WCCUSD Geometry Benchmark 3 Study Guide

WCCUSD Geometry Benchmark 3 Study Guide

Page 1 of 12 MCC@WCCUSD 01/23/15

1 G.GMD.3—Some Things To Know

G.GMD.3

1´ You try: A grain silo is in the shape of a cylinder. The silo has a diameter of 12 feet and a height of 20 feet. Farmer Juan has reaped 10,000 cubic feet of wheat. How many silos would he need to fit all his wheat? Justify your reasoning:

G.GMD.3

The volume formula for a right prism or cylinder is , where B is the area of the base and h is

the height. For a cone or pyramid, the volume formula is

.

For a sphere:

Example: Abigail is putting juice into cone–shaped containers to make popsicles. She has approximately one quarter of a gallon, or 58 cubic inches, of juice to make the popsicles with. Each container has a 3-inch height and 2-inch radius.

Find the maximum number of popsicles that Abigail could make with 58 cubic inches of juice.

Solution: Find the volume for each cone-shaped container:

cubic inches Since Abigail has 58 cu.in. of juice, divide that number by the volume of each container (12.56 cu.in.) to find the number of containers she can fill.

This means that she could fill 4 containers completely, but she doesn’t have enough to fill 5 containers.

Base is a circle, so:



2 G.GPE.4—Some Things To Know

G.GMD.4

2´ You try: Triangle RST is graphed on a coordinate plane with vertices R(–2, –2), S(5, –1), and T(–1, –4). What type of triangle is formed? Justify your reasoning:

G.GMD.4

Use the distance formula and the slope formula to help you determine which geometric shape is created when given the vertices’ coordinates. The distance between two points can be found with the equation .

The slope formula is

Example: Quadrilateral PQRS is graphed on a coordinate plane with vertices located at P(1, 5), Q(3, 4), R(0, –2), and S(–2, –1). What type of quadrilateral is formed?

Solution:

Step 1: Find the length of each side:

Doing the same with the other three sides, we find:

, , . So, both pairs of opposite sides are congruent. Step 2: Find the slope of each line:

Doing the same for the other three sides, we find:

, , Since the adjacent sides’ slopes are negative reciprocals, the lines are perpendicular, making right angles at each of the four vertices.

Step 1 and Step 2 conclusions lead to knowing that the vertices make a rectangle.




--The slope formula is m =y2 − y1x2 − x1

---Slopes of parallel lines are equal. ---Slopes of perpendicular lines are opposite reciprocals.

Example: Given the two points L(3,−2) and M (−4,8) , which of the following statements are true? Select all that apply.

A. The slope of the line LM

is 6.

B. The slope of the line LM

is −107

.

C. The line y = − 16x −3 is parallel to LM

.

D. Line y = 710

x +8 is perpendicular to LM

.

E. The line 10x + 7y = −35 is parallel to LM

. F. Line −x − 6y = 42 is perpendicular to LM

.

Solution: Find the slope of LM

:

--So, A is not correct, but B is correct.

--Parallel lines have equal slopes, so C is not

correct, but D is correct since 710

is the negative

reciprocal of −107

so the lines are perpendicular.

--For E and F, solve each equation for y to determine the slope of the line:

10x + 7y = −35 −x − 6y = 42 7y = −35−10x −6y = 42+ x

y = −5−107x y = −7− 1

6x

E is correct since parallel lines have equal slopes and F is incorrect since the slope of the

perpendicular line must be 710

.

G.GPE.5

3´ You try: Given the two points G(6,−4) and H (−1,3) and the line 2x + 2y = 7 , which of the following statements are true? Select all that apply.

A. The slope of the line GH

is -1.

B. The slope of the line GH

is 17

.

C. The line y = −7x + 4 is parallel to GH

. D. The line y = −x −8 is parallel to GH

.

E. The line −x + y = 6 is perpendicular to GH

. F. The line −x + 7y =12 is perpendicular to

GH

.

Justify your reasoning:

G.GPE.5

m =y2 − y1x2 − x1

=−2−83− (−4)

=−107



4 G.GPE.6—Some Things To Know Example: Given the point P that partitions the directed line segment QR into a ratio of 1:4 with Q(−5,−4) and R(0, 6) , select all statements below that lead to finding the point P.

