Name: ________________________________________________ Date: __________________
Examples to Practice
DRAW AND LABEL DIAGRAMS to help you visualize the problem and solutions.
WRITE: Remember to write what you must give as the answer before you begin.
Use the HSPA Reference sheet.

PRACTICE PROBLEMS with SURFACE AREA AND VOLUME:
1. Find the approximate surface area of this can to the nearest square inch. The diameter of the top is about 6 inches and the height of the cylinder is 8 inches.
2. Given a box with the following dimensions: Length = 12, height = 4, depth = 2
Answer a. ________
Answer b. ________
Answer c. ________

What is the surface area? (The area I would cover with wrapping paper without overlapping). Hint: First label the box with the dimensions given.

What is the volume of this box? (For example, how many 1 inch cubeshaped ice cubes would fit inside this box?

Draw another box with the same volume but a different surface area. (Use whole numbers only.)What are the dimensions of the new box?
MultipleChoice Questions
3. Two boxes have the same volume.
Box A: height = 3” width = 2” length = 6”
Box B: height = 4” width = 3” length = __?”
What is the length of Box B?
A. 2” B. 3” C. 4” D. 9”
4. If a cube has a volume of 64cu.cm, the length of ONE EDGE would =
A. 6 cm. B. 4 cm. C. 8 cm. D. 16 cm.
5. If a cube has a volume of 27cu. Ft., then the AREA of one face would =
A. 3 sq. ft. B. 6 sq. ft. C. 9 sq. ft. D. 12 sq. ft.
6. If a cube has a volume of 8 cubic feet, then what is the PERIMETER of one of its faces?
A. 4 feet B. 8 feet C. 12 feet D. 16 feet
7. If the perimeter of one face of a cube is 20 in., what is the VOLUME of that cube?
A. 40 cu. in. B. 60 cu. in. C. 26 cu. in. D. 125 cu. in.
FIND AREA of SHADED REGION
FigureA FigureB FigureC FigureD
8. FigureA above: Diameter of large circle = 14, diameter of small circle = 10. What is the approximate area of the shaded region? (Use =3.14)
A. 232sq.units B.154sq.units C. 75sq. units D. 6 sq. units
9. FigureB above: One side of the square = 12 centimeters. What is the area of the shaded region?
Answer: _______
10. FigureC above: The square measures 6 feet on each side and its diameter measure approximately 8.5 feet. What is the approximate area of the shaded region?
A. 45 B. 68 C.1516 D. 2021
11. FigureD above: The rectangle has a base of 5 inches and is 8 inches tall. What is the area of the shaded region?
Answer: _______
12. How does the volume of a cone with a radius of 9 and a height of 30, compare to the volume of a cylinder with a radius of 9 and a height of 10?

volume of cylinder is twice as large as the volume of the cone

volume of the cylinder is the same as the area of the cone

volume of the cone is 1/3 the area of the cylinder

volume of the cone is 3 times the area of the cylinder
13. There is a rectangular yard that is 30 feet long and 20 feet wide. In the yard there is a
circular fountain with a radius of 2 feet, and a path (as shown) that is 3 feet wide. The area left over will be covered with grass seed.
What is the approximate area of the grass surface?
width Answer: _______
length
A. 497 sq. feet
B. 510 sq. feet
C. 534 sq. feet
D. 527 sq. feet
14. OpenEnded Question
John has 100 feet of fencing and wants to fence off the largest possible space. John says a circle would be best, but his cousin Jack says a square would give you the largest space. Who is correct? Draw diagrams, show work, and explain your solution.
15. There is a 3 footwide walkway around a small rectangular park. What is the area of the park?

220 sq. feet

2
376 sq. feet

2679 sq. feet

3000 sq. feet
16. Which one does not have an area of 36 sq. units?
A. B. C. D.
a right triangle a square a rectangle a scalene triangle
17. What is the perimeter of an equilateral triangle with one side measuring 3.5 inches?
A. 22.25 B. 10.5 C. 1.05 D. 105
18. What is the area of an isosceles triangle given the following information?
base =16cm., one side =10cm. (Hint: Use the Pythagorean formula to find height.)
A. 48 sq. ft. B. 28 sq. ft. C. 24 sq. ft. D. 60 sq. ft.
19. Which figure below does not have an area of 100?

