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# How do you solve these three geometry problems?

Topic: How do you solve these three geometry problems?
July 18, 2019 / By Korbin
Question: Can someone help me do these problems. They make no sense to me and the teacher didn't explain this part of the homework.

## Best Answers: How do you solve these three geometry problems?

Indigo | 1 day ago
Keep these facts in mind. i. A straight angle has 180°. ii. An inscribed angle is one-half the subtended arc. ________ 1. 8a + 10 = (1/2)260 = 130 || by ii 8a = 120 a = 15° 2. (8a + 10) + (b² + 10) = 180 b² + 10 = 180 - (8a + 10) = 180 - 130 = 50 b² = 40 b = √40 = 2√10° I'll leave the rest to you.
👍 170 | 👎 1
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Originally Answered: Geometry: Solve the following problems?
1. The areas of two similar triangles are 16 and 25. What is the ratio of their corresponding sides? sqrt(16/25) = 4/5 2. The ratio of similitude of two triangles is 3:4. If the perimeter of the smaller triangle is 48 cm, what is the perimeter of the bigger triangle? 48(4/3) =64 cm 3. The shortest side of a triangle is 15 cm while the shortest side of a triangle similar to it is 25 cm. If the perimeter of the smaller triangle is 63 cm, what is the perimeter of the bigger triangle? 63(25/15) = 105 cm
Originally Answered: Geometry: Solve the following problems?
1. 4:5 since areas are proportional to square of lengths 2. 64 3/4 = 48/x 3x=192 x=64 3. 105. 3/5 = 63/x 3x = 315 x=105

Everett
31: The tangent line is a 180 degree angle. the top angle is 180 -(8a+10) degrees The bottom angle is 180-(b2+10) degrees The angles could be solved using simultaneous equations, but there's another way: The top part of the circle is 260 degrees, therefore the bottom part of the circle must include 360-260 degrees. If we look at the circle, a 90 degree angle from the tangent line would bisect the circle and each half would have 180 degrees. Therefore we can say that for every two degrees in the circle that the chord line transsects, would equal one degree in the lower tangent line angle. So, (360 - 260) / 2 = ? degrees? => (360 - 260) / 2 = b^2 + 10 => (360 - 260) / 2 - 10 = b^2 => b = sqrt ((360 - 260) /2 -10) Then solve for a 8a + 10 = 180 - (b^2 +10) => 8a = 180 -10 - b^2 - 10 => 8a = 160 - b^2 => a = 20 - (b^2) / 8 The other problems are handled similarly. Northstar provided the correct answers for question 31. Good luck!
👍 70 | 👎 -3

Cornelius
2) a million. Merwin 2. Bryan 3. Jed 4. JB 5. Allen First step: chanced on first place. JB did no longer are available first, and neither did Brian. considering that Allen got here a place after JB, he ought to no longer have been first, same with Jed who got here 2 places after somebody, for that reason leaving in basic terms Merwin to be first. 2nd step: Jed replaced into 2 places under Merwin, who replaced into in first, so Jed is in 0.33. by using fact the 1st and 0.33 spots crammed, in basic terms the fourth and 5th spots stay as consecutive open spots, permitting Allen and JB to take 5th and fourth respectively. finally the final open spot is going to Bryan via potential of removing.
👍 70 | 👎 -7

Originally Answered: How do solve these Geometry problems?
I'm going to assume that DE is across the middle of the triangle. You would do this: CA is 8. and CD is 2. There, now you have the ratio. You can do CA divided by CD = CE divided by CB. I forget what theorem that is sorry. anyway, you do 8/2 = x/3. Now it is just an easy equation. Solve it, and you will get x=12. the length is 12.

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