Chord Properties and Segments Lengths in Circles

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Presentation transcript:

Chord Properties and Segments Lengths in Circles

If two chords are congruent, then their corresponding arcs are congruent.

Solve for x. 8x – 7 3x + 3 8x – 7 = 3x + 3 x = 2

Find the length of WX.

Find 360 – 100 260 divided by 2 130º

If two chords are congruent, then they are equidistant from the center.

In K, K is the midpoint of RE In K, K is the midpoint of RE. If TY = -3x + 56 and US = 4x, find the length of TY. U T K E x = 8 R S Y TY = 32

If a diameter is perpendicular to a chord, then it also bisects the chord. This results in congruent arcs too. Sometimes, this creates a right triangle & you’ll use Pythagorean Theorem.

IN Q, KL  LZ. If CK = 2x + 3 and CZ = 4x, find x.

MT = 6 AT = 12 In P, if PM  AT, PT = 10, and PM = 8, find AT. P A M

30 Find the length of CE x BD is a radius. CB is a radius. What is the length of the radius? 25 Now double it to find CE. 30 25 x

Find the length of LN. LN = 96 MK and KL are radii. Now double it to find LN. x 50 LN = 96

Segment Lengths in Circles

Go down the chord and multiply Two chords intersect INSIDE the circle Type 1: part part part part Emphasize that the chords are NOT congruent or bisected! Go down the chord and multiply

Example 1: 9 12 6 3 x x 2 2 X = 3 X = 8 x 3 6 2 X = 1

Solve for x. 9 6 x 2 x = 3

Find the length of DB. 8 12 2x 3x A D x = 4 C DB = 20 B

Find the length of AC and DB. x – 4 x 5 C 10 x = 8 AC = 13 DB = 14 B

Sometimes you have to add to get the whole. Two secants intersect OUTSIDE the circle Type 2: Sometimes you have to add to get the whole.

Solve for x. 20 x = 31 7 4 x 7 (20) = (4 + x) 4 140 = 16 + 4x 124 = 4x

x = 11.8 Solve for x. x 5 8 6 6 (6 + 8) = 5 (5 + x) 84 = 25 + 5x

x = 6 Solve for x. 10 x 4 8 x (x + 10) = 8 (8 + 4) x2 +10x = 96

Type 2 (with a twist): Secant and Tangent

x = 36 Solve for x. 242 = 12 (12 + x) 576 = 144 + 12x 432 = 12x x 12

Solve for x. 5 15 x x = 10 x2 = 5 (5 + 15) x2 = 100