A. The scale factor is 15

.

B. The run is 5. C. The rise is 10.

D. −5+ 15(5),−4+ 1

5(10)

"

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&'

Solution: All answers are correct.

A. Since the ratio is 1:4, there are 5 total partitions and the 1 tells you are identifying the first point, which helps identify your scale

factor of 15

.

B. The x-coordinate is increasing by 5, so the run is 5. C. The y-coordinate is increasing by 10, so the rise is 10. D. The values were correctly substituted.

G.GPE.6

4´ You try: A directed line segment is graphed below. Plot point R on the graph so that it splits WY

into segments with a ratio of 1:2.

Explain how you determined your answer.

G.GPE.6

Finding a point on a given line segment that is partitioned by a certain ratio requires knowledge of a few things:

• The partition ratio and the scale factor are directly related (e.g. a partition ratio of 5:1

has a scale factor of .)

• The “run” is the horizontal distance between the segment’s endpoints.

• The “rise” is the vertical distance between the segment’s endpoints.

• The partitioning point’s coordinate is

factor.



5 G.C.2—Some Things To Know

G.C.2

5´ You try: Given Circle A below with EF

tangent to Circle A at point D and tangent to Circle G at point F, if m∠BCD = 62°and 𝑚𝐵𝐶 = 83°, which of the following statements are true?

A. m∠DAB = 62° B. m∠DAB =124° C. m∠ADF =m∠DFG D. m∠EDC = 90° E. 𝑚𝐶𝐵𝐷 = 207° F. 𝑚𝐶𝐷 = 153° G. ∠CDA and ∠CDF are

complementary. Justify your reasoning:

G.C.2

For circles: • A central angle is equal in measure to its

intercepted arc. • An inscribed angle is half the measure of its

intercepted arc. • A line tangent to a point on the circle makes a

right angle with a radius/diameter that meets at the point of tangency.

F

Example: Given Circle C below with tangent to the circle at point B, if which of the following statements are true?

A. 𝑚𝐴𝐵! = 30° B. 𝑚𝐴𝐵! = 60° C. D. E. F.

Solution: 𝑚𝐴𝐵! = 60° is correct because its inscribed angle is 30°. An intercepted arc is twice the measure of its inscribed angle.

is correct because that angle intercepts 𝐴𝐵! , which is 60°. Inscribed angles are half the measure of its intercepted arcs.

There is not enough information to determine the measure of .

is correct because it is a central angle and its intercepted arc has the same measure.

is correct because any tangent line is perpendicular to the radius at the point of tangency.



6 G.C.3—Some Things To Know To circumscribe a circle about a triangle:

1. Construct the perpendicular bisectors of each side. Draw them long enough to intersect each other. (the dashed lines in the figure below)

2. The intersection point of the bisectors is the center of your circle. Open your compass with its point at the circle’s center and draw the circle so that the triangle’s vertices all connect.

To inscribe a circle within a triangle:

1. Construct the angle bisectors of (at least) two angles. The intersection point will be the center of your circle.

2. From this point, construct a line perpendicular to one of the sides of the triangle. This will determine the point of tangency. Draw the circle.

G.C.3

6´ You try: Construct a circle that circumscribes the triangle below. Construct a circle that inscribes the triangle below.

G.C.3

A

B

C

A

B

C

A

B

C

A

B

C




Example: Draw the circle whose equation is

(𝑥 − 4)! + (𝑦 + 2)! = 16.

Solution: Step 1: Determine the center: h is 4 and k is -2 (substituting−2 into 𝑦 − 𝑘 would yield 𝑦 + 2.) Step 2: Determine the radius. Since 𝑟! = 16, the radius must be 4. Step 3: Plot your center on the graph and map out four points that are 4 units away to help draw your circle.

Example: Complete the square to find the center and radius of the circle whose equation is

x2 + y2 + 2x − 6y+1= 0 Solution: The center is at (−1,3) and the radius is 3.

G.GPE.1

7´ You Try: Draw the circle whose equation is

(𝑥 + 1)! + (𝑦 − 2)! = 25. Explain how you determined your answer. You Try: Complete the square to find the center and radius of the circle whose equation is x2 + y2 − 4x +8y+11= 0

G.GPE.1

The equation of a circle is: (𝑥 − ℎ)! + (𝑦 − 𝑘)! = 𝑟!

where r represents the distance of all points from the center of the circle, (ℎ, 𝑘).

