(a) A cylindrical well of radius 1 metre is dug out to a depth of 8 metres. (i) calculate, in m\(^{3}\), the volume of soil dug out ; (ii) if the soil is us...

Answer Details

(a)(i) Volume of soil dug from the well.

The well is a cylinder of radius \(r=1\text{ m}\) and depth \(h=8\text{ m}\).

\[V=\pi r^2 h=\frac{22}{7}\times 1^2\times 8=\frac{176}{7}=25.14\text{ m}^3\]

The volume of soil dug out is \(25.14\text{ m}^3\) (to 2 d.p.).

(a)(ii) Thickness of the new layer of soil.

The soil is spread over a rectangular floor \(4\text{ m}\times 12\text{ m}\).

\[\text{Floor area}=4\times 12=48\text{ m}^2\]

Let the thickness be \(t\). Volume of the layer = floor area \(\times\) thickness, and this equals the volume of soil:

\[48\times t=\frac{176}{7}\]\[t=\frac{176}{7\times 48}=\frac{176}{336}=0.5238\text{ m}\]

Converting to centimetres: \(0.5238\times 100=52.38\text{ cm}\).

Correct to the nearest cm, the thickness is \(52\text{ cm}\).

(b) Quadrilateral \(ABCD\).

From the diagram, \(\angle DAB=90^{\circ}\), \(|AB|=3.3\text{ cm}\), \(|AD|=5.6\text{ cm}\), \(|BC|=3.9\text{ cm}\) and \(|DC|=5.2\text{ cm}\).

(i) Length of \(BD\).

Triangle \(ABD\) is right-angled at \(A\), so by Pythagoras:

\[BD^2=AB^2+AD^2=3.3^2+5.6^2=10.89+31.36=42.25\]\[BD=\sqrt{42.25}=6.5\text{ cm}\]

(ii) Show that \(\angle BCD=90^{\circ}\).

In triangle \(BCD\), \(|BC|=3.9\), \(|DC|=5.2\) and \(|BD|=6.5\). Test the converse of Pythagoras:

\[BC^2+DC^2=3.9^2+5.2^2=15.21+27.04=42.25\]\[BD^2=6.5^2=42.25\]

Since \(BC^2+DC^2=BD^2\), the converse of Pythagoras theorem holds, so the angle opposite \(BD\) is a right angle. Therefore \(\angle BCD=90^{\circ}\).