A rectangle with length 2 and height 50.

A triangle with base 100 and height 2.

A trapezoid with one base 15, other base 25 and height 5.

A square with one side 25.
PRACTICE with AREA and PERIMETER of an IRREGULAR SHAPE
20. Find the length of fencing needed to fence this piece of property. All angles are right angles; then find the area of the property.
Answer: Perimeter = ______ yards
Answer: Area = _________sq. yards
PRACTICE with TRANSFORMATIONS
21. Plot the three points A(1,1), B(3,3) and C(6,2). Connect the points to make a triangle ABC
Now translate this image 4 units down.
What are the new coordinates of this translated image?
A (1,1) becomes A( )
B (3,3) becomes B( )
C (5,1) becomes C( )
22. If this hexagon is translated over the yaxis and then over the xaxis, what quadrant would it be in?
A. I B. II C. III D. IV
23. Sketch a right triangle in quadrantII (shade it in), then reflect it over the yaxis. Show where the right angle is in each triangle.
24. What transformation has occurred from figureA to figureB below?
FigureA FigureB
A. a translation B. a rotation
C. a dilation D. a reflection
25. A circular region rotated 360^{o} around its diameter (serving as an axis) generates a
A. cube B. rectangular parallelepiped C. cone D. sphere
26. What is the height of the dilated right triangle (figureB)?
15
FigureA FigureB A. 45 B. 30
36
12
C. 15 D. 5
27. Which NET will make a rectangular covered box? (without overlapping) _____
The bottoms of the boxes have been shaded in to help you ‘see’ the form.
A. B. C.
28. All of the NETS drawn below are incorrect. Select one and correct it so it can be cut and folded into a box with a cover. Shadein the bottom and top of your corrected net. Please use a ruler and pencil.
PRACTICE with TRIANGLES and ANGLES

If A is a right angle, and mB = 43^{o}, then mC =
A. 43^{o }B. 47^{o }C.^{ }57^{o }D.^{ }900^{ o} E.^{ }86^{o }

If m1=35^{o} and m2=145^{o}, we say these angles are
A. Complementary
B. Supplementary
C. Vertical
D. Perpendicular
E. Equivalent

If an angle measures 60^{o}, then we know its complement will measure
A. 60^{o} B. 120^{o} C. 30^{o} D. 20^{o}

Which one of the triangles shown below (A,B,C,D) is NOT similar to the triangle given to the right? (Note: drawings are not to scale.)
A. B. C. D.
PRACTICE with PARALLEL LINES
33. If line a // b and m1 = 80^{o}, then
A. m2 = 80^{o }B. m7 = 80^{o }
^{ }C. m6 = 80^{o }D. m2 = 80^{o}
34. If line a // b and m3 = 110^{o} which of the following is not true?
A. m8 = 70^{o }B. m7 = 110^{o }
^{ }C. m6 = 70^{o }D. m2 = 110^{o}
35. If line a // b, name a pair of supplementary angles.
. A. 1 and 3^{ }B. 1 and 6^{ }
^{ }C. 7 and 3^{ }D. 8 and 4
36. Which of the following is equal to m1 + m4 +m7 + m8?
A. 120^{o }B. 240^{o} C. 180^{o} D. 360^{o }
Use the figure ABC to the right to answer the following questions.
37. If this is an isosceles triangle, and mABC = 26^{o }then BAC =
A. 26^{o }B. 64^{o }
^{ }C. 77^{o }D. 154^{o}
38. ABC is an isosceles triangle, and mABC = 26^{o }then
the measure of 2 =
A. 26^{o }B. 154^{o }
^{ }C. 72^{o }D. 103^{o}
39. Use the figure to the right. If then
A. triangle AED triangle BEC
^{ }B. triangle AED ~ triangle CEB^{ }
^{ }C.
D. ASA = ASA
40. mAED = mBEC because

triangle AED triangle BEC

triangle AED ~ triangle BEC

Corresponding parts of congruent triangles are equal

Vertical angles are equal
41. Figure ABCD is a parallelogram, therefore all of the following must be true except
A.
B.
C. mADE = mEBC
D.
PRACTICE with OTHER POLYGONS
42. Which of the following is not always true? (Hint: Use a ruler and draw a diagram first.)

