Perimeters and Areas of Similar Figures

Name: _________________________________________________ Period: ___________ Date: ________________
Perimeters and Areas of Similar Figures Assignment
The figures in each pair are similar. Compare the first figure to the second. Find the scale factor and give the ratio of
the perimeters and the ratio of the areas.
𝟏𝟐
1.
2.
𝟐
πŸ”
𝐒𝐂𝐀𝐋𝐄
π…π€π‚π“πŽπ‘
π‘π€π“πˆπŽ πŽπ…
ππ„π‘πˆπŒπ„π“π„π‘π’
π‘π€π“πˆπŽ πŽπ…
𝐀𝐑𝐄𝐀𝐒
𝐒𝐂𝐀𝐋𝐄
π…π€π‚π“πŽπ‘
3.
πŸπŸ–
π‘π€π“πˆπŽ πŽπ…
ππ„π‘πˆπŒπ„π“π„π‘π’
π‘π€π“πˆπŽ πŽπ…
𝐀𝐑𝐄𝐀𝐒
4.
πŸ•. 𝟐
𝟐. πŸ’
πŸ’πŸ“
𝐒𝐂𝐀𝐋𝐄
π…π€π‚π“πŽπ‘
π‘π€π“πˆπŽ πŽπ…
ππ„π‘πˆπŒπ„π“π„π‘π’
π‘π€π“πˆπŽ πŽπ…
𝐀𝐑𝐄𝐀𝐒
πŸ—
𝐒𝐂𝐀𝐋𝐄
π…π€π‚π“πŽπ‘
π‘π€π“πˆπŽ πŽπ…
ππ„π‘πˆπŒπ„π“π„π‘π’
π‘π€π“πˆπŽ πŽπ…
𝐀𝐑𝐄𝐀𝐒
The figures in each pair are similar. The area of one figure is given. Find the area of the other figure
5.
6.
πŸ”π’Ž
πŸ“πŸ’ π’Ž
πŸ’πŸŽ π’Šπ’.
π‘¨πŸ = 𝟏𝟐 π’ŽπŸ
π‘¨πŸ =?
π‘¨πŸ =?
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πŸ– π’Šπ’.
π‘¨πŸ = πŸ‘πŸπ’Šπ’.𝟐
Name: _________________________________________________ Period: ___________ Date: ________________
Perimeters and Areas of Similar Figures Assignment
7.
8.
𝟐𝟏. πŸ” 𝒇𝒕
πŸ‘. πŸ”π’‡π’•
πŸπŸ“ π’„π’Ž
π‘¨πŸ = πŸπŸπŸ”πŸŽπ’‡π’•πŸ
πŸ“ π’„π’Ž
π‘¨πŸ =?
π‘¨πŸ =?
π‘¨πŸ = πŸπŸπŸ“π’„π’ŽπŸ
Find the scale factor and the ratio of perimeters for each pair of similar figures.
9.
10.
π‘¨πŸ = πŸ‘πŸ”π’„π’ŽπŸ
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π‘¨πŸ = πŸπŸ’πŸ’π’„π’ŽπŸ
π‘¨πŸ = πŸ—βˆšπŸ‘π’„π’ŽπŸ
2
π‘¨πŸ = πŸπŸπŸβˆšπŸ‘π’„π’ŽπŸ
Name: _________________________________________________ Period: ___________ Date: ________________
Perimeters and Areas of Similar Figures Assignment
Find the value of x!
11.
12.
πŸπŸ– π’Ž
πŸ‘. πŸ” π’Ž
𝒙
πŸπŸ“ π’Ž
𝒙
πŸ‘πŸ π’Šπ’.
𝟐𝟏 π’Šπ’.
Find the scale factor and the ratio of perimeters for each pair of similar figures.
13. Two regular hexagons with areas:
π‘¨πŸ = πŸ•πŸ π’„π’ŽπŸ
π‘¨πŸ = πŸ“πŸŽ π’„π’ŽπŸ
14. Two regular octagons with areas:
π‘¨πŸ = πŸπŸ’πŸ’ π’ŽπŸ
π‘¨πŸ = πŸ”πŸ’ π’ŽπŸ
15. Two equilateral triangles with areas:
π‘¨πŸ = πŸπŸ—πŸ”βˆšπŸ‘ π’‡π’•πŸ
π‘¨πŸ = πŸπŸπŸ“βˆšπŸ‘ π’‡π’•πŸ
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3
πŸπŸ”πŸ– π’Šπ’.
