course Mth 152 }yVҼassignment #023
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20:20:38 **** 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 --> Angles FOE and AOB are vertical angles so they are equal. The angle of triangle AOB and two sides are equal to the angle of FOE and two sides so the two triangles are congruent.
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20:20:41 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 --> o.k.
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20:21:39 **** Explain the argument you used to show that the triangles were congruent.
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RESPONSE --> If you use the congruence property of Side Angle Side
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20:31:29 **** 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 --> In triangle ABC angle A = 42 and angle C = 90 42 + 90 = 132 180 - 132 = 48 Angle B = 48 In triangle QRP Angle P = 90 90 * 48 = 4320 = 90R R=48 90/90 = Q/42 = 1 = Q/42 Q = 42
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20:31:34 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 --> o.k.
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20:35:57 **** 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 --> 75/ 25 = 3 3 * 20 = 60 3 * 10 = 30 a = 30 b = 60
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20:36:03 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 --> o.k.
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20:56:12 **** 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=7 c=25 b= a^ + b^ = c^ 49 + b^ = 625 625 - 49 = 576 square root of 576 = 24 b = 24
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20:56:17 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 --> o.k.
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21:03:16 **** 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 --> If the two legs of a right triangle have lengths a and b, and the hypotenuse has length c , the a^2 + b^2 = c^2 This theory was proven in the previous problem by squaring the given leg number and subtracting that number from the squared hypotenuse. The answer was the squared number of the other leg.
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21:03:22 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 --> o.k.
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21:13:51 **** 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 --> m = 5 (m^2 + 1) / 2 = (25 + 1) / 2 = 26/2= 13 (m^2 - 1) / 2 = (25 - 1) / 2 = 24 / 2 = 12 5, 12, 13 5^2 + 12^2 = 13^2 25 + 144 = 169
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21:13:57 ** 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 --> o.k.
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21:16:43 **** How did you verify that your result is indeed a Pythagorean Triple?
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RESPONSE --> a^2 + b^2 = c^2 The letters a,b,and c were replaced with the numbers 5,12 and 13, and the theory was proved.
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21:16:49 Student Answer: The numbers checked out when substituted into the Pythagorean Theorem.
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RESPONSE --> o.k.
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21:47:46 **** 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 --> the break is at unknown height x the hypotenuse is at height 10 - x and has the broken part. A triangle is made by the break, the ground and the vertical side. x, 3 , 10-x 10- x is the hypotenuse x^2 + 3^2 = (10 - x ) ^2 x^2 + 9 = 100 - x^2 LOST
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21:47:57 did you obtain your result?
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RESPONSE --> I didn't finish
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21:48:03 ** 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 --> o.k.
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21:48:41 **** How did the Pythagorean Theorem allow you to solve this problem?
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RESPONSE --> again by using the a^2 b^2 = c^2
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21:48:45 I substituted the numbers into the Pythagorean Theorem.
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RESPONSE --> o.k.
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21:50:47 **** 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 am going to have to work on this one. ?????????????????????????
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21:50:51 it?
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RESPONSE --> o.k.
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21:51:17 ** 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 --> Not sure I understand but I will go back and work on this
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21:51:47 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> No surprises but that last problem.
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21:51:52 I had some trouble with the last 2 problems.
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RESPONSE -->
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