ALL YOU NEED TO DO IS DRAW THIS Circle ω1 with center K of radius 4 and circle ω2 of radius 6 intersect at points W and U. If the incenter of 4KW U lies on circle ω2, find the length of W U. (Note: The incenter of a triangle is the intersection of the angle bisectors of the angles of the triangle)

  WU = (14√13)/13 ≈ 6.6564

Step-by-step explanation:

Call the incenter of ∆KWU point A. Call the center of circle ω2 point B.

Then ∠KWA has half the measure of arc WA. ∠AWU is congruent to ∠KWA, so also has half the arc measure. That is, ∠KWU has the same measure as arc WA and ∠KBW.

KB is a perpendicular bisector of chord WU, so ∆KWB is a right triangle, of which WU is twice the altitude to base KB.

The length of KB can be found several ways. One of them is to use the Pythagorean theorem:

  KB² = KW² +WB² = 4² +6² = 52

  KB = √52 = 2√13

The area of triangle KWB is ...

  area ∆KWB = (1/2)KW·WB = (1/2)(4)(6) = 12 . . . . square units

Using KB as the base in the area calculation, we have ...

  area ∆KWB = (1/2)(KB)(WU/2)

  12 = KB·WU/4

  WU = 48/KB = 48/(2√13) = 24/√13

  WU = (24√13)/13 ≈ 6.6564

70

step-by-step explanation:

2*5=10

4*15=60

60+10=

70