In Problems 1 to 8, express the given polar co-ordinates as Cartesian co-ordinates, correct to 3 decimal places.
1. (5, 75◦) 2. (4.4, 1.12 rad) 3. (7, 140◦) 4. (3.6, 2.5 rad) 5. (10.8, 210◦) 6. (4, 4 rad) 7. (1.5, 300◦) 8. (6, 5.5 rad)
9. Figure 21.10 shows 5 equally spaced holes on an 80 mm pitch circle diameter. Calculate their co-ordinates rela- tive to axes 0xand 0yin (a) polar form, (b) Cartesian form.
O x y
Fig. 21.10
22
Areas of plane figures
22.1 Mensuration
Mensurationis a branch of mathematics concerned with the determination of lengths, areas and volumes.
22.2 Properties of quadrilaterals
Polygon
Apolygonis a closed plane figure bounded by straight lines. A polygon which has:
(i) 3 sides is called atriangle (ii) 4 sides is called aquadrilateral (iii) 5 sides is called apentagon (iv) 6 sides is called ahexagon
(v) 7 sides is called aheptagon (vi) 8 sides is called anoctagon
There are five types ofquadrilateral, these being:
(i) rectangle (ii) square (iii) parallelogram (iv) rhombus
(v) trapezium
(The properties of these are given below).
If the opposite corners of any quadrilateral are joined by a straight line, two triangles are produced. Since the sum of the angles of a triangle is 180◦, the sum of the angles of a quadrilateral is 360◦.
In arectangle, shown in Fig. 22.1:
(i) all four angles are right angles,
(ii) opposite sides are parallel and equal in length, and (iii) diagonals AC and BDare equal in length and bisect one
another.
A
D
B
C Fig. 22.1
P Q
S R
Fig. 22.2
In asquare, shown in Fig. 22.2:
(i) all four angles are right angles, (ii) opposite sides are parallel,
(iii) all four sides are equal in length, and
(iv) diagonals PR andQS are equal in length and bisect one another at right angles.
In aparallelogram, shown in Fig. 22.3:
(i) opposite angles are equal,
(ii) opposite sides are parallel and equal in length, and (iii) diagonalsWYandXZbisect one another.
In arhombus, shown in Fig. 22.4:
(i) opposite angles are equal,
(ii) opposite angles are bisected by a diagonal, (iii) opposite sides are parallel,
(iv) all four sides are equal in length, and
(v) diagonalsACandBDbisect one another at right angles.
Areas of plane figures 167
Z Y
X W
Fig. 22.3
A B
D C
b a a
b
Fig. 22.4
E F
H G
Fig. 22.5
In atrapezium, shown in Fig. 22.5:
(i) only one pair of sides is parallel
22.3 Worked problems on areas of plane figures
Table 22.1 (i) Square
Area⫽x2
Area⫽l⫻b
Area⫽b⫻h (ii) Rectangle
(iii) Parallelogram
(iv) Triangle
(v) Trapezium a
h h
h
b b
b b l
x x
Area⫽12⫻b⫻h
(a⫹b)h Area ⫽12
Table 22.1 (Continued) (vi) Circle
(vii) Semicircle
u (viii) Sector of a circle
r
r r
r l d
d
Area⫽πr2 orπd2 4
(u in rads) Area⫽ πd2
πr2 or 8
1 2
Area⫽ u° (πr2) or r2u 360° 12
Problem 1. State the types of quadrilateral shown in Fig. 22.6 and determine the angles markedatol.
A x
x c b B
E
H G
F
J x
x e f
K
M
R
S
U l
N g O
Q P h
j i
k T
L d
40°
35°
75°
115°
65° 52°
C D
a
(i)
(iv) (v)
(ii) (iii)
30°
Fig. 22.6
(i)ABCDis a square
The diagonals of a square bisect each of the right angles, hence
a= 90◦ 2 =45◦ (ii) EFGHis a rectangle
In triangleFGH, 40◦+90◦+b=180◦(angles in a triangle add up to 180◦) from which,b=50◦. Alsoc=40◦(alternate angles between parallel linesEFandHG).
(Alternatively,bandcare 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◦
168 Basic Engineering Mathematics
(iii) JKLMis a rhombus
The diagonals of a rhombus bisect the interior angles and opposite internal angles are equal.
Thus∠JKM=∠MKL=∠JMK=∠LMK=30◦, hencee=30◦
In triangleKLM, 30◦+∠KLM+30◦=180◦ (angles in a triangle add up to 180◦), hence∠KLM=120◦.
The diagonalJLbisects∠KLM, hence f = 120◦
2 =60◦ (iv) NOPQis a parallelogram
g=52◦(since opposite interior angles of a parallelogram are equal).
In triangleNOQ,g+h+65◦=180◦(angles in a triangle add up to 180◦), from which,
h=180◦−65◦−52◦=63◦
i=65◦(alternate angles between parallel linesNQandOP).
j=52◦+i=52◦+65◦=117◦(external angle of a triangle equals the sum of the interior opposite angles).
