I need to find the length of AB. Chord AB and diameter CD bisect at E. DE=13 and EC=4.?

Answer 1

Hi,

There are 3 triangles involved and all are right angled triangles, where the Pythagoras theorem Can be applied as follows.

As triangle CBD is formed by joining the end points of diameter CD with point B on the circle's circumference , it is rt. angled triangle and So, CD^2 = BC^2 + BD^2 (1)

The Triangle BED is also a rt. angled triangle, because the diameter CD can bisect the chord AB, only if it is a perpendicular bisector of AB. Here So, BD^2 = EB^2 + ED^2 (2)

As BC^2 = CE^2 + EB^2 (3)

EB^2 = BC^2 - CE^2 From (3)
EB^2 = CD^2 - BD^2 - CE^2 From (1)
EB^2 = CD^2 - (EB^2 + ED^2 ) - CE^2 From (2)
2EB^2 = CD^2 - (ED^2 + CE^2)
EB^2 = 1/2{CD^2 - (ED^2 + CE^2)}
AB = 2EB = 2sqrt[1/2{CD^2 - (ED^2 + CE^2)}]
AB = 2sqrt[1/2{17^2 - (13^2 + 4^2)}] = 14.42

If the formula in the last line is already given in your lesson use it, instead of deriving from first principles, as I have done.

Good luck

Answer 2

Let O be the center of the circle. Then OC = OB = 6.5 and OE = 6.5 - 4 = 2.5. Now we can find half the length of chord AB, which is EB, using the Pythagorean Theorem:

(EB)² = (OB)² - (OE)²
(EB)² = (6.5)² - (2.5)²
(EB)² = 42.25 - 6.25
(EB)² = 36
EB = 6

Since AB = 2EB, AB = 2 (6) = 12.

The length of chord AB is 12.

Answer 3

what's that? I'm only 10 and i want to get points so that y i answering.