(a)
In the diagram, < RTS = 28°, < VRM = 46°, MQ is a tangent to the circle VRSTU at the point R. Find < VUS.
(b) A cylinder tin, 7cm high, is closed at one end. If its total surface area is 462\(cm^{2}\), calculate its radius. [Take \(\pi = \frac{22}{7}\)].
In the diagram, < RTS = 28°, < VRM = 46°, MQ is a tangent to the circle VRSTU at the point R. Find < VUS.
(b) A cylinder tin, 7cm high, is closed at one end. If its total surface area is 462\(cm^{2}\), calculate its radius. [Take \(\pi = \frac{22}{7}\)].
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Correct Answer: Option n
Explanation:
(a)
In the diagram above, \(\alpha = 28°\) (angles in alternate segment)
\(46° + \beta + \alpha = 180°\) (angles on a straight line)
\(46° + \beta + 28° = 74° + \beta = 180°\)
\(\beta = 180° - 74° = 106°\)
\(x + \beta = 180°\) (opposite angles of a cyclic quadrilateral)
\(x + 106° = 180°\)
\(x = 180° - 106° = 74°\)
Hence, < VUS = 74°.
(b) Total surface area = curved surface area + area of closed end
= \(2\pi rh + \pi r^{2}\)
= \(\pi r(2h + r)\)
\(462 = \frac{22r}{7}((2 \times 7) + r)\)
\(\frac{462 \times 7}{22} = r(14 + r)\)
\(147 = 14r + r^{2}\)
\(r^{2} + 14r - 147 = 0\)
\(r^{2} - 7r + 21r - 147 = 0\)
\(r(r - 7) + 21(r - 7) = 0\)
\((r + 21)(r - 7) = 0\)
\(r = \text{-21 or 7}\)
The radius cannot be negative, therefore, r = 7 cm.
(a)
In the diagram above, \(\alpha = 28°\) (angles in alternate segment)
\(46° + \beta + \alpha = 180°\) (angles on a straight line)
\(46° + \beta + 28° = 74° + \beta = 180°\)
\(\beta = 180° - 74° = 106°\)
\(x + \beta = 180°\) (opposite angles of a cyclic quadrilateral)
\(x + 106° = 180°\)
\(x = 180° - 106° = 74°\)
Hence, < VUS = 74°.
(b) Total surface area = curved surface area + area of closed end
= \(2\pi rh + \pi r^{2}\)
= \(\pi r(2h + r)\)
\(462 = \frac{22r}{7}((2 \times 7) + r)\)
\(\frac{462 \times 7}{22} = r(14 + r)\)
\(147 = 14r + r^{2}\)
\(r^{2} + 14r - 147 = 0\)
\(r^{2} - 7r + 21r - 147 = 0\)
\(r(r - 7) + 21(r - 7) = 0\)
\((r + 21)(r - 7) = 0\)
\(r = \text{-21 or 7}\)
The radius cannot be negative, therefore, r = 7 cm.