Given the equation of a circle in the standard form, complete the square for both x- and y-terms to find the center of the circle.

Move constant to right side.

Commute x- and y-terms together.

Complete the square.

Factor.

x2 + y2 + 2x − 6y+1= 0x2 + y2 + 2x − 6y = −1x2 + 2x + y2 − 6y = −1

(x2 + 2x + ___)+ (y2 − 6y+ ___) = −1

(x2 + 2x + 1( )2 )+ (y2 − 6y+ (3)2 ) = −1+ 1( )2 + (3)2

(x +1)2 + (y−3)2 = 9




Example: Is the point (−3, 4) inside, on or outside the circle (𝑥 − 6)! + (𝑦 + 1)! = 100?

Solution: The center of the circle is (6,−1) . Substituting this point and the point in question into the Distance Formula:

d = (−3− 6)2 + (4− (−1))2 d = (−9)2 + (5)2 d = 81+ 25 d = 106 ≈10.3 → Example: Given the endpoints of the diameter of a circle, (−4,−2) and (4,8) find the equation of the circle.

Solution: Find the center of the circle using the midpoint formula: −4+ 42

, −2+82

"

#$

%

&'→

02, 62

!

"#

$

%&→ 0,3( )→Center

Find the radius:

d = (4− 0)2 + (8−3)2 d = (4)2 + (5)2

d = 16+ 25 d = 41→Radius The equation is x2 + (y−3)2 = 41

G.GPE.4

8´ You Try: Is the point (5,−8) inside, on or outside the circle (𝑥 − 3)! + (𝑦 + 5)! = 81?

You Try: Given the endpoints of the diameter of a circle, (10,−5) and (−2,−9) find the equation of the circle.

G.GPE.4

End of Study Guide

The Distance Formula ( ) can be used to determine whether a given point is inside, on or outside a given circle. Use the circle’s center coordinates and the given point in question and substitute into the Distance Formula.

Use the Midpoint Formula to

find the center of a circle given the endpoints of the diameter.

This is a little more than the radius of 10 units, so the point is outside the circle.



You Try Solutions 1´ You try: A grain silo is in the shape of a

cylinder. The silo has a diameter of 12 feet and a height of 20 feet. Farmer Juan has reaped 10,000 cubic feet of wheat. How many silos would he need to fit all his wheat? Justify your reasoning:

He reaped 10,000 cu. ft. of wheat. So, dividing this by the amount each silo holds results in: 10,000÷ 2260.8 ≈ 4.4

which means he would need 5 silos to hold his wheat.

GMD.3

2´ You try: Triangle RST is graphed on a coordinate plane with vertices R(–2, –2), S(5, –1), and T(–1, –4). What type of triangle is formed?

Justify your reasoning: Find the side lengths:

No sides have the same measure, so the triangle is scalene.

Now check the slopes of each side: mRS =

−2− (−1)−2− 5

=−1−7

=17

mST =−1− (−4)5− (−1)

=36=12

mRT =−2− (−4)−2− (−1)

=2−1

= −2

Since the slopes 12

and −2 are opposite reciprocals,

those two lines are perpendicular, forming a right angle.

G.GMD.4

If d = 12, then r = 6.

h = 20

Base is a circle, so

cu.ft. (each silo holds this amount)

So the triangle formed is a right scalene triangle.



3´ You try: Given the two points G(6,−4) and H (−1,3) , which of the following statements are true? Select all that apply.

A. The slope of the line GH

is -1.

B. The slope of the line GH

is 17

.

C. The line y = −7x + 4 is parallel to GH

. D. The line y = −x −8 is parallel to GH

.

E. The line −x + y = 6 is perpendicular to GH

. F. The line −x + 7y =12 is perpendicular to

GH

.

Justify your reasoning: The slope of the line containing the two points is:

m =−4−36− (−1)

=−77= −1

So A. is correct. D. is also correct since the slope of that line is also −1. Solving −x + y = 6 for y, we find y = x + 6, which has a slope of 1. This is the opposite reciprocal of -1, so the line is perpendicular to GH

. So, E. is correct.

