© 2007 - 09 by s-squared, inc. all rights reserved. 1.find the distance, d, and the midpoint, m,...
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© 2007 - 09 by S-Squared, Inc. All Rights Reserved.
1. Find the distance, d, and the midpoint, m, between(4, − 2) and (2, 6)
Distance Formula: d = (x2 – x1)2 + (y2 – y1)2
Substitute
Simplify
(6 – (− 2))2 + (2 – 4)2
(6 + 2)2 + (2 – 4)2
(8)2 + (− 2)2
64 + 4
68 4 • 17 2 17
Algebra I Concept Test # 20 – Geometry Practice Test
﴾ , ﴿2 3
1. Find the distance, d, and the midpoint, m, between(4, − 2) and (2, 6)
Midpoint Formula: m = ﴾ , ﴿ 2
y1 + y2x1 + x2
2
Substitute
Simplify
﴾ , ﴿ 2
− 2 + 6 4 + 2
2m =
﴾ , ﴿ 2
4 6
2m =
m =
Algebra I Concept Test # 20 – Geometry Practice Test
2. Find the perimeter and the area of the given rectangle:
6 cm
14 cm
Area = length • widthPerimeter = 2•length + 2•width
Area = 14 • 6Substitute
Area = 84 cm2
Perimeter = 2 • 14 + 2 • 6
Perimeter = 28 + 12
Perimeter = 40 cm
Algebra I Concept Test # 20 – Geometry Practice Test
3. The perimeter of the rectangle below is 54 cm. Find the length.
Algebra I Concept Test # 20 – Geometry Practice Test
9 cm
lPerimeter = 2•length + 2•width
width = 9 length = l
Perimeter = 54
Formula
Identify variables
Substitute54 = 2 • l + 2 • 9
Simplify54 = 2l + 18
– 18
Subtract – 18
36 = 2lDivide
2 2
18 = l 18cm = length
x2 + 52 = (x + 1)2
x2 + 25 = (x + 1)(x + 1)
4. Find the two unknown side lengths of the given triangle:
x + 1
x
5 ft
Hint: Use Pythagorean’s Theorem to find x.
a2 + b2 = c2
Note: Identify a, b and c
Substitute
b = 5 a = x
c = x + 1
Simplify
Subtract – x2
– 1
25 = 2x + 1
24 = 2x
x2 + 25 = x2 + 2x + 1 – x2
Subtract – 1
Divide 2 2
12 = xNote: Side lengths are x
and x + 1
x = 12 ft x + 1 = 13 ft
Algebra I Concept Test # 20 – Geometry Practice Test
5. Find the perimeter and the area of the given rectangle:
yd
yd
Volume = length • width • height
Volume = • • Substitute
Volume = or yds3
4
35
4
2
1yds.2 4
3
5
4
2
12
Volume = • •4
3
5
4
2
5
Note: Turn mixed number to improper fraction
1
1
1
1
Volume = 2
3
23
2
11
Reduce
Algebra I Concept Test # 20 – Geometry Practice Test
6. Find the circumference and the area of the given circle:
7 m
Circumference = 2πr where r is the radius and let π = 3.14
Substitute Circumference = 2 • 3.14 • 7
Circumference = 6.28 • 7
Circumference = 43.96 m
Note: Identify r
r = 7
Simplify
Algebra I Concept Test # 20 – Geometry Practice Test
6. Find the circumference and the area of the given circle:
7 m
Area = πr2 where r is the radius and let π = 3.14
Substitute Area = 3.14 • 72
Area = 3.14 • 49
Area = 153.86 m2
Note: Identify r
r = 7
Simplify
Algebra I Concept Test # 20 – Geometry Practice Test
7. Find the area and perimeter of the given shape:
3 in
Since the altitude divides the base of the triangle, which is also a side length of the square, in half each segment is 4 inches.
8 in
8 in
Hint: The altitude divides the base of the triangle in half
4 in 4 inAll side lengths of a square are equal so the other side length is 8 inches.
8 inUse the Pythagorean Theorem to find the legs of the triangle.
42 + 32 = c2
16 + 9 = c2
25 = c2Substitute
Simplify
Square Root
5 = c25 = c2
5 in 5 in
Algebra I Concept Test # 20 – Geometry Practice Test
7. Find the area and perimeter of the given shape:
3 in
8 in
8 in
Hint: The altitude divides the base of the triangle in half
4 in 4 in
8 in
5 in 5 in
Note: Perimeter is distance around
Perimeter = 8 + 8 + 8 + 5 + 5
Perimeter = 16 + 8 + 5 + 5
Perimeter = 24 + 5 + 5
Perimeter = 29 + 5
Perimeter = 34 in
Algebra I Concept Test # 20 – Geometry Practice Test
Note: Area of the shape is the sum of the area of the triangle and the area of the square.
7. Find the area and perimeter of the given shape:
3 in
8 in
8 in
Hint: The altitude divides the base of the triangle in half
4 in 4 in
8 in
5 in 5 in
Area(square) = 82
Area(square) = 64 in2
Area of the square is s2
Area of the triangle is b • h21
base = 8 height = 3
Area(triangle) = (8)(3) 21
Area(triangle) = (4)(3)
Area(triangle) = 12 in2
Area(shape) = 64 + 12
Area(shape) = 76 in2
Algebra I Concept Test # 20 – Geometry Practice Test