Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

## Tuesday, November 6, 2018

### Geometry Problem 1398: Isosceles Triangle, Circle, Radius, Perpendicular, Measurement

Labels:
circle,
geometry problem,
isosceles,
measurement,
perpendicular,
radius,
tangent,
triangle

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AB = 2^2/1 = 4 = BC

ReplyDeletex = 8^2/BC = 16

But there appears to be something wrong since in the diagram AB + BC < AC

Sumith Peiris

Moratuwa

Sri Lanka

Thanks for your comment. Problem has been updated.

ReplyDeleteProof of updated version, same method

ReplyDeleteAB = d^2/e and BC = f^2/x

Since AB = BC the result follows

Sumith Peiris

Moratuwa

Sri Lanka

f2 =BC*CF = BC*x

ReplyDeleted2 = AE*AB =BC*e

f2/d2 =x/e

x = e * (f2)/(d2)

Florin Popa , Comanesti, Romania

Tangency rule: f^2=x*(x-BF) and d^2=e*(e+EB)

ReplyDeleteBAC is isoscele, therefore: d^2=e*(x-BF)

Dividing side by side:

f^2/d^2=(x*(x-BF))/(e*(x-BF))

Result is: x=e*(f^2/d^2)