226 Basic Engineering Mathematics
- Determine the area of an equilateral triangle
of side 10.0cm. - If paving slabs are produced in 250mm by
250mm squares, determine the number of
slabsrequiredtocoveranareaof2m^2.
Here are some further worked problems on finding the
areas of common shapes.Problem 11. Find the area of a circle having a
radius of 5cmAreaof circle=πr^2 =π( 5 )^2 = 25 π= 78 .54cm^2Problem 12. Find the area of a circle having a
diameter of 15mmAreaof circle=πd^2
4=π( 15 )^2
4=225 π
4= 176 .7mm^2Problem 13. Find the area of a circle having a
circumference of 70mmCircumference,c= 2 πr,henceradius,r=c
2 π=70
2 π=35
πmmArea of circle=πr^2 =π(
35
π) 2
=352
π= 389 .9mm^2 or 3 .899cm^2Problem 14. Calculate the area of the sector of a
circle having radius 6cm with angle subtended at
centre 50◦Area of sector=θ^2
360(πr^2 )=50
360(π 62 )=50 ×π× 36
360= 15 .71cm^2Problem 15. Calculate the area of the sector of a
circle having diameter 80mm with angle subtended
at centre 107◦ 42 ′If diameter=80mm then radius,r=40mm, andarea of sector=107 ◦ 42 ′
360(π 402 )=10742
60
360(π 402 )=107. 7
360(π 402 )=1504mm^2 or15.04cm^2Problem 16. A hollow shaft has an outside
diameter of 5.45cm and an inside diameter of
2 .25cm. Calculate the cross-sectional area of the
shaftThe cross-sectional area of the shaft is shown by the
shaded part in Figure 25.22 (often called anannulus).d 5
2.25 cm
D 5 5.45 cmFigure 25.22Area of shaded part=area of large circle – area of
small circle=πD^2
4−πd^2
4=π
4(D^2 −d^2 )=π
4( 5. 452 − 2. 252 )=19.35cm^2Now try the following Practice ExercisePracticeExercise 98 Areas of common
shapes (answers on page 351)- A rectangular garden measures 40m by 15m.
A 1m flower border is made round the two
shorter sides and one long side. A circular
swimming poolof diameter 8m is constructed