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Area
Lesson 16
MFM1P – Foundations of Mathematics Unit 4 – Lesson 16
Copyright © 2005, Durham Continuing Education Page 2 of 62
Lesson Sixteen Concepts Overall Expectations ¾ Determine, through investigation, the optimal values of various measurements of
rectangles; ¾ Solve problems involving the measurements of two-dimensional shapes and
volumes of three-dimensional figures. Specific Expectations ¾ Determine the maximum area of a rectangle with a given perimeter by constructing a
variety of rectangles, using a variety of tools; ¾ Determine the minimum perimeter of a rectangle with a given area by constructing a
variety of rectangles, using a variety of tools; ¾ Solve problems that require maximizing the area of a rectangle for a fixed perimeter
of minimizing the perimeter of a rectangle for a fixed area ¾ Solve problems involving the areas and perimeters of composite two-dimensional
shapes;
Area and Surface Area Area Area is the number of square units needed to cover a region.
MFM1P – Foundations of Mathematics Unit 4 – Lesson 16
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Formulas to be used to calculate area.
2A rπ=
MFM1P – Foundations of Mathematics Unit 4 – Lesson 16
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Example Find the area of each shape. a) 10 cm 6 cm 8 cm d = 9 cm
b)
c)
12 cm. 8 c 7 cm.
MFM1P – Foundations of Mathematics Unit 4 – Lesson 16
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Solution
a) 10 cm 6 cm 8 cm
2cm24A248
A
2)8)(6(
A
2bh
A
=
=
=
=
d = 9 cm
b)
5.4r29
r
2d
r
=
=
=
2
2
2
cm59.63A
)5.4)(14.3(A
rA
≈
=
π=
c)
2cm84A
)7)(12(AlwA
=
==
12 cm. 8 c 7 cm.
Remember to use BEDMAS. Brackets before multiplication.
.cm 48cm6cm8 211 =× The 2cm comes from the multiplying
of the cm together.
MFM1P – Foundations of Mathematics Unit 4 – Lesson 16
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d) 10 m 22 m Total area = area of rectangle + area of ½ circle.
2t
t
t
2
t
2
t
21t
m25.259A
25.39220A2
5.78220A
2)5)(14.3(
)22)(10(A
2r
lwA
2A
AA
=
+=
+=
+=
π+=
+=
Example Find the area. 5 m 10 m
15 m 8m 13 m
This object should be divided into two easily calculated pieces. The rectangle should be the first ( 1A ) and ½ of the
circle ( 2A ) should be the other.
MFM1P – Foundations of Mathematics Unit 4 – Lesson 16
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Solution Find the area. 5 m 10 m
15 m 1A 8m 3A 2A 13 m Need to divide the object into easily calculated shapes. tA = Total Area 321t AAAA ++= Area of Rectangle One Area of Rectangle Two
2
1
1
1
cm75A
)15)(5(AlwA
=
==
2
3
2
2
cm130A
)10)(13(AlwA
=
==
Area of Triangle Total Area
23
3
3
3
cm25A2
50A
2)10)(5(
A
2bh
A
=
=
=
=
2
t
t
321t
cm230A
2513075AAAAA
=
++=
++=
By deductive reasoning the length of the side of the rectangle is 20–10 =10 cm.
MFM1P – Foundations of Mathematics Unit 4 – Lesson 16
Copyright © 2005, Durham Continuing Education Page 8 of 62
Support Questions 1. Calculate the area for each of the following objects.
a) b) c) 8cm 5cm 10 cm
4 cm 6cm 5cm 3 cm d) e) 4 cm f) 8cm
7 cm 3.5 cm
MFM1P – Foundations of Mathematics Unit 4 – Lesson 16
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Support Questions 2. Calculate the shaded area for each of the following objects. a)
25 cm 11 cm
b) 8 cm
1.5 cm
11 cm
MFM1P – Foundations of Mathematics Unit 4 – Lesson 16
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Surface Area Surface Area is a measure of the area on the surface of a three-dimensional object.
Formulas to be used to calculate surface area.
rh2r2.A.S
rh2.A.Sr.A.S
r.A.S
2total
side
2base
2top
π+π=
π=
π=
π=
2r4.A.S π=
2total
2total
2base
triangle
bbs2.A.S
b2bs
4.A.S
b.A.S2bs
.A.S
+=
+⎟⎠⎞
⎜⎝⎛=
=
=
)lhlwwh(2.A.S ++=
lbls22
bh2.A.S ++⎟
⎠⎞
⎜⎝⎛=
MFM1P – Foundations of Mathematics Unit 4 – Lesson 16
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)rs(r.A.S
r.A.S
rs.A.S
total
2base
cone
+π=
π=
π=
Example Find the surface area of each figure. 10 m a) 8 m b) 22 cm 8cm 7cm c) 8 m 12 m 6m 7.41 m
MFM1P – Foundations of Mathematics Unit 4 – Lesson 16
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Solution Find the surface area of each figure. 10 m a) 8 m
2
2
2
m2.408A.S
2.251157A.S)8)(5(2)5)(14.3(2A.S
rh2r2A.S
=
+=π+=
π+π=
b) 22 cm 8cm 7cm
[ ]( )
2cm772.A.S
)386(2.A.S176154562.A.S
)8)(22()7)(22()8)(7(2.A.S)lhlwwh(2.A.S
=
=++=
++=++=
This is still area so the units are squared.
MFM1P – Foundations of Mathematics Unit 4 – Lesson 16
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c) 8 m 12 m 6m 7.41 m
2m46.308.A.S
7219246.44.A.S
)6)(12()8)(12(22
)41.7)(6(2.A.S
lbls22
bh2.A.S
=
++=
++⎟⎠⎞
⎜⎝⎛=
++⎟⎠⎞
⎜⎝⎛=
Support Questions 3. Calculate the surface area for each of the following objects.
a) b)
There are 2 ls’s because two of the rectangles that make the sides of the prism have the same dimensions
There are two
s'2
bhbecause of the
triangles on each end.
7 cm
8cm
11 cm
6 cm
5 cm
MFM1P – Foundations of Mathematics Unit 4 – Lesson 16
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Support Questions
MFM1P – Foundations of Mathematics Unit 4 – Lesson 16
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Key Question #16 1. Calculate the area for each of the following objects. (8 marks)
a) b) c) 6cm 2cm 9 m
5 cm 3cm 7 m 2 cm d) e) 5.5 cm f) 1.25 cm
9 cm 6 cm
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Key Question #16
2. Calculate the surface area for each of the following objects. (8 marks) 3 m
a) 7 m
b) 20 cm
9cm 5cm
c) 7 m 11 m 5m
d)
5 m
6.54 m
9 cm
11 cm
12.82 cm
MFM1P – Foundations of Mathematics Unit 4 – Lesson 16
Copyright © 2005, Durham Continuing Education Page 17 of 62
Key Question #16 3. Determine the minimum amount of packaging needed to completely cover a
triangular prism Oblerone bar with these dimensions: length 22.5 cm; triangular face has edges 4.5 cm and height 4.0. Express the surface area to the nearest square centimetre. (4 marks)
4. Look at the formula for the surface area of a rectangular prism. How does the
surface area change when the length is doubled? (3 marks) 5. Calculate the surface area of the outside of the solid. (4 marks) 6 cm 18 cm 35 cm 26 cm 6. Calculate the shaded area. (4 marks)
5.5 cm
12 cm