+279 I Assume that Maybe I can Copy and Paste the Links Coerrectly Now. But Just incase, Page 817 :)
Numers 2, 3, 4, 5, 7, 9, If You Can, and Only if You Want To! :) I am Very Grateful of All Help.
+128578 2) x = A radial line that is perpendicular to a chord bisects that chord. Thus (1/2) of 12 = 6. So we have a right triangle with two "legs" 4 and 6 and the radius "x" forms the hypoteneuse....so...... √(^6^2 + 4^2) = √(52) = 2√13
3) z° = 60° We have a 30-60-90 triangle.......the angle that intercepts arc measuring "z°" is a 30°
angle, so it intercepts an arc of twice its measure!!
4) x = 10.5 Products of segments of intersecting chords in a circle...... (7*3) = (2 * x)...
21 = 2x........x = 10.5
5) To find x°, we have to use the theorem that says that the angle formed by drawing two secants from a point outside a circle = (1/2) (the difference of the measures of the arcs they intercept)
40° = (1/2)(106° -x) multiply both sides by 2
80° =106° - x .... add x to both sides and subtract 80 from both sides
x = 106 - 80 = 26°
To find y° ....its measure is just 1/2 of the arc that reamains after we subtract the two known arcs from 360....thus ... 360 - 106 - 145 = 109° ....and 1/2 of that is just 54.5°
7) I had to refresh my memory on this one.......it's the "secant-secant" rule that says that a whole secant times its external part = the other whole secant times its external part....I'm assuming that the "20" is the whole length of that one secant...so we have
20*10 = 12*x
200 = 12*x
x = 200/12 = 100/6 = 50/3 = 16 + 2/3 or 16.66
9) The measure of arc AB is 65° as well......(I'm assuming that either 1/2 of the length of both bisected chords is 8, or that a 'partial raidus' of length 8 bisects the chords...either way, they're both congruent chords, interceptig arcs with equal measures)
These problems aren't well-labeled, IMHO!!!
+128578 2) x = A radial line that is perpendicular to a chord bisects that chord. Thus (1/2) of 12 = 6. So we have a right triangle with two "legs" 4 and 6 and the radius "x" forms the hypoteneuse....so...... √(^6^2 + 4^2) = √(52) = 2√13
3) z° = 60° We have a 30-60-90 triangle.......the angle that intercepts arc measuring "z°" is a 30°
angle, so it intercepts an arc of twice its measure!!
4) x = 10.5 Products of segments of intersecting chords in a circle...... (7*3) = (2 * x)...
21 = 2x........x = 10.5
5) To find x°, we have to use the theorem that says that the angle formed by drawing two secants from a point outside a circle = (1/2) (the difference of the measures of the arcs they intercept)
40° = (1/2)(106° -x) multiply both sides by 2
80° =106° - x .... add x to both sides and subtract 80 from both sides
x = 106 - 80 = 26°
To find y° ....its measure is just 1/2 of the arc that reamains after we subtract the two known arcs from 360....thus ... 360 - 106 - 145 = 109° ....and 1/2 of that is just 54.5°
7) I had to refresh my memory on this one.......it's the "secant-secant" rule that says that a whole secant times its external part = the other whole secant times its external part....I'm assuming that the "20" is the whole length of that one secant...so we have
20*10 = 12*x
200 = 12*x
x = 200/12 = 100/6 = 50/3 = 16 + 2/3 or 16.66
9) The measure of arc AB is 65° as well......(I'm assuming that either 1/2 of the length of both bisected chords is 8, or that a 'partial raidus' of length 8 bisects the chords...either way, they're both congruent chords, interceptig arcs with equal measures)
These problems aren't well-labeled, IMHO!!!
+279 What Does "IMHO" Mean? :)
Sorry, I did Not Create The Problems. And I Apologize for any Trouble it Caused?
Thank You for All the Answers and Thorough Explanation! :)
+128578 natasza...I've made a change to one of these answers ....No. 5....I saw what it wanted....I realized that my answer to the y° part was incorrect and I also saw that it wanted an answer for "x°"...look at it again!!
IMHO = In My Humble Opinion
NO!!...I'm not blaming you.....it's just hard to determine some of the labellings...not your fault !!!
