# Geometry Semester 2 Model Problems (CA Essential Standards) Part 2

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Geometry Semester 2 Model Problems (CA Essential Standards) Part 2
PowerPoint by Kenneth C Utt John Kimball High -- TUSD

Model Problem #33 44o + 95o + xo = 180o 139o + xo = 180o
In the diagram shown, mCBD = 95˚. What is the measure of CDB ? 44o + 95o + xo = 180o 139o + xo = 180o xo = 180o – 139o xo = 41o 95o

Model Problem #34 180 – (90 + 31) = MPN 180 – 121 = 59o
In the diagram shown, P is a point on ML. What is the measure of the angle marked X? 180 – ( ) = MPN 180 – 121 = 59o 180 – ( ) = LPQ 180 – 114 = 66o 180 – ( ) = PQL 180 – 156 = 24o 180 – 24 = x = 156o 59o 66o 24o

Model Problem #35 A’ =(-1,-3) B’ = (-4,-4) C’ = (-3,-2)
If ABC is rotated 90˚clockwise about the origin to form A’B’C’, what would be the coordinates of A’? A’ =(-1,-3) B’ = (-4,-4) C’ = (-3,-2)

Model Problem #36 J’ = (-2 + 4,-1 – 2) = (2,-3)
The coordinates of the vertices of ∆JKL, are J(-2,-1), K(1,3), L(4,-3). If ∆JKL is translated 2 units down and 4 units to the right to create , what are the coordinates of the vertices of ∆J’K’L’? J’ = (-2 + 4,-1 – 2) = (2,-3) K’ = (1 + 4, 3 – 2) = (5, 1) L’ = (4 + 4, -3 – 2) = (8,-5)

Model Problem #37 If quadrilateral DEFG is reflected across the y- axis, it would create quadrilateral D’E’F’G’. What are the coordinates of point G’ ? G’ = (6,2)

Model Problem #38 (a + b)2 c2 + 4(a ∙ b) ÷ 2
A diagram from a proof of the Pythagorean Theorem is shown. Write an equation that represents the area of the entire square in two ways. On the left side, express the area as the product of the length and the width. On the right, represent the sum of the areas of the triangles and the smaller square. Then use the equation to prove the theorem. (a + b)2 c2 + 4(a ∙ b) ÷ 2

Model Problem #39 112 = 62 + b2 121 = 36 + b2 85 = b2 √85 = b
A right triangle’s hypotenuse has length 11. If one leg has length 6, what is the length of the other leg? 112 = 62 + b2 121 = 36 + b2 85 = b2 √85 = b

Model Problem #40 √4100 = 10√41 = 2 ∙10 √41 502 + 402 = h2 20√41
In a basketball game, a player from the home team threw the ball from corner C to a player standing at point E. (E is the midpoint of AD). Then the player at point E threw the ball to a player at corner B. If the court was 80 feet long and 50 feet wide, how far was the ball thrown? (Leave in simplified radical form) √4100 = 10√41 = 2 ∙10 √41 20√41 = h2 = h2 √4100 = h

Model Problem #41 Sin A = 0.4 = BC/40 (40)(0.4) = BC 16 = BC
In the figure shown, sin A  0.4, cos A  0.5, and tan A  What is the approximate length of BC? Sin A = 0.4 = BC/40 (40)(0.4) = BC 16 = BC

Model Problem #42 5 4 32 + 42 = h2 9 + 16 = h2 √25 = 5 = h sin A = 4/5
In the figure below, if tan A = 4/3, what are sin A and cos A? 5 4 = h2 = h2 √25 = 5 = h sin A = 4/5 cos A = 3/5 3

Model Problem #43 sin 65o = ?/30’ 0.9 = x/30’ 27’ = x
A ladder is leaned against a wall at an angle of 65° to the ground. How far off the ground does the ladder touch the wall? sin 65o = ?/30’ 0.9 = x/30’ 27’ = x

Model Problem #44 Triangle JKL is shown in the diagram. Which equation should be used to find the length of LJ? A sin 28o = LJ/54 B sin 28o = 54/LJ C cos 28o = LJ/54 D cos 28o = 54/LJ sin cos

tan 70o = 2.75 cos 70o = 0.34 sin 70o = 0.94 Model Problem #45
On a swing set, on engineer used a support bar that was 20 feet long. If the support bar forms a 70° angle to the ground, how far apart will the support bars be at the base? sin 70o = 0.94 cos 70o = 0.34 tan 70o = 2.75 tan 70o = 2.75 cos 70o = 0.34 sin 70o = 0.94 cos 70o = Adj/20 0.34 = Adj/20 6.8 = adj 6.8 ∙ 2 = 13.6

Model Problem #46 In the diagram, mB = 75o and AC = 11.9 in. Which equation could be used to find BC? A x = 11.9(tan 75o) B x = 11.9(sin 75o) C x = 11.9/tan 75o C x = 11.9/sin 75o tan 75o = 11.9 / x tan 75o (x) = 11.9 x = 11.9 / tan 75o

Model Problem #47 The right triangle in the diagram has one side with a length of 5√3. What is the length of the side marked x? A 5 B 15 C 5√6 D 10√3 x 5√3 60o

Model Problem #48 Arc BD = 124o 124o – 60o = 64o
In the circle shown, the measure of BC= 60o, and the measure of ABD = 62o. What is the measure of CD? Arc BD = 124o 124o – 60o = 64o

Model Problem #49 4 ∙ 9 = x ∙ 12 36 = 12x 3 = x
In the circle shown, DF and CE are chords intersecting at G. If DG = 9, FG = 4, and EG = 12, what is the length of CG? 9 4 ∙ 9 = x ∙ 12 36 = 12x 3 = x 12 x 4