Name: _________________________________________________ Period: ___________ Date: ________________
Perimeters and Areas of Similar Figures Assignment
Solve each problem!
16. The area of a regular octagon is πŸ’πŸ“ π’ŽπŸ . What is the
𝟏
area of a regular octagon with sides πŸ‘ the length of
sides of the larger octagon?
17. The similarity ratio of two similar polygons is πŸ‘: πŸ“.
The perimeter of the larger polygon is πŸπŸ“πŸŽ π’„π’Ž. What
is the perimeter of the smaller polygon?
18. The lengths of the sides of two triangles are in the
similarity ratio of πŸ’: πŸ“. If the area of the larger
triangle is πŸπŸπŸ“ π’ŽπŸ , find the area of the smallest
triangle.
19. The ratio of the perimeters of two similar hexagons is
πŸ“: πŸ–.The area of larger hexagon is πŸ‘πŸπŸŽ π’Šπ’.𝟐 What is
the area of the smaller hexagon?
20. Two similar parallelograms have areas πŸ•πŸ π’ŽπŸ and
πŸ‘πŸ π’ŽπŸ . The height of the larger parallelogram is
𝟏𝟐 π’Ž. What are the lengths of the bases of both
parallelograms?
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Name: _________________________________________________ Period: ___________ Date: ________________
Perimeters and Areas of Similar Figures Assignment
ANSWERS
The figures in each pair are similar. Compare the first figure to the second. Find the scale factor and give the ratio of
the perimeters and the ratio of the areas.
𝟏𝟐
1.
2.
𝟐
πŸ”
𝒂
𝟐
𝟏
=
=
𝒃 𝟏𝟐 πŸ”
𝐒𝐂𝐀𝐋𝐄
π…π€π‚π“πŽπ‘
𝑷𝒂 𝟏
=
𝑷𝑩 πŸ”
π‘π€π“πˆπŽ πŽπ…
ππ„π‘πˆπŒπ„π“π„π‘π’
𝒂
πŸ”
𝟏
=
=
𝒃 πŸπŸ– πŸ‘
𝑨𝒂 𝟏𝟐
𝟏
= 𝟐=
𝑨𝑩 πŸ”
πŸ‘πŸ”
𝐒𝐂𝐀𝐋𝐄
π…π€π‚π“πŽπ‘
π‘π€π“πˆπŽ πŽπ…
𝐀𝐑𝐄𝐀𝐒
3.
πŸπŸ–
𝑷𝒂 𝟏
=
𝑷𝑩 πŸ‘
π‘π€π“πˆπŽ πŽπ…
ππ„π‘πˆπŒπ„π“π„π‘π’
𝑨𝒂 𝟏𝟐 𝟏
=
=
𝑨𝑩 πŸ‘πŸ πŸ—
π‘π€π“πˆπŽ πŽπ…
𝐀𝐑𝐄𝐀𝐒
4.
πŸ•. 𝟐
𝟐. πŸ’
πŸ’πŸ“
𝒂 πŸ•. 𝟐 πŸ‘
=
=
𝒃 𝟐. πŸ’ 𝟏
𝐒𝐂𝐀𝐋𝐄
π…π€π‚π“πŽπ‘
𝑷𝒂 πŸ‘
=
𝑷𝑩 𝟏
π‘π€π“πˆπŽ πŽπ…
ππ„π‘πˆπŒπ„π“π„π‘π’
πŸ—
𝒂 πŸ’πŸ“ πŸ“
=
=
𝒃
πŸ—
𝟏
𝑨𝒂 πŸ‘πŸ πŸ—
=
=
𝑨𝑩 𝟏𝟐 𝟏
𝐒𝐂𝐀𝐋𝐄
π…π€π‚π“πŽπ‘
π‘π€π“πˆπŽ πŽπ…
𝐀𝐑𝐄𝐀𝐒
𝑷𝒂 πŸ“
=
𝑷𝑩 𝟏
π‘π€π“πˆπŽ πŽπ…
ππ„π‘πˆπŒπ„π“π„π‘π’
𝑨𝒂 πŸ“πŸ πŸπŸ“
=
=
𝑨𝑩 𝟏𝟐
𝟏
π‘π€π“πˆπŽ πŽπ…
𝐀𝐑𝐄𝐀𝐒
The figures in each pair are similar. The area of one figure is given. Find the area of the other figure
5.
6.
πŸ”π’Ž
πŸ“πŸ’ π’Ž
πŸ’πŸŽ π’Šπ’.