(v) RSTU is a trapezium
35◦+k=75◦(external angle of a triangle equals the sum of the interior opposite angles), hencek=40◦
∠STR=35◦(alternate angles between parallel linesRUand ST).
l+35◦=115◦(external angle of a triangle equals the sum of the interior opposite angles), hence
l=115◦−35◦=80◦
Problem 2. A rectangular tray is 820 mm long and 400 mm wide. Find its area in (a) mm2, (b) cm2, (c) m2.
(a) Area=length×width=820×400=328 000 mm2 (b) 1 cm2=100 mm2. Hence
328 000 mm2= 328 000
100 cm2=3280 cm2 (c) 1 m2=10 000 cm2. Hence
3280 cm2= 3280
10 000m2=0.3280 m2
Problem 3. Find (a) the cross-sectional area of the girder shown in Fig. 22.7(a) and (b) the area of the path shown in Fig. 22.7(b).
50 mm 5 mm
8 mm
6 mm
2 m 25 m
20m
75mm
A
B
C 70 mm
(a) (b)
Fig. 22.7
(a) The girder may be divided into three separate rectangles as shown.
Area of rectangleA=50×5=250 mm2 Area of rectangleB=(75−8−5)×6
=62×6=372 mm2 Area of rectangleC=70×8=560 mm2
Total area of girder=250+372+560=1182 mm2 or 11.82 cm2
(b) Area of path=area of large rectangle − area of small rectangle
=(25×20)−(21×16)=500−336=164 m2
Problem 4. Find the area of the parallelogram shown in Fig. 22.8 (dimensions are in mm).
A
D C E
B 15 h
25 34 Fig. 22.8
Area of parallelogram=base×perpendicular height. The per- pendicular heighthis found using Pythagoras’ theorem.
BC2=CE2+h2 i.e. 152=(34−25)2+h2
h2=152−92=225−81=144 Hence,h=√
144=12 mm (−12 can be neglected).
Hence, area ofABCD=25×12=300 mm2
Areas of plane figures 169
Problem 5. Figure 22.9 shows the gable end of a building.
Determine the area of brickwork in the gable end.
5 m 5 m
6 m B
D A
C
8 m Fig. 22.9
The shape is that of a rectangle and a triangle.
Area of rectangle=6×8=48 m2 Area of triangle=12×base×height.
CD=4 m, AD=5 m, hence AC=3 m (since it is a 3, 4, 5 triangle).
Hence, area of triangleABD=12×8×3=12 m2 Total area of brickwork=48+12=60 m2
Problem 6. Determine the area of the shape shown in Fig. 22.10.
5.5 mm
27.4 mm
8.6 mm Fig. 22.10
The shape shown is a trapezium.
Area of trapezium=12(sum of parallel sides)(perpendicular distance between them)
=12(27.4+8.6)(5.5)
=12×36×5.5=99 mm2
Problem 7. Find the areas of the circles having (a) a radius of 5 cm, (b) a diameter of 15 mm, (c) a circumference of 70 mm.
Area of a circle=πr2or πd2 4
(a) Area=πr2=π(5)2=25π=78.54 cm2 (b) Area=πd2
4 =π(15)2 4 =225π
4 =176.7 mm2
(c) Circumference,c=2πr, hence r= c
2π = 70 2π = 35
πmm Area of circle=πr2=π
35 π
2
=352
=389.9 mm2or3.899 cmπ 2
Problem 8. Calculate the areas of the following sectors of circles:
(a) having radius 6 cm with angle subtended at centre 50◦ (b) having diameter 80 mm with angle subtended at centre
107◦42
(c) having radius 8 cm with angle subtended at centre 1.15 radians.
Area of sector of a circle= θ2
360(πr2) or1
2r2θ(θin radians).
(a) Area of sector
= 50
360(π62)= 50×π×36
360 =5π =15.71 cm2 (b) If diameter=80 mm, then radius,r=40 mm, and area of
sector
= 107◦42
360 (π402)= 10742 60
360 (π402)= 107.7 360 (π402)
=1504 mm2 or 15.04 cm2 (c) Area of sector=1
2r2θ=1
2×82×1.15=36.8 cm2 Problem 9. A hollow shaft has an outside diameter of 5.45 cm and an inside diameter of 2.25 cm. Calculate the cross-sectional area of the shaft.
The cross-sectional area of the shaft is shown by the shaded part in Fig. 22.11 (often called anannulus).
d⫽ 2.25 cm d⫽5.45 cm Fig. 22.11
Area of shaded part=area of large circle−area of small circle
= πD2 4 −πd2
4 = π
4(D2−d2)= π
4(5.452−2.252)
=19.35 cm2
170 Basic Engineering Mathematics
Now try the following exercise