(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 (stackrelfrown{BDF} = 66°) and (stackrelfrown{DBC} = 57°), calculate, (i) (stackrelfrown{EBF}) and (ii) (stackrelfrown{BGF}).
Explanation
(a) TSA of a closed cylinder = (2pi r^{2} + 2pi rh)
= (2pi r (r + h))
(2 times frac{22}{7} times 3.5 (3.5 + h) = 209)
(22(3.5 + h) = 209 implies 77 + 22h = 209)
(22h = 209 – 77 = 132)
(h = frac{132}{22} = 6 cm)
(b) (< EDB = 90°) (angle subtended at the circumference by a semi-circle)
(therefore < EDF = 90° – 66° = 24°)
(< EDF = < EBF = 24°) (angles in the same segment)
(ii)
(< DEB = < DBC = 57°) (Alternate segment theorem)
(therefore < BGF = 180 – (57 + 24) = 180 – 81)
(< BGF = 99°)