Mathematics 10 — Exercise Review · Q2

Full question

A clock radius is 6 cm and its 2-to-10 chord is given as 10 cm. Find the perpendicular centre distance using those dimensions.

Step-by-step solution

1. Draw the perpendicular from the centre O to the chord AB, meeting at M.

2. A perpendicular from the centre bisects a chord, so AM=MB=AB/2.

3. Divide the full chord length by 2.

   $$\mathrm{MB}=\frac{10}{2}$$

4. Calculate the half-chord length.

   $$\mathrm{MB}=5$$

5. The radius is the hypotenuse. Rearrange r²=d²+(L/2)² to d²=r²−(L/2)².

   $$\sqrt{6^2-5^2}$$

6. Calculate this operation before combining the remaining terms.

   $$6^2=36$$

7. Calculate this operation before combining the remaining terms.

   $$5^2=25$$

8. Calculate this operation before combining the remaining terms.

   $$\left(-1\right)\times 25=-25$$

9. Calculate this operation before combining the remaining terms.

   $$36-25=11$$

10. Use the positive value because lengths and distances are nonnegative.

    $$\sqrt{11}$$

Answer

The final answer is shown below.