(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 i...

Answer Details

(a) Cylindrical pipe, length \(28\text{ m}=2800\text{ cm}\), internal radius \(r=3.5\text{ cm}\), external radius \(R=5\text{ cm}\).

(i) Volume of metal used.

The metal is the hollow shell between the outer and inner cylinders:

\[V_{\text{metal}}=\pi(R^2-r^2)\times L\]\[R^2-r^2=5^2-3.5^2=25-12.25=12.75\text{ cm}^2\]\[V_{\text{metal}}=\frac{22}{7}\times 12.75\times 2800\]\[=\frac{22}{7}\times 35700=22\times 5100=112200\text{ cm}^3\]

The volume of metal is \(112200\text{ cm}^3\).

(ii) Volume of water the pipe holds when full.

This is the inner cylinder's volume:

\[V_{\text{water}}=\pi r^2 L=\frac{22}{7}\times 3.5^2\times 2800\]\[=\frac{22}{7}\times 12.25\times 2800=\frac{22}{7}\times 34300=22\times 4900=107800\text{ cm}^3\]

Convert to litres using \(1\text{ litre}=1000\text{ cm}^3\):

\[\frac{107800}{1000}=107.8\text{ litres}\]

The pipe holds \(107.8\) litres (to 1 d.p.).

(b) Tangent \(MP\) at \(M\), chord \(LN\parallel MP\): show \(\triangle LMN\) is isosceles.

By the tangent-chord (alternate segment) theorem, the angle between tangent \(MP\) and chord \(MN\) equals the angle in the alternate segment standing on \(MN\):

\[\angle PMN=\angle MLN \quad\text{...(1)}\]

Since \(LN\parallel MP\) and \(MN\) is a transversal, alternate angles are equal:

\[\angle PMN=\angle MNL \quad\text{...(2)}\]

From (1) and (2):

\[\angle MLN=\angle MNL\]

In triangle \(LMN\) the base angles at \(L\) and \(N\) are equal, so the sides opposite them are equal, i.e. \(|MN|=|ML|\).

Therefore triangle \(LMN\) is isosceles. (Q.E.D.)