#$&* course Mth 152 Dec 5 - 11:20pm 023. ``q Query 23
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Given Solution: `a SAS: Angle AOB and Angle FOE are equal because they are vertical angles, so we have 2 sides and the included angle of triangle AOB equal, respectively, to 2 sides and the included angle of triangle FOE. Thus, the Side-Angle-Side property holds that triangle AOB is congruent to triangle FOE. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* Question: `q Query 9.4.18 ACB and QPR similar triangles, C and P rt angles, A=42 deg **** List the measures of the three angles of each triangle and explain how you obtained each. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ACB and QPR are similar triangles C and P = right angles A = 42 degrees ACB We know A = 42 C has to equal 90 degrees since it’s a right angle. If we have two angles then we can guess the third by adding the two angles and subtracting them from 180 degrees. So the missing angle is 48 degrees. QPR P = 90 degrees Since QPR and ACB are similar triangles we can try to find the missing angles by relating the other triangle’s angles. 90 (48) = 90 4320 = 90 The missing angle R = 48 Now since we know what two of the angles are we can add them and then subtract them by 180. P = 42 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* Question: `q 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: PG 574 One triangle has the sides 10, 20, 25 The similar triangle has sides a, b, and 75 We need to find a and b To find side A: 75 (10) = 25 750 = 25 30 To find side B: 75/25 = b/20 1500/25 = 25b/25 b = 60 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a To find a, 75 (10) = 25a 750 = 25a a= 30 To find b, 75/25 = b/20 1500/25 = 25b/25 so b = 60. a = 30, b = 60 and c = 75. These values are triple the values of the similar triangle. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* Question: `q Query 9.4.42 rt triangle a = 7, c = 25, find b **** What is the length of side b and how did you obtain it? **** What does the Pythagorean Theorem say about the triangle as given and how did you use this Theorem to find the length of b? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Right triangle. A=7/C=25/ We need to find B Since we need to only know one side we can use the Pythagorean Theory. 49+ b^2 = 625 b^2 = 576 b =24 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* Question: `q Query 9.4.60 m, (m^2 +- 1) / 2 gives Pythagorean Triple **** What Pythagorean Triple is given by m = 5? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: m, m^2-1/2, m^2+1/2 m = 5 (m^2+1)/2= (5^2 +1)/2 = 2=26 2=13 (m^2 - 1) / 2 = (5^2 - 1 ) / 2 = 24 / 2 = 12 5, 13, and 12 Check: 5^2 + 12^2 = 25 + 144 = 169 13^2 = 169 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a ** If m = 5 then (m^2 + 1) / 2 = (5^2 + 1 ) / 2 = 26 / 2 = 13 (m^2 - 1) / 2 = (5^2 - 1 ) / 2 = 24 / 2 = 12 So the Pythagorean triple is 5, 12, 13. We can verify this: 5^2 + 12^2 should equal 13^2. 5^2 + 12^2 = 25 + 144 = 169. 13^2 = 169. The two expressions are equal so this is indeed a Pythagorean triple. ** **** How did you verify that your result is indeed a Pythagorean Triple? Student Answer: The numbers checked out when substituted into the Pythagorean Theorem. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* Question: `q Query 9.4.75 10 ft bamboo broken, upper end touches ground 3 ft from stem. **** How high is the break, and how did you obtain your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Query 9.4.84 isosceles triangle perimeter 128 alt 48 **** What is the area of the triangle and how did you find it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The perimeter of the isosceles triangle ABC (with AB=BC) is 128 in. The altitude BD is 48in. What is the area of ABC? confidence rating #$&*: / ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a ** This problem is algebraically demanding. Your text might have a slicker way to do this, but the following works: If the equal sides are x then the base is 128 - 2 x. The altitude forms a right triangle with half the base and one of the equal sides. The sides of this right triangle are therefore 48, 1/2 (128 - 2x) = 64 - x, and x. The right angle is formed between base and altitude so x is the hypotenuse. We therefore have 48^2 + (64 - x)^2 = x^2 so that 48^2 + (64 - x) ( 64 - x) = x^2 or 48^2 + 64 ( 64-x) - x(64 - x) = x^2 or 48^2 + 64^2 - 64 x - 64 x + x^2 = x^2 or 48^2 + 64^2 - 128 x + x^2 = x^2. Subtracting x^2 from both sides we get 48^2 + 64^2 - 128 x = 0. Adding 128 x to both sides we get 48^2 + 64^2 = 128 x. Multiplying both sides by 1/128 get have (48^2 + 64^2) / 128 = x. Evaluating this expression we end up with x = 50. The base of the triangle is therefore 128 - 2x = 128 - 2 * 50 = 128 - 100 = 28. So its area is 1/2 b h = 1/2 * 28 * 48 = 672. ** DRV &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This is very complicated. I follow and understand it when I read it but not really. I couldn’t of written it myself. ------------------------------------------------ Self-critique Rating: 1 ********************************************* Question: `q Query 9.4.84 isosceles triangle perimeter 128 alt 48 **** What is the area of the triangle and how did you find it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The perimeter of the isosceles triangle ABC (with AB=BC) is 128 in. The altitude BD is 48in. What is the area of ABC? confidence rating #$&*: / ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a ** 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This is very complicated. I follow and understand it when I read it but not really. I couldn’t of written it myself. ------------------------------------------------ Self-critique Rating: 1 #*&!