basic_engineering_mathematics0

168 Basic Engineering Mathematics
(iii) JKLM is a rhombus
The diagonals of a rhombus bisect the interior angles and
opposite internal angles are equal.
Thus ∠JKM = ∠MKL = ∠JMK = ∠LMK = 30 ◦ ,
hence e = 30 ◦
In triangle KLM,30 ◦ + ∠KLM + 30 ◦ = 180 ◦ (angles in a
triangle add up to 180 ◦ ), hence ∠KLM = 120 ◦ .
The diagonal JL bisects ∠KLM, hence
f = 120◦
2
= 60 ◦
(iv) NOPQ is a parallelogram
g = 52 ◦ (since opposite interior angles of a parallelogram
are equal).
In triangle NOQ, 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 lines NQ and OP).
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), hence k = 40 ◦
∠STR = 35 ◦ (alternate angles between parallel lines RU and
ST).
l + 35 ◦ = 115 ◦ (external angle of a triangle equals the sum
of the interior opposite angles), hence
l = 115 ◦ − 35 ◦ = 80 ◦
8mm
5mm
Fig. 22.7
50 mm
A
B
C
70 mm
(a)
6mm
75 mm
2m
25 m
(a) The girder may be divided into three separate rectangles as
shown.
Area of rectangle A = 50 × 5 = 250 mm 2
Area of rectangle B = (75 − 8 − 5) × 6
= 62 × 6 = 372 mm 2
Area of rectangle C = 70 × 8 = 560 mm 2
Total area of girder = 250 + 372 + 560 = 1182 mm 2 or
11.82 cm 2
(b) Area of path = area of large rectangle − area of small
rectangle
(b)
= (25 × 20) − (21 × 16) = 500 − 336 = 164 m 2
Problem 4. Find the area of the parallelogram shown in
Fig. 22.8 (dimensions are in mm).
A
B
20 m
Problem 2. A rectangular tray is 820 mm long and 400 mm
wide. Find its area in (a) mm 2 , (b) cm 2 , (c) m 2 .
(a) Area = length × width = 820 × 400 = 328 000 mm 2
(b) 1 cm 2 = 100 mm 2 . Hence
Fig. 22.8
15
h
D 25 C E
34
328 000 mm 2 =
(c) 1 m 2 = 10 000 cm 2 . Hence
328 000
100
cm 2 = 3280 cm 2
3280 cm 2 = 3280
10 000 m2 = 0.3280 m 2
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).
Area of parallelogram = base × perpendicular height. The perpendicular
height h is found using Pythagoras’ theorem.
BC 2 = CE 2 + h 2
i.e. 15 2 = (34 − 25) 2 + h 2
h 2 = 15 2 − 9 2 = 225 − 81 = 144
Hence, h = √ 144 = 12 mm (−12 can be neglected).
Hence, area of ABCD = 25 × 12 = 300 mm 2