MSE2183
ASSIGNMENT 3
FULL SOLUTIONS
Due date: 02 September 2024
Unique Number: 693219
Complete Solutions
Geometry
UNISA
2024
,Question 1
Name and Surname:
Student Number:
My posts in Forum 4
Other students’
responses /
comments on
My comments on
other students’ posts
NB: Discuss fully.
(A zero mark will be
awarded for
responses/comments
like
‘I fully agree with your
statement’; ‘Good
work’; ‘indeed’ and
more other
superficial/insignifica
nt
responses/comments)
,SOLUTION
Question 2
2.1 Perimeter 1 = 2L + 2B = 2(a) + 2(b) = 2a + 2b
Perimeter 2 = 2(2a) + 2(b/2) = 4a + b
Now , if 4a + b = 2a + 2b
2a – b = 0
b = 2a
Hence the two rectangles have the same perimeter whenever b = 2a , where a and b
are real numbers.
, Area of rectangle 1 = Length x Breadth
=axb
= ab
Area of rectangle 2 = Length x Breadth
= 2a x (b/2)
= ab
Hence the two rectangles have the same area for all real values of a and b.
2.2 Perimeter of a dodecagon = 12S
= 12( 7cm)
= 84 cm
2.3 Perimeter of an equilateral Octagon = 8S
42 = 8S
S = 42/8
S = 21/4
S = 5.25 cm
2.4
Perimeter of figure = 2(x -3) + 2(x + 8) + 16
94 = 2x – 6 + 2x + 16 + 16
94 -26 = 4x
68 = 4x
x = 68/4
ASSIGNMENT 3
FULL SOLUTIONS
Due date: 02 September 2024
Unique Number: 693219
Complete Solutions
Geometry
UNISA
2024
,Question 1
Name and Surname:
Student Number:
My posts in Forum 4
Other students’
responses /
comments on
My comments on
other students’ posts
NB: Discuss fully.
(A zero mark will be
awarded for
responses/comments
like
‘I fully agree with your
statement’; ‘Good
work’; ‘indeed’ and
more other
superficial/insignifica
nt
responses/comments)
,SOLUTION
Question 2
2.1 Perimeter 1 = 2L + 2B = 2(a) + 2(b) = 2a + 2b
Perimeter 2 = 2(2a) + 2(b/2) = 4a + b
Now , if 4a + b = 2a + 2b
2a – b = 0
b = 2a
Hence the two rectangles have the same perimeter whenever b = 2a , where a and b
are real numbers.
, Area of rectangle 1 = Length x Breadth
=axb
= ab
Area of rectangle 2 = Length x Breadth
= 2a x (b/2)
= ab
Hence the two rectangles have the same area for all real values of a and b.
2.2 Perimeter of a dodecagon = 12S
= 12( 7cm)
= 84 cm
2.3 Perimeter of an equilateral Octagon = 8S
42 = 8S
S = 42/8
S = 21/4
S = 5.25 cm
2.4
Perimeter of figure = 2(x -3) + 2(x + 8) + 16
94 = 2x – 6 + 2x + 16 + 16
94 -26 = 4x
68 = 4x
x = 68/4
