Geometry problem 6 by Dhaval Furia

Geometry Level pending

In a circle of radius $11$ cm, $CD$ is a diameter and $AB$ is a chord of length $20.5$ cm. If $AB$ and $CD$ intersect at a point $E$ inside the circle and $CE$ has length $7$ cm, then the difference of the lengths of $BE$ and $AE$ , in cm, is $_____$

0.5

3.5

2.5

1.5

1 solution

Ron Gallagher
Jun 12, 2020

Let O be the center of the circle. Draw radii OA and OB, both of which have length 11. We see angles EAO and EBO are congruent (they are opposite congruent sides). Since OC is a radius, it has length 11. We are given CE has length 7. Hence, OE has length 4. Let x = AE and y = EB. Apply the Law of cosines to triangle OEA to get

16 = x^2 + 121 - 22 cos(EAO)

Apply the Law of Cosines to OEB to get

16 = y^2 + 121 - 22 cos(EBO)

Subtracting these two equations and using the fact that y^2 - x^2 = 20.5*(y-x) yields

Cos(EBO) = 20.5/22

Lastly, apply the Law of Cosines to one final time to triangle EBO to get

16 = y^2 + 121 - 2 y *20.5/22

Solving this equation yields y = 10.5 or y = 10. Since x + y = 20.5, this means x = 10 or x = 10.5. In either event, the difference between x and y is 1/2

Geometry problem 6 by Dhaval Furia

1 solution

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