G.GPE.5

4´ You try: Directed line segment is graphed below. Plot point R on the graph so that it splits WY

into segments with a ratio of 1:2.

Explain how you determined your answer. The starting point is W (−2,−4) . First find the rise and run. The rise is 9 and the run is 6. The ratio segmentation is 1:2, which gives a

scale factor of 13

.

Now, use a + k(run),b + k(rise)( ) to find the

point in question:

−2+ 13(6),−4+ 1

3(9)

"

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%

&'

−2+ 2,−4+3( )

G.GPE.6

This is the point that will partition the given line segment into a ratio of 1:2.

R



5´ You try: Given Circle A below with EF

tangent to Circle A at point D and tangent to Circle G at point F, if m∠BCD = 62° and 𝑚𝐵𝐶 = 83°, which of the following statements are true?

A. m∠DAB = 62° B. m∠DAB =124° C. m∠ADF =m∠DFG D. m∠EDC = 90° E. 𝑚𝐶𝐵𝐷 = 207° F. 𝑚𝐶𝐷 = 153° G. ∠CDA and ∠CDF are complementary.

Justify your reasoning:

If m∠BCD = 62°, then 𝑚𝐵𝐷 = 124° since an intercepted arc is twice its inscribed angle. This means that m∠DAB =124° since a central angle has the same measure as its intercepted arc. So B is correct. Both ∠ADF and ∠DFG are right angles since they are created by a tangent line and a radius of each circle, which form perpendicular lines. So C is correct. D is not correct because DC is not part of a radius. If 𝑚𝐵𝐷 = 124° and 𝑚𝐵𝐶 = 83° then 𝑚𝐶𝐵𝐷 = 207° since this major arc is the sum of the two minor arcs. E is correct. 𝑚𝐶𝐷 = 153° because this is what remains of a complete circle of 360° . F is correct. ∠CDA and ∠CDF are complementary because they form a right angle. G is correct.

G.C.2

6´ You try: Construct a circle that circumscribes the triangle below. Construct the perpendicular bisectors of at least two of the sides. Their intersection is the center of the circle (circumcenter). Open your compass such that the point is on this center and the pencil point is on a vertex of the triangle. Draw your circle. Construct a circle that inscribes the triangle below. Bisect two of the angles. Then use the intersection point (which is the circle’s center) to construct a line perpendicular to a side. This will determine one of the points of tangency. Use your compass to draw your circle.

G.C.3



7´ You Try: Draw the circle whose equation is (𝑥 + 1)! + (𝑦 − 2)! = 25.

Explain how you determined your answer. Based on the equation, the center is at (−1,2) and the radius is 5. You Try: Complete the square to find the center and radius of the circle whose equation is x2 + y2 − 4x +8y+11= 0 .

x2 + y2 − 4x +8y+11= 0

x2 + y2 − 4x +8y = −11x2 − 4x + y2 +8y = −11

x2 − 4x + (__)+ y2 +8y+ (__) = −11x2 − 4x + (−2)2 + y2 +8y+ (4)2 = −11+ (−2)2 + (4)2

(x − 2)2 + (y+ 4)2 = −11+ 4+16(x − 2)2 + (y+ 4)2 = 9

G.GPE.1

8´ You Try: Is the point (9,−12) inside, on or outside the circle (𝑥 − 3)! + (𝑦 + 5)! = 81?

The center of the circle is (3,−5) and the radius is 9. So the distance from the center to the point in question must be compared with 9. d = (9−3)2 + (−12− (−5))2

d = (6)2 + (−7)2

d = 36+ 49 d = 85 ≈ 9.2 This distance is a little more than 9 units. So the point lies outside the circle. You Try: Given the endpoints of the diameter of a circle, (10,−5) and (−2,−9)find the equation of the circle. Find the center of the circle using the midpoint formula: 10+ (−2)

2, −5+ (−9)

2"

#$

%

&'→

82, −142

"

#$

%

&'→ 4,−7( )

Find the radius using one of the points on the circle and the center: d = (10− 4)2 + (−5− (−7))2

d = (6)2 + (2)2

d = 36+ 4 d = 40

G.GPE.4

The equation is (𝑥 − 4)! + (𝑦 + 7)! = 40.

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WCCUSD Geometry Benchmark 3 Study Guide