(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})].
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
= (2pi 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.