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2002 WAEC Mathematics Theory  The diagram is a portion of a right circular solid cylinder of radius 7 cm...

Mathematics
WAEC 2002

 The diagram is a portion of a right circular solid cylinder of radius 7 cm and height 15 cm. The centre of the base of the cylinder is Q, while that of the top is B, where \(\stackrel\frown{ABC} = \stackrel\frown{PQR} = 120°\). Calculate, correct to one decimal place:

(a) The volume 

(b) the total surface area of the solid. [Take \(\pi = \frac{22}{7}\)].

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Explanation

(a) Area of the major sector = \(\frac{\theta}{360°} \times \pi r^{2}\)

= \(\frac{360 - 120}{360} \times \frac{22}{7} \times 7 \times 7\)

= \((\frac{2}{3} \times 154) cm^{2}\)

Volume = \(\text{Area of the major sector} \times \text{height (cross section)}\)

= \(\frac{2}{3} \times 154 \times 15 = 1540 cm^{3}\)

(b) Curved surface area = \(\frac{240}{360} \times 2\pi rh\)

= \(\frac{240}{360} \times 2 \times \frac{22}{7} \times 7 \times 15\)

= \(440 cm^{2}\)

Area of rectangular surface = \(2 \times l \times b\)

= \(2 \times 7 \times 15 = 210 cm^{2}\)

Area of top and bottom of cylinder (major sector) 

= \(2 \times \frac{240}{360} \times \frac{22}{7} \times 7 \times 7\)

= \(205.33 cm^{2}\)

\(\therefore \text{Total surface area} = 440 + 205.33 + 210 = 855.33 cm^{2}\)


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WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
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WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC offline past questions - with all answers and explanations in one app - Download for free