The diagonals of a rhombus are perpendicular to each other.

Opposite sides of rhombus are parallel and to each other.

All angles of a rhombus are equal.

The diagonals of a square are perpendicular to each other.
43. Opposite sides of a trapezoid are always .

True B. False C. Not enough information given
44. The sum of the interior angles of a hexagon is
(Hint: If you don’t remember the formula, draw a diagram.)
A. 720^{o} B. 360^{o} C. 1,080^{o} D. 1440^{o}
45. The sum of the exterior angles of a pentagon is
A. 180^{o} B. 360^{o} C. 1,080^{o} D. 1440^{o}
46. OpenEnded Question:
Michelle says you cannot have 2 obtuse angles in a triangle. Janet says you can.
Who is correct? Explain your reasoning.
47. OpenEnded Question:
In this diagram, one person is standing at point A, one person is at point B, and both are looking at a helicopter at point C. The helicopter is 1,000 feet above the ground on which the two people are standing. D is the point on the ground directly below the helicopter.

If the angle of elevation at point A is 60^{o }and the angle of elevation at point B is 20^{o}, approximately how many feet apart are A and B?

How much further is the distance from B to the helicopter than from A to the helicopter?
48. Given one rectangle inscribed in a larger rectangle. What is the area of the smaller rectangle (ABCD)?

25 Sq. units

36 sq. units

49 sq. units

100 sq. units

None of the above
49. In the figure shown, ABC is an equilateral triangle. Also, AC = 3 and DB = BE = 1. Find the perimeter of quadrilateral ACED. (Figure not drawn to scale.)
A. 6 B. 6 ½
C. 7 D. 7 ½
E. 8
50. How many degrees in one of the exterior angles of this regular hexagon shown?
(Figure not drawn to scale.)

120^{o}

270^{o}

360^{o}

450^{o}

None of the above
5
A
1. In the figure shown, the area of the inscribed circle is “A.” What is the approximate length of a side of the square? (3.14 is used instead of )
(Figure not drawn to scale.)
A. B. C. D. E.
52. In the cube ABCDEFGH with side AB = 2, what is the length of the diagonal AF?

