Question

A semi circular sheet of metal of diameter 28cm is bent into an open conical cup.What is the approximate depth of the curve?

Answer 1

Answer on Question #77140 – Math – Geometry

Question

A semicircular sheet of metal of diameter $28\mathrm{cm}$ is bent into an open conical cup. What is the approximate depth of the curve?

Solution

$$\begin{array}{l} D = 28\mathrm{cm} \Rightarrow r = \frac{1}{2} D = \frac{1}{2} 28 = 14\mathrm{cm}. \\ \text{Circumference of semicircle} = \pi * r \Rightarrow \text{circumference of semicircle} = \\ = \frac{22}{7} * 14 = 44\mathrm{cm}. \end{array}$$

Circumference of base of cone = circumference of semicircle :

$$2\pi R = 44$$ $$R = \frac{44}{2\pi} = \frac{44}{2 * \frac{22}{7}} = \frac{44 * 7}{44} = 7\mathrm{cm}.$$

By the Pythagorean Theorem:

(radius of the semicircular sheet)$^2$ = (depth of the curve)$^2$ + $R^2$

$$14^2 = (\text{depth of the curve})^2 + 7^2$$ $$196 = (\text{depth of the curve})^2 + 49$$ $$(\text{depth of the curve})^2 = 147$$ $$\text{depth of the curve} = \sqrt{147} = 7\sqrt{3} \approx 12.12\mathrm{cm}$$

Answer: $12.12\mathrm{cm}$.

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