Answer:
1. = 3√10 2.
= 5 3.
= 6√2
Step-by-step explanation:
Use <u>Intersecting Secant-Tangent Theorem</u>:
If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.
1. In this circle, DB is a tangent and DA is a secant. They intersect at point D.
So DB² = DC x DA
² = 5 x (5 + 13)
² = 90
= √90 = 3√10
2. In this circle, FE is a tangent and FH is a secant. They intersect at point F.
So FE² = FG x FH
10² = x 20
100 = 20
= 5
3. In this circle, NM is a tangent and NP is a secant. They intersect at point N.
So NM² = NO x NP
² = 4 x (4 + 14)
² = 72
= √72 = 6√2