Trigonometry and Geometry - Solved Assignments and Essays | Desklib

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TABLE OF CONTENTS
Table of Contents.............................................................................................................................1
ASSESSMENT TASK 4.................................................................................................................2
Question 1........................................................................................................................................2
A..................................................................................................................................................2
B...................................................................................................................................................2
Question 2........................................................................................................................................3
Question 3........................................................................................................................................4
A. Perimeter of patch...................................................................................................................4
B. Perimeter of patch...................................................................................................................4
C. Cost of edging.........................................................................................................................5
D. Volume of soil........................................................................................................................5
E. Surface area.............................................................................................................................5
F. Composite soil required...........................................................................................................5
Question 4........................................................................................................................................6
Question 5........................................................................................................................................8
Question 6........................................................................................................................................8
(i) Missing side............................................................................................................................9
(ii) Missing angle.........................................................................................................................9
(iii) Checking answer.................................................................................................................10
Question 7......................................................................................................................................10
1
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ASSESSMENT TASK 4
Question 1
A
x = 3 meter and y = 5 meter as opposite sides of rectangle are equal
X= 5 cm
y can be calculated using pythagorus theorem
y2 = 52 – 2.52
y2 = 25 – 6.25
y = 4.33 cm
B.
Radius of circle is half of the diameter. Given that diameter is 2.75 cm, radius is equals to 1.375
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Question 2
Shape
Write the name of the
shape
Perimeter
Write correct units
Area
Write correct units
a
Rectangle
Formula: 2 [l +b] where l is length
and b is breadth of rectangle
Perimeter = 2 * [3+5]
Perimeter = 16 meter
Formula: l*b
Area = 3 *5
Area = 15 meter 2
b Triangle Formula: a+b+c
where a,b and c are sides of the
triangle
Answer: Perimeter = 5+5+5 = 15 cm
Formula: 0.5 * base * height
Answer: 0.5 * 5* 4.33
area = 10.82 cm 2
c
circle
Formula: 2* π *r
where r is the radius of circle
Answer:
Given diameter = 2.75 cm
radius r = 1.375 cm
Perimeter = 2 * 3.14 * 1.375
Perimeter = 8.63 cm
Formula: π *r * r
Answer:
Area = 3.14 * 1.375 * 1.375
area = 5.93 cm2
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Question 3
A. Perimeter of patch
The vegetable patch will be in shape of rectangle with length 6 meter and width 2 meter.
Perimeter of rectangle = 2 [Length + width]
= 2 * [2+6]
Perimeter = 16 meter
B. Perimeter of patch
New length of rectangular part = 4 meter
Diameter of semicircle = 2 meter
Radius of semicircle = 1 meter
The two semicircles each of diameter 2 meter makes one circle. Thus circumference of circle is
= 2 * 3.14 * r
4
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Perimeter of circle = 2 * 3.14 * 1 = 6.28 meter
Total perimeter of new patch = Perimeter of circle + perimeter of two sides of rectangle
= 6.28 + 4 + 4
Perimeter = 14.28 meter
C. Cost of edging
Unit cost of edging = $3.97 per meter
Perimeter of vegetable patch = 14.28 meter
Thus edging will be purchased for 15 meter.
Cost of edging vegetable patch = 3.97 * 15 = $ 59.55
D. Volume of soil
Depth = 200 mm = 0.2 meter
Given length and width of rectangle are 6 and 2 meter respectively
Volume of rectangle = length * width * depth
Volume = 6 * 2 * 0.2
Volume = 2.4 meter 3
Volume of semi-circle = 0.5 * π r2h
Radius of semi-circle = 1 meter
Volume of semi-circle = 0.5 * 3.14 * 1 *1 * 0.2 = 0.314
As there are two semi-circles so volume of both = 2 * 0.314 = 0.628 meter 3
Total volume = 2.4 + 0.628 = 3.02 meter 3
Volume of soil needed to remove = 3.02 meter 3
E. Surface area
Surface area for rectangular part- A=2 (wl+ hl+ hw)
= 2*(2*4+0.2*4+0.2*2)
= 18.4 m2
Surface area for circular part-
A=2πrh+2πr2
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=2π * 1 * 0.2+ 2π * 12
=7.53982 m2
Total Surface area= 18.4 + 7.53982 = 25.93982 m2
F. Composite soil required
Volume to be filled = 3.02 meter 3
Ratio of composite and excavated soil = 3:1
Thus volume of composite soil which will be needed = [3.02 * 3] /4 = 2.26 meter 3
Question 4
Shape Name Volume (units)
a
Square based
pyramid Area of the base= 7 cm x 7 cm =
49 cm2
= 49 cm2 x 15 cm x 1/3
= 735 cm3 x 1/3
= 245 cm3
b Triangular
prism
Volume = area of base * length of
prism
Prism length = 12 cm
Area of base triangle = 0.5 * base *
height
Base = 4 cm and height = 6 cm
Area = 0.5 * 4 * 6 = 12 cm2
Volume of triangular prism = 12 *
12 = 144 cm3
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c Square based
pyramid
Volume = 1/3 * area of base *
height
Height = 15 cm
area of base = 6 *8 = 48 cm2
Volume = [1/3] * 48 * 15
Volume = 240 cm3
d cylinder Volume = π r2h
Given:
Height h = 20 cm
Diameter = 20 cm
radius = 20/2 = 10 cm
Volume = 3.14 * 10*10*20
Volume = 6280 cm3
e cone
Volume = π r2h / 3
Radius = 4 cm
Height = 15 cm
Volume = 3.14 * 4*4*15 /3
Volume = 251.2 cm3
f Volume = base area * height
b = 4 cm
h = 70 mm = 7 cm
Base area = 0.5 * 4 * 7 = 14
Height of prism = 12 cm
Volume = 14*12 = 168
7
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b = 4cm
h = 70 mm
H = 12 cm
Triangular
prism
Volume = 168 cm3
Question 5
Height of the tree = AB
Length of ladder AC = 9 meter
Distance between base of trunk and ladder = BC = 4 meter
The arrangement forms a right angle triangle
Using Pythagoras theorem:
AC2 = AB2 + BC2
AB2 = AC2 - BC2
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AB2 = 92 - 42
AB2 = 81 – 16
AB2 = 65
AB = 8.06 meter
Thus cat is at height of 8.06 meter
Question 6
(i) Missing side
Using the law of sines:
8 /sin 25 = x/sin 90
Sin 90 = 1
8 /sin 25 = x
x = 18.95 cm
Using Pythagoras theorem:
x2 = y2 + 82
y2 = (18.95)2 -64
y2 = (18.95)2 -64
y2 = 295.1
y = 17.17 cm
(ii) Missing angle
As per trigonometric ratio; in right angle triangle cos x = base / height
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From figure it can be seen that:
Cos a = 8 / 18.95
Cos a = 0.422
Angle a = 65.04°
Approximately angle = 65 degree
(iii) Checking answer
The answer can be verified by using the property that sum of angles of a triangle is 180.
Thus for given question all three angles are 90, 25 and 65 which sum up to 180 degrees. It proves
that answer is correct.
Question 7
Using Law of Sines,
So,
10
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= B =
B= 42°
Find missing angle 2
C= 180- A+B = 180 – 76 + 42
C= 62°
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