2

2

2

4

2
53. Which is the longest side in triangle ABC?
A. B. C.
54. Which is the longest side in triangle ACD?
A. B. C.
5
5. Which is the longest side in the entire figure?
A. B. C. D. E.
Explain why.
ANSWER KEY: SOLUTIONS to Spatial Sense and Geometry HSPA PREP examples.
1. First you should sketch the parts of a can the way they would look if you could lay them flat. A can (really a cylinder) is made of two circles and a rectangle.
Diameter=6
^{ 8 } ^{ } ^{Area}rectangle = l w = (circumference)(height)=(d)(h) = [(3.14)(6)](8) =150.72 sq. in.
(Notice that the circumference of the circle is really the same as the length of the rectangle.)
^{Area}circle = r^{2} = (3.14) (?)^{2} I don’t know the radius; but, I do not the diameter is 6, therefore the radius is 3.
^{ }^{Area}circle = r^{2 }= (3.14) (3)^{2} = (3.14)(9) = 28.26 sq. inches (area of one circle)
The total area of the can is the area of the rectangle + the area of two circles.
150.72 + 2(28.26) = 207.24 or approximately ~ 207 in.^{2}
2a. lateral surface area = 160sq.units (12x4)(2)+(12x2)(2) = (2x4)(2) = 160
2b. volume = 96 cubic units (12)(4)(2) = 96
2c. Answers will vary but they show 3 sides whose product is 96.
Volumes: (3)(4)(8)=96 (4)(4)(6)=96 (2)(2)(24)=96 (1)(2)(48)=96
Surf.areas: 62 112 56 290
2(3)(4)+2(3)(8)+2(4)(8) 2(4)(4)+2(4)(4)+2(4)(6)
3. B3 (3)(2)(6) = 36; (4)(3)(?)=36 (12)(?)=36 (12)(3)=36
4. B4 (side)^{3}=64, side=4 (the length of one edge)
5. C9 Volume=27, (side)^{3}=27, side=3 Area of one face = (side)^{2}=3^{2}=9
6. B 8 Volume = (side)^{3}=8, side=2; Perimeter of one face = 4(2) = 8
7. D 125 Perimeter=20, Perimeter=4(side); one side=20/4=5 Volume=(side)^{3}= 5^{3}=125.
8. C ~75 Area large circle = r^{2 } = (3.14)(7^{2}) = (3.14)(49) = 153.86
Area small circle = r^{2 } = (3.14)(5^{2}) = (3.14)(25) = 78.50
Area larger circle less area small circle = area shaded region = 75.36
9. ~20.96 SQ. cm. area sq.=(12)(12)=144; area circle=(3.14)(6)^{2}=113.04
area shaded region = squarecircle = 144 – 113.04 = 20.96
10. D 2021 Area circle = r^{2 } = (3.14)(4.25)^{2} = (3.14)(18.06) = 56.7
Area square = (6)(6) = 36; area shaded region = 56.7 less 36 = 20.7 or between 2021
11. 20= area shaded region
rectangle=(5)(8)=40 – area triangle = ½ ((5)(8)=20; area shade region = 4020 =20
12. B They are the same.
Volume cone = 1/3(area base)(height); Volume cylinder=(area base)(height)
Volume cone = 1/3(r^{2})h Volume of cylinder = (r^{2})h
Volume this cone = 1/3(3.14)(9)^{2}(30) Volume this cylinder = (3.14)(9)^{2}(10)
Volume this cone = (3.14)(9)^{2}(10) Volume this cylinder = (3.14)( 9)^{2}(10)
13. A 497 is the approximate area of the grass part of the park.
area rectangle=(30)(20) = 600; area path = (3)(30) = 90; area circle = (3.14)(2)(2) =12.56
area shaded region = 600 less 90, less 12.56 = 497.44 sq.
14. Area of circle would be greater than area of square using 100 ft. of fencing.
Use the circumference of the circle to find the radius of the circle, then find area.
Circumference = 100 =d; 100 = (3.14)(d); 31.847 = d; 15.9235 = r
Area of circle =r^{2}; (3.14)(15.9235)^{2} is approximately 796 sq. feet
Use the perimeter of the square to find the side, then find area.
Perimeter = 100 = 4(side), so side = 25 feet
Area of square = (side)^{2}; (25)^{2 }= 625 sq. feet
15. B 2,376sq.ft.
shorter side=50  (3+3) = 44; longer side = 60  (3+3) = 54; area = (44)(54) = 2,376
16. The rectangle does not have an area of 36 sq. units
Area right triangle = ½(9)(8) = ½(72) = 36;
Area square = (6)(6) = 36
Area rectangle = (18)(3) = 54
Area acute triangle = ½(2)(36) = 36
17. B 10.5 Perimeter of an equilateral triangle = 3(side) = 3(3.5) = 10.5
18. A 48 sq.cm.
Find height: a^{2} + b^{2 }= c^{2}; 8^{2} + b^{2 }=10^{2}; ^{ }64 + b^{2}=100; b(the height)=6; (or
just recognize that it is a 3,4,5 right triangle with each dimension doubled)
Area of triangle = sq. centimeters
19. D The square Area square = s^{2}= (25)(25)=625
Area rectangle=(50)(2)=100
Area triangle==100 Area trapezoid=
20. Perimeter = 7+3+5+12+15+2+17+17=78 yards; Area = 73+122+172=79 square yards
21. A(1,1) becomes (1,3); B(2,3) becomes (2,1); C(6,2) becomes C(6, 2)
22. D quadrant IV
23. Sketches will vary. Reflected triangle should be flipped leftright and located in quadrant I
24. C Dilation; only the size changed.
25. D Sphere
26. D Dilated shapes are in the same proportion as the original shape; they are similar.
In this case the smaller shape is 1/3 the size of the original shape.
Let be the height of the dilated right triangle.
; therefore x = 5
27. A
28. Answers will vary
Practice with TRIANGLES
29. B 47^{ o} 180^{ o} – 90^{ o} – 43^{ o} = 47^{ o}
30. B Supplementary, 35^{o} + 145^{ o} = 180^{ o}
31. C 60^{o} + 30^{o} = 90^{ o}
32. C All others are in the same ratio: 3:4:5
Practice with PARALLEL LINES
33. B m7 = 80^{ o}
34. D m3 +m2 must = 180^{ o} ; 110°+110°180°
35. B 1 and 6
36. D 360^{o}
37. C C=77^{o}; 180  26=154; 154/2 = 77^{ o}
38. D 103^{o }; mBCA=77, 18077=103^{ o}
39. B triangle AED triangle CEB
40. D Vertical angles are =.
41. D Diagonals of a parallelogram are not necessarily congruent.
Practice with OTHER POLYGONS
42. C All s of a rhombus are NOT =, only opposite s are =.
43. B False; opposite sides of a trapezoid are not always equal.
44. A 720^{ o} is the sum of the interior angles of a hexagon. A hexagon is a 6 sided polygon. Formula to find degrees in exterior angle = (number of sides 2) (180^{o}). Therefore (62)(180) = (4)(180) = 720
Or
Draw a hexagon (a 6sided polygon) and count how many triangles you can draw from one vertex. Count the number of triangles and multiply by 180.
45. B The sum of the exterior angles of any polygon is 360^{o}.
46. Michelle is correct. An obtuse angle is one that measures >90^{ o}.
The sum of 2 angles >90 will be > 180. Since 3 angles in a triangle must add to 180^{o}, you cannot have 2 angles that exceed 180^{ o}.
47. Let x = distance from A to D. Tan 60^{ o} = , so x = 577.35 feet.
Let y = distance from D to B. Tan 20^{ o} = , so y = 2,747.48 feet.

