Hspa Prep Day 2

Spatial Sense and Geometry HSPA PREP with SOLUTIONS 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. ______

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

b. What is the volume of this box? (For example, how many 1 inch cube-shaped ice cubes would fit inside this box?

c. Draw another box with the same volume but a different surface area. (Use whole numbers only.)What are the dimensions of the new box?

of 17 4/4/2018 Spatial Sense and Geometry HSPA PREP with SOLUTIONS Multiple-Choice 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

Figure-A Figure-B Figure-C Figure-D

8. Figure-A 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

of 17 4/4/2018 Spatial Sense and Geometry HSPA PREP with SOLUTIONS 9. Figure-B above: One side of the square = 12 centimeters. What is the area of the shaded region?

Answer: ______

10. Figure-C 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. 4-5 B. 6-8 C.15-16 D. 20-21

11. Figure-D 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?

A. volume of cylinder is twice as large as the volume of the cone B. volume of the cylinder is the same as the area of the cone C. volume of the cone is 1/3 the area of the cylinder D. 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? path width Answer: ______

length A. 497 sq. feet B. 510 sq. feet C. 534 sq. feet D. 527 sq. feet

14. Open-Ended 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.

of 17 4/4/2018 Spatial Sense and Geometry HSPA PREP with SOLUTIONS

15. There is a 3 foot-wide walkway around a small rectangular park. What is the area of the park? 60 A. 220 sq. feet B. 2376 sq. feet C. 2679 sq. feet 50 D. 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

8 6 3 2 9 18 36

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. A rectangle with length 2 and height 50. B. A triangle with base 100 and height 2. C. A trapezoid with one base 15, other base 25 and height 5. D. A square with one side 25.

of 17 4/4/2018 Spatial Sense and Geometry HSPA PREP with SOLUTIONS 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 3 17 Answer: Area = ______sq. yards 5 7 12

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 y-axis and then over the x-axis, what quadrant would it be in?

A. I B. II C. III D. IV

23. Sketch a right triangle in quadrant-II (shade it in), then reflect it over the y-axis. Show where the right angle is in each triangle.

of 17 4/4/2018 Spatial Sense and Geometry HSPA PREP with SOLUTIONS 24. What transformation has occurred from figure-A to figure-B below? Figure-A Figure-B

A. a translation B. a rotation

C. a dilation D. a reflection

25. A circular region rotated 360o 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 (figure-B)?

Figure-A Figure-B A. 45 B. 30 15 C. 15 D. 5 36 12

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. Shade-in the bottom and top of your corrected net. Please use a ruler and pencil.

PRACTICE with TRIANGLES and ANGLES C 29. If  A is a right angle, and m B = 43o, then m  C =

A. 43o B. 47o C. 57o D. 900 o E. 86o A B

30. If m  1=35o and m 2=145o, we say these angles are

A. Complementary B. Supplementary C. Vertical 2 1 D. Perpendicular E. Equivalent

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

32. 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.) 5 4

3 A. B. C. D. 2.5 10 8.6 15 2 8 7 12

1.5 6 7 9

PRACTICE with PARALLEL LINES 1 4 2 3 33. If line a // b and m 1 = 80o, then a 8 5 A. m 2 = 80o B. m  7 = 80o 6 7 C. m  6 = 80o D. m 2 = 80o b

of 17 4/4/2018 Spatial Sense and Geometry HSPA PREP with SOLUTIONS 34. If line a // b and m 3 = 110o which of the following is not true? A. m 8 = 70o B. m  7 = 110o C. m  6 = 70o D. m 2 = 110o 35. If line a // b, name a pair of supplementary angles. . A.  1 and  3 B.  1 and  6 1 4 2 C.  7 and  3 D.  8 and  4 3 a 5 36. Which of the following is equal to m 1 + m  4 +m 7 + m  8? 8 6 7 A. 120o B. 240oC. 180oD. 360o b

Use the figure ABC to the right to answer the following questions. B 37. If this is an isosceles triangle, AB  BC and m ABC = 26o then  BAC = A. 26o B. 64o C. 77o D. 154o 2 A C

38. ABC is an isosceles triangle, AB  BC and m ABC = 26o then B the measure of  2 =

A. 26o B. 154o C. 72o D. 103o 2 A C

39. Use the figure to the right. If AD // BC then

A. triangle AED  triangle BEC B. triangle AED ~ triangle CEB A B C. AE  EC E D. ASA = ASA D 40. m  AED = m BEC because C

A. triangle AED  triangle BEC B. triangle AED ~ triangle BEC C. Corresponding parts of congruent triangles are equal D. Vertical angles are equal

41. Figure ABCD is a parallelogram, therefore all of the following must be true except

A. AE  EC A B B. DE  EB E

D C Page 8 of 17 4/4/2018 Spatial Sense and Geometry HSPA PREP with SOLUTIONS C. m  ADE = m EBC D. AC  DB

PRACTICE with OTHER POLYGONS

42. Which of the following is not always true? (Hint: Use a ruler and draw a diagram first.)

A. The diagonals of a rhombus are perpendicular to each other. B. Opposite sides of rhombus are parallel and  to each other. C. All angles of a rhombus are equal. D. The diagonals of a square are perpendicular to each other.

43. Opposite sides of a trapezoid are always  .

A. 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. 720o B. 360oC. 1,080o D. 1440o

45. The sum of the exterior angles of a pentagon is A. 180o B. 360oC. 1,080o D. 1440o

46. Open-Ended Question: Michelle says you cannot have 2 obtuse angles in a triangle. Janet says you can.

Who is correct? Explain your reasoning.

of 17 4/4/2018 Spatial Sense and Geometry HSPA PREP with SOLUTIONS 47. Open-Ended 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. C

1,000 feet A B D

. If the angle of elevation at point A is 60o and the angle of elevation at point B is 20o, 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)?

