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

good luck!

cutie~

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^