## - Sample Math Practice Problems

The math problems below can be generated by MathScore.com, a math practice program for schools and individual families. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. In the main program, all problems are automatically graded and the difficulty adapts dynamically based on performance. Answers to these sample questions appear at the bottom of the page. This page does not grade your responses.

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### Complexity=1

Find the area of the highlighted region. Gives answers to the nearest hundreth. Use 3.14 for pi
 1.   The triangle inside is a right triangle. Area: 2.   The figure outside is a square. Area:

### Complexity=2

Find the area of the highlighted region. Gives answers to the nearest hundreth. Use 3.14 for pi
 1.   The figure outside is a square. Area: 2.   The triangle inside is equilateral. Area:

### Complexity=3

Find the area of the highlighted region. Gives answers to the nearest hundreth. Use 3.14 for pi
 1.   The figure inside is a square. Area: 2.   The triangle inside is a right triangle. Area:

### Complexity=1

Find the area of the highlighted region. Gives answers to the nearest hundreth. Use 3.14 for pi
1
The triangle inside is a right triangle.
Area:
Solution
Area = Area(Circle) - Area(Triangle)
All triangles with the longest side as the diameter of a circle are right triangles.
Diameter2 = Hypotenuse2 = (side 1)2 + (side 2)2 = 42 + 32 = 16 + 9 = 25
Area of circle = πr2 = πd2 /4 = 3.14 × 25/4 = 19.625
Area of triangle = bh/2 = 4 × 3 / 2 = 6
Area = Area of Circle - Area of triangle = 19.625 - 6 = 13.63
2
The figure outside is a square.
Area:
Solution
Area = Area(Square) - Area(Circle)
Side length of square = Diameter = 3
Area of circle = πr2 = πd2/4 = 3.14 × 32/4 = 7.065
Area of square = s2 = 32 = 9
Area = Area(square) - Area(circle) = 9 - 7.065 = 1.94

### Complexity=2

Find the area of the highlighted region. Gives answers to the nearest hundreth. Use 3.14 for pi
1
The figure outside is a square.
Area:
Solution
Area = Area(Square) - Area(Circle)
Diameter of circle = 2 × 5 = 10
Side length of square = Diameter = 10
Area of circle = πr2 = 3.14 × 52 = 78.5
Area of square = s2 = 102 = 100
Area = Area(square) - Area(circle) = 100 - 78.5 = 21.5
2
The triangle inside is equilateral.
Area:
Solution
Area = Area(circle) - Area(Triangle)
Area(circle) = πr2 = 3.14 × 92 = 254.34
Side length of triangle = √3 × radius = √3 × 9 = 15.59
Area of equilateral triangle = √(3)s2 /4 = √(3) × 15.592/4 = 105.22
Area = Area of circle - Area of triangle = 254.34 - 105.22 = 149.12

### Complexity=3

Find the area of the highlighted region. Gives answers to the nearest hundreth. Use 3.14 for pi
1
The figure inside is a square.
Area:
Solution
Area = Area(Circle) - Area(Square)
Diagonal of square = 2 × 7 =14
Side of square = Diagonal/(√2)
Area of square = s2 = Diagonal2 /2 = 142/2 = 98
Area of circle = πr2 = 3.14 × 72 = 153.86
Area = Area of circle - Area of square = 153.86 - 98 = 55.86
2
The triangle inside is a right triangle.
Area:
Solution
Area = Area(Circle) - Area(Triangle)
All triangles with the longest side as the diameter of a circle are right triangles.
Diameter2 = Hypotenuse2 = (side 1)2 + (side 2)2 = 32 + 122 = 9 + 144 = 153
Area of circle = πr2 = πd2 /4 = 3.14 × 153/4 = 120.105
Area of triangle = bh/2 = 3 × 12 / 2 = 18
Area = Area of Circle - Area of triangle = 120.105 - 18 = 102.11