Model Problem #50 1 = ½(104o – 38o) ½(66o) 33o
In the circle shown, what is the measure of angle 1? 1 = ½(104o – 38o) ½(66o) 33o

Model Problem #51 180o -65o = 115o 180o – (115o + 50o) 180o – 165o 15o
LM is tangent to a circle, whose center is C, at point M. MQ is a diameter. If mQNP = 65˚ and mNPM = 50˚, what is mPMR? 180o -65o = 115o 180o – (115o + 50o) 180o – 165o 15o 90o – 15o 75o 15o 115o

Model Problem #52 Square = 4 ∙10 = 40 Circle = 10 = 10 10/40 1/4
A square is circumscribed about a circle. What is the ratio of the circumference of the circle to the perimeter of the square? A ¼ B ½ C 2/ D /4 Square = 4 ∙10 = 40 Circle = 10 = 10 10/40 1/4 10

Model Problem #53 10 ∙  ∙ 6 = 60 ins.
A cylinder rolls across a table top for 10 complete revolutions. If the diameter of the base is 6 inches, how far did the cylinder travel? (Leave the answer in terms of ). 10 ∙  ∙ 6 = 60 ins.

Model Problem #54 3 x 6 = 18 4 x 6 = 24 32 + 42 = h2 25 = h2
The prism shown has a base in the shape of a right triangle. What is the lateral surface area of the prism? 3 x 6 = 18 4 x 6 = 24 = h2 25 = h2 5 x 6 = 30 = 72 cm2

Model Problem #55 4 ( 8 + 6 ) ÷ 2 = 28 = Base 28 x 10 = 280 mm3
What is the volume of the prism shown? 4 ( ) ÷ 2 = 28 = Base 28 x 10 = 280 mm3

Model Problem #56 32 - 22 + 12 9 - 4 + 1 6 or 18.84
A target for a yard game is made with areas that are alternately painted white and gray, as shown in the diagram. The inner circle is white and has a radius of 1 inch. Each of the other three rings has a radius 1 inch more than the ring before it. What is the area of the white portion of the target? 32 - 22 + 12 9 - 4 + 1 6 or 18.84

Model Problem #57 40 – (12 + 12) = 16 16 ÷ 2 = 8 8 ∙ 12 = 96 ft2
A rectangle that is 12 feet wide has a perimeter of 40 feet. What is the area of the rectangle? 40 – ( ) = 16 16 ÷ 2 = 8 8 ∙ 12 = 96 ft2

Model Problem #58 22 + b2 = 42 4 + b2 = 16 b2 = 12 b = √4 ∙ 3 or 2√3
Each side of a triangle measures 4 m. What is the area of the triangle? (Leave the answer in simplified radical form). 22 + b2 = 42 4 + b2 = 16 b2 = 12 b = √4 ∙ 3 or 2√3 4 ∙ 2√3 ÷ 2 4√3 m2

Model Problem #59 10 ÷ 2 = 5 & 8 ÷ 2 = 4 4 ∙ (5 ∙ 4) ÷ 2 =
Quadrilateral ABCD is a rhombus. If AC = 10 inches and BD = 8 inches, what is the area of ABCD? 4 10 ÷ 2 = 5 & 8 ÷ 2 = 4 4 ∙ (5 ∙ 4) ÷ 2 = 2 ∙ 20 = 40 in2 5

Model Problem #60 4 (3 + 9) ÷ 2 2 (3 + 9) 2 ∙ 12 = 24 cm2
The diagram shows a trapezoid with a height of 4 cm. What is the area of the trapezoid? 4 (3 + 9) ÷ 2 2 (3 + 9) 2 ∙ 12 = 24 cm2

Model Problem #61 2 x 3 x 3 = 18 4 x 6 x 6 = 144 cm3
The volume of a right rectangular prism is calculated to be 18 cubic centimeters. If the length, the width, and the height of the prism are all doubled, what would be the volume of the new prism? 2 x 3 x 3 = 18 4 x 6 x 6 = 144 cm3

Model Problem #62 V = Bh V = 52 ∙ 4 V = 25 ∙ 4 V = 100 V = 314 cm3
The cylinder shown has a height of 4 cm and the diameter of the base is 10 cm. What is the volume of the cylinder? V = Bh V = 52 ∙ 4 V = 25 ∙ 4 V = 100 V = 314 cm3

Model Problem #63 8 x 8 = 64 4 x 8 x 6 ÷ 2 = + 96 160 cm2
The pyramid shown has a square base that measures 8 cm on each side. The slant height of the pyramid is 6 cm. What is the surface area of the prism? 8 x 8 = 4 x 8 x 6 ÷ 2 = + 96 160 cm2

Model Problem #64 Given: AB. What is the first step in constructing the perpendicular bisector to AB? a. Draw a line segment connecting points E and F. b. From point C, draw an arc that intersects the line at points A and B. c. Draw a line segment connecting points A and B. d. From points A and B, draw equal arcs that intersect at points E and F.

Model Problem #65 Darla is constructing an equilateral triangle. Which of the following could be her first step? A B C D

Model Problem #66 Marsha is using a straightedge and compass to do the construction shown. Which statement best describes the construction Martha is doing? a. a line through P parallel to line l by constructing two lines perpendicular to the same line b. a line through P parallel to line l by copying an angle c. a line through P perpendicular to line l d. a line through P congruent to line l

Model Problem #67 Amina is bisecting an angle. Which of the construction diagrams shown below best represents the beginning of Amina’s construction? A B C D

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