π‘¨πŸ = 𝟏𝟐 π’ŽπŸ
π‘¨πŸ =?
π‘¨πŸ =?
𝒂
πŸ”π’Ž
𝟏
=
=
𝒃 πŸ“πŸ’ π’Ž
πŸ—
π‘¨πŸ 𝟏𝟐
𝟏
=
=
π‘¨πŸ πŸ—πŸ πŸ–πŸ
πŸπŸπ’ŽπŸ
𝟏
=
π‘¨πŸ
πŸ–πŸ
π‘¨πŸ = πŸ–πŸ βˆ— πŸπŸπ’ŽπŸ
π‘¨πŸ = πŸ—πŸ•πŸ π’ŽπŸ
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πŸ– π’Šπ’.
π‘¨πŸ = πŸ‘πŸπ’Šπ’πŸ
𝒂
πŸ’πŸŽ π’Šπ’. πŸ“
=
=
𝒃
πŸ– π’Šπ’.
𝟏
π‘¨πŸ πŸ“πŸ πŸπŸ“
=
=
π‘¨πŸ 𝟏𝟐
𝟏
π‘¨πŸ
πŸπŸ“
=
πŸ‘πŸ π’Šπ’.𝟐
𝟏
π‘¨πŸ = πŸ‘πŸ π’Šπ’.𝟐 βˆ— πŸπŸ“
π‘¨πŸ = πŸ–πŸŽπŸŽ π’Šπ’.𝟐
5
Name: _________________________________________________ Period: ___________ Date: ________________
Perimeters and Areas of Similar Figures Assignment
7.
8.
𝟐𝟏. πŸ” 𝒇𝒕
πŸ‘. πŸ”π’‡π’•
πŸπŸ“ π’„π’Ž
π‘¨πŸ = πŸπŸπŸ”πŸŽπ’‡π’•πŸ
πŸ“ π’„π’Ž
π‘¨πŸ =?
π‘¨πŸ = πŸπŸπŸ“π’„π’ŽπŸ
π‘¨πŸ =?
𝒂 πŸπŸ“ πŸ“
=
=
𝒃
πŸ“
𝟏
π‘¨πŸ πŸ“πŸ πŸπŸ“
=
=
π‘¨πŸ 𝟏𝟐
𝟏
π‘¨πŸ
πŸπŸ“
=
𝟐
πŸπŸπŸ“ π’„π’Ž
𝟏
𝒂 𝟐𝟏. πŸ” 𝒇𝒕 πŸ”
=
=
𝒃
πŸ‘. πŸ” 𝒇𝒕
𝟏
𝟐
π‘¨πŸ πŸ”
πŸ‘πŸ”
= 𝟐=
π‘¨πŸ 𝟏
𝟏
πŸπŸπŸ”πŸŽ π’‡π’•πŸ πŸ‘πŸ”
=
π‘¨πŸ
𝟏
πŸπŸπŸ”πŸŽπ’‡π’•πŸ
π‘¨πŸ =
πŸ‘πŸ”
π‘¨πŸ = πŸπŸ“ βˆ— πŸπŸπŸ“ π’„π’ŽπŸ
π‘¨πŸ = πŸ“πŸ”πŸπŸ“ π’„π’ŽπŸ
π‘¨πŸ = πŸ”πŸŽ π’‡π’•πŸ
Find the scale factor and the ratio of perimeters for each pair of similar figures.
9.
10.
π‘¨πŸ = πŸ‘πŸ”π’„π’ŽπŸ
π‘¨πŸ = πŸπŸ’πŸ’π’„π’ŽπŸ
π‘¨πŸ = πŸ—βˆšπŸ‘π’„π’ŽπŸ
π‘¨πŸ
πŸ‘πŸ” π’„π’ŽπŸ
𝟏
=
=
𝟐
π‘¨πŸ πŸπŸ’πŸ’ π’„π’Ž
πŸ’
π‘·πŸ βˆšπ‘¨πŸ √𝟏 𝟏
=
=
=
π‘·πŸ βˆšπ‘¨πŸ βˆšπŸ’ 𝟐
𝒂 π‘·πŸ 𝟏
=
=
𝒃 π‘·πŸ 𝟐
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π‘¨πŸ = πŸπŸπŸβˆšπŸ‘π’„π’ŽπŸ
π‘¨πŸ
πŸ—βˆšπŸ‘ π’„π’ŽπŸ
πŸ—
=
=
𝟐
π‘¨πŸ πŸπŸπŸβˆšπŸ‘ π’„π’Ž
𝟏𝟐𝟏
π‘·πŸ βˆšπ‘¨πŸ
πŸ‘
βˆšπŸ—
=
=
=
π‘·πŸ βˆšπ‘¨πŸ √𝟏𝟐𝟏 𝟏𝟏
𝒂 π‘·πŸ
πŸ‘
=
=
𝒃 π‘·πŸ 𝟏𝟏
6
Name: _________________________________________________ Period: ___________ Date: ________________
Perimeters and Areas of Similar Figures Assignment
Find the value of x !
