(a) A cylinder with radius 3.5 cm has its two ends... | WAEC SSCE - General Mathematics - 2007 (Essay)

Question 1 Report

(a) A cylinder with radius 3.5 cm has its two ends closed, if the total surface area is \(209 cm^{2}\), calculate the height of the cylinder. [Take \(\pi = \frac{22}{7}\)].

(b) In the diagram, O is the centre of the circle and ABC is a tangent at B. If \(\stackrel\frown{BDF} = 66°\) and \(\stackrel\frown{DBC} = 57°\), calculate, (i) \(\stackrel\frown{EBF}\) and (ii) \(\stackrel\frown{BGF}\).

Answer

(a) Height of the closed cylinder

For a cylinder closed at both ends, the total surface area is

\[TSA = 2\pi r(r+h)\]

With \(r = 3.5\text{ cm}\), \(TSA = 209\text{ cm}^2\) and \(\pi = \tfrac{22}{7}\):

\[2\times\frac{22}{7}\times 3.5\,(3.5+h) = 209\]

Now \(2\times\dfrac{22}{7}\times 3.5 = 22\), so

\[22(3.5+h) = 209 \;\Rightarrow\; 3.5+h = \frac{209}{22} = 9.5\]\[h = 9.5 - 3.5 = 6\text{ cm}\]

The height of the cylinder is \(6\text{ cm}\).

(b) Circle with tangent ABC at B

From the diagram, \(EB\) is a diameter (it passes through the centre O). The given angles are the inscribed angle \(\angle BDF = 66^\circ\) at D and the tangent-chord angle \(\angle DBC = 57^\circ\) at B. G is the point where chord \(FD\) crosses the diameter \(EB\).

Finding the arcs

The inscribed angle \(\angle BDF\) stands on arc \(BF\) (not containing D), so

\[\text{arc }BF = 2\times 66^\circ = 132^\circ\]

The tangent-chord angle \(\angle DBC\) equals half the intercepted arc \(BD\):

\[\text{arc }BD = 2\times 57^\circ = 114^\circ\]

Since \(EB\) is a diameter, each semicircle is \(180^\circ\):

\[\text{arc }EF = 180^\circ - \text{arc }BF = 180^\circ - 132^\circ = 48^\circ\]\[\text{arc }ED = 180^\circ - \text{arc }BD = 180^\circ - 114^\circ = 66^\circ\]

(Check: \(114+66+48+132 = 360^\circ\). Correct.)

(i) \(\angle EBF\)

\(\angle EBF\) is the inscribed angle at B standing on arc \(EF\):

\[\angle EBF = \tfrac{1}{2}\,\text{arc }EF = \tfrac{1}{2}\times 48^\circ = 24^\circ\]

(ii) \(\angle BGF\)

G is the intersection of chords \(EB\) and \(FD\). The angle between two chords equals half the sum of the two arcs it intercepts (arc \(BF\) and the vertically opposite arc \(ED\)):

\[\angle BGF = \tfrac{1}{2}\big(\text{arc }BF + \text{arc }ED\big) = \tfrac{1}{2}(132^\circ + 66^\circ) = \tfrac{1}{2}\times 198^\circ = 99^\circ\]

Therefore \(\angle EBF = 24^\circ\) and \(\angle BGF = 99^\circ\).

Read lesson note on Circles (WAEC)

Read lesson note on Areas (WAEC)

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