8 B 6 A. 25 Sq. units 8 6 B. 36 sq. units C C. 49 sq. units A

D. 100 sq. units 8 6

E. None of the above 6 D 8

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 A. 6 B. 6 ½

C. 7 D. 7 ½ D

E. 8 C E B

50. How many degrees in one of the exterior angles of this regular hexagon shown? (Figure not drawn to scale.) A. 120o B. 270o C. 360o xo D. 450o xo E. None of the above xo

xo xo xo

51. 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

A 2A 2 A 2A A. B. C. D. 2 E. 3.14 3.14 3.14 3.14 3.14

52. In the cube ABCDEFGH with side AB = 2, what is the length of the diagonal AF?

A. 2 E F B. 2 2 C. 2 3 D C D. 4 E. 2 5 H G

of 17 4/4/2018 A B Spatial Sense and Geometry HSPA PREP with SOLUTIONS

B 53. Which is the longest side in triangle ABC? 60o A. AB B. BC C. CA

54. Which is the longest side in triangle ACD? A 100o A. AD B. DC C. CA 30o 20o 55. Which is the longest side in the entire figure? 60o C A. AB B. BC C. CA D. AD E. DC 90o Explain why. D

of 17 4/4/2018 Spatial Sense and Geometry HSPA PREP with SOLUTIONS 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

Arearectangle = 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.)

Areacircle =  r2 = (3.14) (?)2 I don’t know the radius; but, I do not the diameter is 6, therefore the radius is 3. Areacircle =  r2 = (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=32=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= 53=125.

8. C ~75 Area large circle =  r2 = (3.14)(72) = (3.14)(49) = 153.86 Area small circle =  r2 = (3.14)(52) = (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 = square-circle = 144 – 113.04 = 20.96

10. D 20-21 Area circle =  r2 = (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 20-21

11. 20= area shaded region rectangle=(5)(8)=40 – area triangle = ½ ((5)(8)=20; area shade region = 40-20 =20

12. B They are the same. Volume cone = 1/3(area base)(height); Volume cylinder=(area base)(height) Volume cone = 1/3( r2)h Volume of cylinder = ( r2)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 = r2; (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: a2 + b2 = c2; 82 + b2 =102; 64 + b2=100; b(the height)=6; (or b c=10 just recognize that it is a 3,4,5 right triangle with each dimension doubled) a=8 (base)(height) (16)(6) Area of triangle =   48 sq. centimeters 2 2 19. D The square Area square = s2= (25)(25)=625 Area rectangle=(50)(2)=100 bh (100)(2) Area triangle=  =100 2 2 (b  b ) (15  25) Area trapezoid= h 1 2  5  5(20)  100 2 2

20. Perimeter = 7+3+5+12+15+2+17+17=78 yards; Area = 73+122+172=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

of 17 4/4/2018 Spatial Sense and Geometry HSPA PREP with SOLUTIONS 23. Sketches will vary. Reflected triangle should be flipped left-right 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 x be the height of the dilated right triangle. 12 x 1 x  or  ; therefore x = 5 36 15 3 15

28. Answers will vary

Practice with TRIANGLES

29. B 47 o 180 o – 90 o – 43 o = 47 o 30. B Supplementary, 35o + 145 o = 180 o 31. C 60o + 30o = 90 o 32. C All others are in the same ratio: 3:4:5

Practice with PARALLEL LINES 33. B m  7 = 80 o 34. D m3 +m2 must = 180 o ; 110°+110°180° 35. B  1 and  6 36. D 360o 37. C C=77o; 180 - 26=154; 154/2 = 77 o 38. D 103o ; mBCA=77, 180-77=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) (180o). Therefore (6-2)(180) = (4)(180) = 720

Draw a hexagon (a 6-sided 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 360o.

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 180o, you cannot have 2 angles that exceed 180 o.

1,000 47. Let x = distance from A to D. Tan 60 o = , so x = 577.35 feet. x 1,000 Let y = distance from D to B. Tan 20 o = , so y = 2,747.48 feet. y  Therefore x + y = 577.35 + 2,747.48 = 3,325 feet. (A and B are 3,325 feet apart.)

AC  AD2  CD2  (577.35)2  (1000)2 1154 feet (distance from A to C)

BC  BD2  CD2  (2747.48)2  (1000)2  2924 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 142 = 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.

of 17 4/4/2018 Spatial Sense and Geometry HSPA PREP with SOLUTIONS 50. E None of the above. The exterior angle of any regular polygon is 360 360   60o the number of sides 6

51. D a side of the square is 2r, where r is the radius of the circle. Solving A A A=(3.14)r2 for r gives r  , so the side of the square = 2 3.14 3.14

52. C 2 3 Note that in a cube, each edge measures the same, therefore AB=2, BG=2 and then AG= 8 (the hypotenuse of right triangle ABG with both sides = 2.) AG= 8 , FG=2, then AF is the hypotenuse of right triangle AGF and 2 (segment AF )2 = 22 +  8 =4 + 8 = 12; then segment AF = 12  4 3  2 3

53. B. BC It is the side opposite the largest angle (100o) in triangle ABC.

54. C. CA It is the side opposite the largest angle (90o) in triangle ACD.

55. B. BC It is the side opposite the largest angle in both triangles. It is opposite A , which measures100o.

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