Question

Two parallel chords through a circle, one 16cm and one 30cm, are 23cm apart
What is the radius of the circle?

Answer 1

Question #30840

Two parallel chords through a circle, one 16cm and one 30cm, are 23cm apart. What is the radius of the circle?

Solution.

Let AB=16, CD=30 and MK=23. Observe that $MK \perp AB$ and $MK \perp CD$ . The triangles $\triangle AOB$ and $\triangle COD$ are isosceles, since $AO = BO = OC = OD = R$ . It follows that $MO$ and $KO$ are medians of $\triangle AOB$ and $\triangle COD$ , respectively and so $AM = MB = \frac{1}{2} AB = 8$ and $CK = KD = \frac{1}{2} CD = 15$ . For the right triangles $\triangle AMO, \triangle COK$ , we have

$$O M = \sqrt {R ^ {2} - A M ^ {2}} = \sqrt {R ^ {2} - 8 ^ {2}},$$ $$O K = \sqrt {R ^ {2} - C K ^ {2}} = \sqrt {R ^ {2} - 1 5 ^ {2}}.$$

Since $OM + OK = MK = 23$ , then $\sqrt{R^2 - 8^2} + \sqrt{R^2 - 15^2} = 23$ .

$$R ^ {2} - 6 4 + 2 \sqrt {(R ^ {2} - 6 4) (R ^ {2} - 2 2 5)} + R ^ {2} - 2 2 5 = 5 2 9,$$ $$2 R ^ {2} + 2 \sqrt {(R ^ {2} - 6 4) (R ^ {2} - 2 2 5)} = 8 1 8,$$ $$\sqrt {(R ^ {2} - 6 4) (R ^ {2} - 2 2 5)} = 4 0 9 \cdot R ^ {2},$$ $$R ^ {4} - 2 8 9 R ^ {2} + 6 4 \cdot 2 2 5 = 4 0 9 ^ {2} - 2 \cdot 4 0 9 R ^ {2} + R ^ {4}.$$

Thus, we obtain the equation

$$5 2 9 R ^ {2} = 1 5 2 8 8 1,$$ $$R ^ {2} = 2 8 9.$$

Finally, $R = 17$

Answer. $R = 17$