸøøö‘‘œ¢Í«xFТ¼¤assignment #023
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18:25:29 **** query 9.4.6 ABC, DEF transversed by EOB at rt angles; OB = EO; show triangles ABO and DOF congruent.
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RESPONSE --> BO = OE (given) OB is perpendicular to AC (given) OE is perpendicular to DF (given) AOB = FOE (vertical angles are equal)
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18:25:55 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.
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RESPONSE --> I was headed in the right direction.
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18:27:17 **** Explain the argument you used to show that the triangles were congruent.
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RESPONSE --> I used the fact that vertical angles are equal and the SAS property, which states that if two sides and the included angle of one triangle are equal to two sides and the included angle of a second triangle, then the triangles are congruent.
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18:36:07 **** 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.
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RESPONSE --> Angles C & P, A & Q, B & R are equivalent. For the first triangle, 2 angles are given. Angle A is 42 deg and Angle C is 90 deg. Therefore, Angle P is 90 deg and Angle Q is 42 deg. I can figure the missing angle because the three angles of a triangle = 180 degrees. 48 + 90 + x = 180 138 + x = 180 138 - 138 + x = 180 - 138 x = 42 deg. Angles B & R are 42 deg.
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18:37:44 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.
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RESPONSE --> I think I got the 42 deg and 48 deg mixed up, but I had the correct answer on my work paper.
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18:39:55 **** 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?
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RESPONSE --> Side a = 30 Side b = 60 In comparing the triangles, I can see that the longest side of the smaller triangle is 25. The longest side of the big triangle is 75. This is 3 times different. I multiplied the other sides by 3 to obtain the sides of the larger triangle.
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18:41:18 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.
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RESPONSE --> Correct answer. Good to see it written out mathematically though, as I wasn't sure how to do that.
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18:43:23 **** 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?
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RESPONSE --> a^2 + b^2 = c^2 7^2 + b^2 = 25^2 49 + b^2 = 625 49-49 + b^2 = 625 - 49 b^2 = 576 b = 24 7^2 + 24^2 = 25^2 49 + 576 = 625
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18:43:32 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.
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RESPONSE --> Correct answer.
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18:45:32 **** What does the Pythagorean Theorem say about the triangle as given and how did you use this Theorem to find the length of b?
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RESPONSE --> The Pythagorean Theorem states that the sum of the squares of the lengths of the legs of a triangle are equal to the square of teh hypotenuse of the triangle or a^2 + b^2 = c^2. I inserted the known values into the formula and solved for b.
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18:45:42 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.
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RESPONSE --> Correct answer.
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18:47:42 **** query 9.4.60 m, (m^2 +- 1) / 2 gives Pythagorean Triple **** What Pythagorean Triple is given by m = 5?
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RESPONSE --> 5, 5^2-1/2, 5^2+1/2 5, 12, and 13?
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18:48:05 ** 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. **
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RESPONSE --> Correct answer, but I didn't prove it.
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18:50:04 **** How did you verify that your result is indeed a Pythagorean Triple?
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RESPONSE --> I didn't, but should have verified by finding that a^2 + b^2 = c^2 5^2 + 12^2 = 13^2 25 + 144 = 169
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18:50:15 Student Answer: The numbers checked out when substituted into the Pythagorean Theorem.
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RESPONSE --> Correct answer.
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19:40:27 **** query 9.4.75 10 ft bamboo broken, upper end touches ground 3 ft from stem. **** How high is the break, and how
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RESPONSE --> I am not sure how to solve this problem.
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19:40:33 did you obtain your result?
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RESPONSE --> No
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19:43:46 ** 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. **
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RESPONSE --> I was thinking something along these lines, but hadn't quite gotten it yet.
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19:44:57 **** How did the Pythagorean Theorem allow you to solve this problem?
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RESPONSE --> a^2 + b^2 = c^2 Putting in the known values and solving for x.
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19:45:07 I substituted the numbers into the Pythagorean Theorem.
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RESPONSE --> Correct answer.
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19:50:58 **** query 9.4.84 isosceles triangle perimeter 128 alt 48 **** What is the area of the triangle and how did you find
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RESPONSE --> I'm not sure about this one either, but I'll give it a try. AC = BD Area of a triangle = 1/2bh A = 24 * 48 = 1,152 in^3
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19:51:07 it?
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RESPONSE --> Not sure.
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19:55:44 ** 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
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RESPONSE --> I was way off.
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19:55:59 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Some of those final problems were a little tough for me.
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