Basic Engineering Mathematics, Fifth Edition

Areas of common shapes 221

In triangleFGH,40◦+ 90 ◦+b= 180 ◦, since the

angles in a triangle add up to 180◦, from which

b= 50 ◦.Also,c= 40 ◦(alternate angles between

parallel linesEFandHG). (Alternatively,bandc

are complementary; i.e., add up to 90◦.)

d= 90 ◦+c(external angle of a triangle equals

the sum of the interior opposite angles), hence

d= 90 ◦+ 40 ◦= 130 ◦ (or ∠EFH= 50 ◦ and

d= 180 ◦− 50 ◦= 130 ◦).

(c) JKLMis a rhombus

The diagonals of a rhombus bisect the interior

angles and the opposite internal angles are equal.

Thus,∠JKM=∠MKL=∠JMK=∠LMK= 30 ◦,

hence,e= 30 ◦.

In triangle KLM,30◦+∠KLM+ 30 ◦= 180 ◦

(the angles in a triangle add up to 180◦), hence,

∠KLM= 120 ◦. The diagonalJLbisects∠KLM,

hence,f=

120 ◦

2

= 60 ◦.

(d) NOPQis a parallelogram

g= 52 ◦since the opposite interior angles of a

parallelogram are equal.

In triangleNOQ,g+h+ 65 ◦= 180 ◦(the angles

in a triangle add up to 180◦), from which

h= 180 ◦− 65 ◦− 52 ◦= 63 ◦.

i= 65 ◦(alternate angles between parallel lines

NQandOP).

j= 52 ◦+i= 52 ◦+ 65 ◦= 117 ◦ (the external

angle of a triangle equals the sum of the interior

opposite angles). (Alternatively, ∠PQO=h=

63 ◦; hence,j= 180 ◦− 63 ◦= 117 ◦.)

(e) RSTUis a trapezium

35 ◦+k= 75 ◦(external angle of a triangle equals

the sum of the interior opposite angles), hence,

k= 40 ◦.

∠STR= 35 ◦ (alternate angles between parallel

linesRUandST).l+ 35 ◦= 115 ◦(external angle

of a triangle equals the sum of the interioropposite

angles), hence,l= 115 ◦− 35 ◦= 80 ◦.

Now try the following Practice Exercise

PracticeExercise 96 Common shapes

(answers on page 351)

1. Find the anglespandqin Figure 25.8(a).


2. Find the anglesrandsin Figure 25.8(b).


3. Find the angletin Figure 25.8(c).

758

388

1258

628

958

(^478578)
(^408)
p q s
r
t
(a) (b) (c)
Figure 25.8

25.3 Areas of common shapes

The formulae forthe areas of common shapes are shown

in Table 25.1.

Here are some worked problems to demonstrate how

the formulae are used to determine the area of common

shapes.

Problem 2. Calculate the area and length of the

perimeter of the square shown in Figure 25.9

4.0 cm

4.0 cm

Figure 25.9

Area of square=x^2 =( 4. 0 )^2 = 4 .0cm× 4 .0cm

= 16 .0cm^2

(Note the unit of area is cm×cm=cm^2 ; i.e., square

centimetres or centimetres squared.)

Perimeter of square= 4 .0cm+ 4 .0cm+ 4 .0cm

+ 4 .0cm=16.0cm

Problem 3. Calculate the area and length of the

perimeter of the rectangle shown in Figure 25.10

7.0 cm

4.5 cm

A

D C

B

Figure 25.10