224 Basic Engineering Mathematics
Problem 7. A rectangular tray is 820mm long
and 400mm wide. Find its area in (a) mm^2 (b) cm^2
(c) m^2
(a)Area of tray=length×width= 820 × 400
=328000mm^2
(b) Since 1cm=10mm,1cm^2 =1cm×1cm
=10mm×10mm=100mm^2 ,or
1mm^2 =
1
100
cm^2 = 0 .01cm^2
Hence,328000mm^2 = 328000 × 0 .01cm^2
=3280cm^2.
(c) Since 1m=100cm,1m^2 =1m×1m
=100cm×100cm=10000cm^2 ,or
1cm^2 =
1
10000
m^2 = 0 .0001m^2
Hence,3280cm^2 = 3280 × 0 .0001m^2
= 0 .3280m^2.
Problem 8. The outside measurements of a
picture frame are 100cm by 50cm. If the frame is
4cm wide, find the area of the wood used to make
the frame
A sketch of the frame is shown shaded in Figure 25.15.
100 cm
50 cm 42 cm
92 cm
Figure 25.15
Area of wood=area of large rectangle−area of
small rectangle
=( 100 × 50 )−( 92 × 42 )
= 5000 − 3864
=1136cm^2
Problem 9. Find the cross-sectional area of the
girder shown in Figure 25.16
5mm
50 mm
B
C
A
8mm
70 mm
75 mm
6mm
Figure 25.16
Thegirdermaybedividedintothreeseparaterectangles,
as shown.
Area of rectangleA= 50 × 5 =250mm^2
Area of rectangleB=( 75 − 8 − 5 )× 6
= 62 × 6 =372mm^2
Area of rectangleC= 70 × 8 =560mm^2
Total area of girder= 250 + 372 + 560
=1182mm^2 or 11 .82cm^2
Problem 10. Figure 25.17 shows the gable end of
a building. Determine the area of brickwork in the
gable end
6m
5m 5m
A
C D
B
8m
Figure 25.17
The shape is that of a rectangle and a triangle.
Area of rectangle= 6 × 8 =48m^2
Area of triangle=
1
2
×base×height
CD=4mandAD=5m, henceAC=3m (since it is a
3, 4, 5 triangle – or by Pythagoras).
Hence,area of triangleABD=
1
2
× 8 × 3 =12m^2.