11
.
πŸ‘. πŸ” π’Ž
12
.
πŸπŸ– π’Ž
𝒙
πŸπŸ“ π’Ž
𝒙
πŸ‘πŸ π’Šπ’.
𝟐𝟏 π’Šπ’.
𝒂 πŸ‘. πŸ” π’Ž 𝟏
=
=
𝒃
πŸπŸ– π’Ž πŸ“
𝒙
𝟏
=
πŸπŸ“π’Ž πŸ“
𝒙=πŸ“π’Ž
πŸπŸ”πŸ– π’Šπ’.
𝒂
𝟐𝟏 π’Šπ’.
𝟏
=
=
𝒃 πŸπŸ”πŸ– π’Šπ’. πŸ–
𝒙
𝟏
=
πŸ‘πŸ π’Šπ’. πŸ–
𝒙 = πŸ’ π’Šπ’.
Find the scale factor and the ratio of perimeters for each pair of similar figures.
13. Two regular hexagons with areas:
π‘¨πŸ = πŸ•πŸ π’„π’ŽπŸ
π‘¨πŸ = πŸ“πŸŽ π’„π’ŽπŸ
π‘¨πŸ πŸ•πŸπ’„π’ŽπŸ πŸ‘πŸ”
=
=
π‘¨πŸ πŸ“πŸŽπ’„π’ŽπŸ πŸπŸ“
π‘·πŸ βˆšπŸ‘πŸ” πŸ”
=
=
π‘·πŸ βˆšπŸπŸ“ πŸ“
𝒂 π‘·πŸ πŸ”
=
=
𝒃 π‘·πŸ πŸ“
14. Two regular octagons with areas:
π‘¨πŸ = πŸπŸ’πŸ’ π’ŽπŸ
π‘¨πŸ = πŸ”πŸ’ π’ŽπŸ
π‘¨πŸ πŸπŸ’πŸ’ π’ŽπŸ πŸ—
=
=
π‘¨πŸ
πŸ”πŸ’ π’ŽπŸ
πŸ’
π‘·πŸ βˆšπŸ— πŸ‘
=
=
π‘·πŸ βˆšπŸ’ 𝟐
𝒂 π‘·πŸ πŸ‘
=
=
𝒃 π‘·πŸ 𝟐
π‘¨πŸ πŸπŸ—πŸ”βˆšπŸ‘ π’‡π’•πŸ πŸπŸ—πŸ”
=
=
π‘¨πŸ πŸπŸπŸ“βˆšπŸ‘ π’‡π’•πŸ πŸπŸπŸ“
π‘·πŸ βˆšπŸπŸ—πŸ” πŸπŸ’
=
=
π‘·πŸ βˆšπŸπŸπŸ“ πŸπŸ“
𝒂 π‘·πŸ πŸπŸ’
=
=
𝒃 π‘·πŸ πŸπŸ“
15. Two equilateral triangles with areas:
π‘¨πŸ = πŸπŸ—πŸ”βˆšπŸ‘ π’‡π’•πŸ
π‘¨πŸ = πŸπŸπŸ“βˆšπŸ‘ π’‡π’•πŸ
Copyright © GeometryCoach.com
7
Name: _________________________________________________ Period: ___________ Date: ________________
Perimeters and Areas of Similar Figures Assignment
Solve each problem!
16. The area of a regular octagon is πŸ’πŸ“ π’ŽπŸ . What is the
𝟏
area of a regular octagon with sides πŸ‘ the length of
sides of the larger octagon?
π‘¨πŸ = πŸ’πŸ“ π’ŽπŸ π‘¨πŸ = ?