Therefore x + y = 577.35 + 2,747.48 = 3,325 feet.
(A and B are 3,325 feet apart.)
feet (distance from A to C)
feet (distance from B to C)

BC – AC 1,770 feet further from B than A to the helicopter.
(Distance from B) less the (Distance from A) = 2,924 – 1,154 =1,770
48. D 100 sq. units Notice that each triangle is a 6, 8, 10 right triangle. Therefore, the each side of the smaller square is 10 units long. Area of a rectangle is l w = (10)(10)=100
OR Large rectangle is a square with side length 14. Area of square is 14^{2} = 196
area of one of the triangles is ½(6)(8) = 24. 196 – 4(24) = 100
49. E 8 Starting with AC, the lengths of the 4 sides of ACED are 3+2+1+2.
Use what you know about similar triangles to help you here.
50. E None of the above. The exterior angle of any regular polygon is
60^{o}
51. D a side of the square is 2r, where r is the radius of the circle. Solving A=(3.14)r^{2} for r gives , so the side of the square =
52. C 2 Note that in a cube, each edge measures the same, therefore AB=2, BG=2 and then AG= (the hypotenuse of right triangle ABG with both sides = 2.)
AG=, FG=2, then AF is the hypotenuse of right triangle AGF and
(segment)^{2 }= 2^{2} + =4 + 8 = 12;
then segment =
53. B. It is the side opposite the largest angle (100^{o}) in triangle ABC.
54. C. It is the side opposite the largest angle (90^{o}) in triangle ACD.
55. B. It is the side opposite the largest angle in both triangles. It is opposite
, which measures100^{o}.
Page of 4/21/2016
