Basic Engineering Mathematics, Fifth Edition

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.