𝟏
π’”πŸ = π’”πŸ
πŸ‘
𝒂 π’”πŸ
=
𝒃 π’”πŸ
𝒂
π’”πŸ
πŸ‘
=
=
𝟏
𝒃
𝟏
𝒔
πŸ‘ 𝟏
π‘¨πŸ π’‚πŸ πŸ‘πŸ πŸ—
=
=
=
π‘¨πŸ π’ƒπŸ 𝟏𝟐 𝟏
πŸ’πŸ“ π’ŽπŸ πŸ—
=
π‘¨πŸ
𝟏
π‘¨πŸ = πŸ“ π’ŽπŸ
17. The similarity ratio of two similar polygons is πŸ‘: πŸ“.
The perimeter of the larger polygon is πŸπŸ“πŸŽ π’„π’Ž. What
is the perimeter of the smaller polygon?
𝒂 πŸ‘
=
𝒃 πŸ“
18. The lengths of the sides of two triangles are in the
similarity ratio of πŸ’: πŸ“. If the area of the larger
triangle is πŸπŸπŸ“ π’ŽπŸ , find the area of the smallest
triangle.
𝒂 πŸ’
=
𝒃 πŸ“
19. The ratio of the perimeters of two similar hexagons is
πŸ“: πŸ–.The area of larger hexagon is πŸ‘πŸπŸŽ π’Šπ’.𝟐 What is
the area of the smaller hexagon?
π‘·πŸ πŸ“
=
π‘·πŸ πŸ–
π‘·πŸ = πŸπŸ“πŸŽ π’„π’Ž π‘·πŸ =?
π‘·πŸ 𝒂
=
π‘·πŸ 𝒃
π‘·πŸ
πŸ‘
=
πŸπŸ“πŸŽ π’„π’Ž πŸ“
πŸπŸ“πŸŽ π’„π’Ž βˆ— πŸ‘
π‘·πŸ =
πŸ“
π‘·πŸ = πŸ—πŸŽ π’„π’Ž
π‘¨πŸ = πŸπŸπŸ“ π’ŽπŸ
π‘¨πŸ 𝒂
=
π‘¨πŸ 𝒃 𝟐
π‘¨πŸ
πŸ’πŸ
=
πŸπŸπŸ“ π’ŽπŸ πŸ“πŸ
π‘¨πŸ
πŸπŸ”
=
𝟐
πŸπŸπŸ“ π’Ž
πŸπŸ“
π‘¨πŸ = πŸ–πŸŽ π’ŽπŸ
π‘¨πŸ = πŸ‘πŸπŸŽ π’Šπ’.𝟐
𝒂 πŸ“
=
𝒃 πŸ–
π‘¨πŸ π’‚πŸ
=
π‘¨πŸ 𝒃 𝟐
π‘¨πŸ
πŸ“πŸ
=
πŸ‘πŸπŸŽ π’Šπ’.𝟐 πŸ–πŸ
πŸ‘πŸπŸŽ π’Šπ’.𝟐 βˆ— πŸπŸ“
πŸ”πŸ’
π‘¨πŸ = πŸπŸπŸ“ π’Šπ’.𝟐
π‘¨πŸ =
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π‘¨πŸ =?
𝟐
π‘¨πŸ =?
Name: _________________________________________________ Period: ___________ Date: ________________
Perimeters and Areas of Similar Figures Assignment
20. Two similar parallelograms have areas πŸ•πŸ π’ŽπŸ and
πŸ‘πŸ π’ŽπŸ . The height of the larger parallelogram is
𝟏𝟐 π’Ž. What are the lengths of the bases of both
parallelograms?
π‘¨πŸ = πŸ•πŸ π’ŽπŸ π‘¨πŸ = πŸ‘πŸ π’ŽπŸ
π’ƒπŸ =? π’ƒπŸ =?
π’‰πŸ = 𝟏𝟐 π’Ž
π‘¨πŸ = π’ƒπŸ βˆ— π’‰πŸ
πŸ•πŸ π’ŽπŸ = π’ƒπŸ βˆ— 𝟏𝟐 π’Ž
π’ƒπŸ = πŸ” π’Ž
π‘¨πŸ 𝒃 𝟏 𝟐
=
π‘¨πŸ 𝒃 𝟐 𝟐
πŸ•πŸ
πŸ”πŸ
= 𝟐
πŸ‘πŸ
π’ƒπŸ
πŸ•πŸ
πŸ‘πŸ”
= 𝟐
πŸ‘πŸ
π’ƒπŸ
πŸ‘πŸ βˆ— πŸ‘πŸ”
𝟐
π’ƒπŸ =
πŸ•πŸ
π’ƒπŸ 𝟐 = πŸπŸ”
π’ƒπŸ = βˆšπŸπŸ”
π’ƒπŸ = πŸ’ π’Ž
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