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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.



Formulas to be used to calculate area.

2A rπ=



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.



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.



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.



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.



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



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



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 ++⎟

⎠⎞

⎜⎝⎛=



)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



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.



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



Support Questions



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



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



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

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