Two Parallel Chords in a Circular Plate
A metalworker is cutting a circular plate to hold two parallel slots. The longer slot is a chord of length 24 cm and the shorter slot is a chord of length 10 cm. The perpendicular distance between the two chords is 7 cm. Both chords must lie entirely inside the plate, and the centre of the plate lies on the perpendicular bisector of each chord. A helper says the radius must be 13 cm. Another helper argues that the chords could be on opposite sides of the centre, which would change the radius. The workshop also has a standard plate of radius 12 cm. The metalworker needs to know whether the standard plate is large enough and whether the 13 cm answer is the only possible radius under the given measurements.
1. Let d1 be the distance from the centre to the 24 cm chord and d2 the distance from the centre to the 10 cm chord. For chords on the same side of the centre, set up equations and find the radius R.
Read the worked answer
The perpendicular from the centre to a chord bisects the chord. So the 24 cm chord has half-length 12 cm, and the 10 cm chord has half-length 5 cm.
By Pythagoras, R² = d1² + 12² and R² = d2² + 5². Equating gives d1² + 144 = d2² + 25, so d2² − d1² = 119.
The longer chord is closer to the centre, so d1 < d2. If the chords are on the same side of the centre, their separation is d2 − d1 = 7. Factor the difference of squares: (d2 − d1)(d2 + d1) = 119. Substituting d2 − d1 = 7 gives 7(d2 + d1) = 119, so d2 + d1 = 17. Solving with d2 − d1 = 7 gives d2 = 12 and d1 = 5.
Then R² = 5² + 12² = 25 + 144 = 169, so R = 13 cm.
2. Test the second helper’s claim that the chords could be on opposite sides of the centre with separation 7 cm. Does this give a valid radius?
Read the worked answer
If the chords were on opposite sides of the centre, the distances would satisfy d1 + d2 = 7. The same chord calculation gives d2² − d1² = 119, so (d2 − d1)(d1 + d2) = 119. Substituting d1 + d2 = 7 gives 7(d2 − d1) = 119, so d2 − d1 = 17.
Now solve d2 + d1 = 7 and d2 − d1 = 17. Adding gives 2d2 = 24, so d2 = 12, and then d1 = −5. A distance from the centre cannot be negative, so this configuration is impossible. Therefore the chords cannot be 7 cm apart on opposite sides of the centre for these chord lengths. The second helper’s alternative does not produce a different radius; the 13 cm result is forced.
For comparison, if the separation were 17 cm, opposite-side placement would give d1 = 5 and d2 = 12, still with R = 13 cm.
3. Is the standard 12 cm plate large enough? State the minimum radius and describe where the centre must lie relative to the two chords.
Read the worked answer
No. The calculation gives R = 13 cm, so a plate of radius 12 cm is too small; the outer chord would not fit entirely inside the plate.
The centre must lie 5 cm from the 24 cm chord and 12 cm from the 10 cm chord, on the same side of both chords in the valid configuration. The 10 cm chord is the farther one from the centre. The minimum radius is therefore 13 cm, and the standard 12 cm plate cannot do the job without cutting off part of the outer slot.