Answer:
8 cm
Explanation:
A circle with center B and radius 13 cm has three distinct points, F, D and E, on its circumference so that BF and BE and D is on the minor arc FE. Point A is on BF so that DA is perpendicular to BF. The point C is on BE so that ABCD is a rectangle and the distance from C to E is 1cm. Determine the distance from A to F.
Solution:
Since the radius of the circle is 13 cm, hence:
BF = BD = BE = 13 cm.
For line segment BE; BE = BC + CE
BE = BC + CE
13 = BC + 1
BC = 13 - 1
BC = 12 cm
AD = BC = 12 cm (opposite sides of a rectangle are equal to each other)
Triangle ABD, is a right angled triangle because ∠B = 90°. We can use Pythagoras theorem to find AB:
BD² = AB² + AD²
13² = AB² + 12²
169 = AB² + 144
AB² = 169 - 144
AB² = 25
AB = √25 = 5
For line segment BF, we can use the line segment addition postulate to find the distance from A to F. Let x be the distance from A to F:
BF = AB + x
x = BF - AB
x = 13 - 5
x = 8 cm