(a) A cylindrical pipe is 28 metres long. Its internal radius is 3.5 cm and external radius 5 cm. Calaulate : (i) the volume, in cm\(^{3}\), of metal used in making the pipe ; (ii) the volume of water in litres that the pipe can hold when full, correct to 1 decimal place. [Take \(\pi = \frac{22}{7}\)]
(b) In the diagram, MP is a tangent to the circle LMN at M. If the chord LN is parallel to MP, show that the triangle LMN is isosceles.
(b) In the diagram, MP is a tangent to the circle LMN at M. If the chord LN is parallel to MP, show that the triangle LMN is isosceles.
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Correct Answer: Option n
Explanation:
(a)(i) Volume of outer pipe = \(\pi r^{2} h\)
= \(\frac{22}{7} \times (5.0)^{2} \times 28 \times 100\)
Volume of inner pipe = \(\frac{22}{7} \times (3.5)^{2} \times 28 \times 100\)
Volume of metal used = Volume of outer pipe - Volume of inner pipe
= \(22 \times 400 \times (5.0^{2} - 3.5^{2})\)
= \(22 \times 400 \times (5.0 - 3.5) \times (5.0 + 3.5)\)
= \(112,200 cm^{3}\)
(ii) Vol of water the pipe can hold = vol of inner pipe
= \(\frac{22}{7} \times (3.5)^{2} \times 28 \times 100\)
= \(107,800 cm^{3}\)
= 107.8 litres.
(b) < PMN = < LNM (alternate angles)
< PMN = < LMN (angles in alternate segment)
\(\therefore\) < LNM = < NLM, then \(\Delta\) LMN is isosceles.
(a)(i) Volume of outer pipe = \(\pi r^{2} h\)
= \(\frac{22}{7} \times (5.0)^{2} \times 28 \times 100\)
Volume of inner pipe = \(\frac{22}{7} \times (3.5)^{2} \times 28 \times 100\)
Volume of metal used = Volume of outer pipe - Volume of inner pipe
= \(22 \times 400 \times (5.0^{2} - 3.5^{2})\)
= \(22 \times 400 \times (5.0 - 3.5) \times (5.0 + 3.5)\)
= \(112,200 cm^{3}\)
(ii) Vol of water the pipe can hold = vol of inner pipe
= \(\frac{22}{7} \times (3.5)^{2} \times 28 \times 100\)
= \(107,800 cm^{3}\)
= 107.8 litres.
(b) < PMN = < LNM (alternate angles)
< PMN = < LMN (angles in alternate segment)
\(\therefore\) < LNM = < NLM, then \(\Delta\) LMN is isosceles.