A circle with center B and radius 1cm 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 and 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.

8 cm

Step-by-step 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

can you rephrase the question ? it is unclear : -)

step-by-step explanation:

the answer is two. hope this !