#### You've reached the end of your free preview.

Want to read all 17 pages?

**Unformatted text preview: **10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry) 1.2.5 Week 2 Quiz
Score for this quiz: 90 out of 100
Submitted Oct 15 at 4:55pm
This attempt took about 1 hour.
(?iframe_request=true) Question 1: 10 pts
Find the area and perimeter. 15.21 cm2 ; 15.6 cm
100% of points
21.15 cm2 ; 15.6 cm
0% of points
5.32 cm2 ; 5.8 cm
0% of points
15.21 cm2 ; 11.8 cm
0% of points The ﬁgure has two pairs of opposite equal sides, so, by the conditions for parallelorgams, the quadrilateral is a
parallelogram.
One angle of this parallelogram is a right angle, so, by the conditions for rectangles, this parallelogram is a
rectangle.
It is given that the rectangle has all four side equal. By the deﬁnition, a square is a quadrilateral with four right
angles and four congruent sides. Thus, the given ﬁgure is a square.
The ﬁgure is a square with the length of one side 3. 9 cm. 3.9 cm 1/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry) Use the formula for the perimeter of a square.
P = 4s Substitute 3. 9 cm for s .
P = 4(3. 9) Simplify. P = 15. 6 cm Therefore, the perimeter of the square is 15. 6 cm.
Use the formula for the area of a square.
2 A = s Substitute 3. 9 cm for s .
2 A = 3. 9 Simplify. A = 15. 21 cm 2 Therefore, the area of the square is 15. 21 cm .
2 10 / 10 Question banks: edit ( ?
iframe_request=true#&iframe_request=true) Show quiz names ( ) (?iframe_request=true) Question 2: 10 pts
Identify the distance, to the nearest tenth, between J and K (−3, 4) (7, −9) . 9.2
0% of points
1.5
0% of points
16.4
100% of points
14.7
0% of points 2/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry) y
J 4
7 x −3 −9 K Use the distance formula d = Substitute x , 1 , ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
2
2
‾
(x2 − x1 ) + (y2 − y1 )
√ to ﬁnd the length of the line segment. = −3 x2 = 7 y1 = 4
2 , and y 2 into the distance formula. = −9 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
K J = √‾
(7
− (−3)) + (−9 − 4)‾ Simplify.
2
2
‾‾‾‾‾‾‾‾‾‾‾‾
K J = √‾
(10)
+ (−13)‾ Take the squares.
K J = √‾
100
+
169
‾‾‾‾
‾‾‾
‾‾ Simplify.
K J = √269
‾‾‾‾ Use a calculator to approximate.
K J ≈ 16. 4 Therefore, the distance between J (−3, 4) and K (7, −9) is approximately 16. 4 units. 10 / 10 Question banks: edit ( ?
iframe_request=true#&iframe_request=true) Show quiz names ( ) (?iframe_request=true) Question 3: 10 pts
The area of a rectangle is 108. 35 in2 and the length is 7. 8 in . Find the width.
27.78 in.
0% of points
21.44 in.
0% of points
13.89 in.
100% of points
6.95 in.
0% of points 3/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry) A=108.35 in 2 w 7.8 in. Use the formula for the area of a rectangle.
A = lw Substitute 108. 35 for A and 7. 8 for l.
108. 35 = 7. 8w Divide both sides by 7. 8.
108.35
7.8 = w Simplify.
13. 89 ≈ w Therefore, the width of the rectangle is approximately 13. 89 in .
10 / 10 Question banks: edit ( ?
iframe_request=true#&iframe_request=true) Show quiz names ( ) (?iframe_request=true) Question 4: 10 pts
The coordinates of the vertices of △ PQR are P (2, −1) , Q (4, 2) , and R (6, 0). Identify the perimeter of △
Round each side length to the nearest tenth before adding. PQR . 13.4
0% of points
17.9
0% of points
21.3
0% of points
10.5
100% of points 4/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry) y Q (4, 2)
P (2, −1) R(6, 0) Use the distance formula d
To ﬁnd PQ , substitute x 1 = x ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
2
2
‾
(x2 − x1 ) + (y2 − y1 )
√ to ﬁnd the length of each line segment. , , , and y = −1 , , and y = 2 = 4 x2 = 2 y1 = 2 2 into the distance formula. 2 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
PQ = √‾
(2
− 4) + (−1 − 2)‾ Simplify.
2 2 ‾‾‾‾‾‾‾‾‾‾‾
PQ = √‾
(−2)
+ (−3)‾ Take the squares.
PQ = √‾
4‾‾‾
+ 9
‾ Simplify.
PQ = √13
‾
‾‾ Use a calculator to approximate.
PQ ≈ 3. 6 To ﬁnd QR , substitute x 1 2 , = 6 x2 = 4 y1 = 0 2 into the distance formula. 2 ‾‾
‾‾‾‾‾‾‾‾‾‾‾‾‾‾
QR = √‾
(4
−
6) + (2 − 0)‾ Simplify.
2 2 ‾‾‾‾‾‾‾‾‾‾
QR = √‾
(−2)
+ (2)‾ Take the squares.
QR = √‾
4‾‾‾
+ 4
‾ Simplify.
QR = √8
‾ Use a calculator to approximate.
QR ≈ 2. 8 To ﬁnd RP, substitute x 1 , , , and y = 2 x2 = 6 y1 = −1 2 = 0 into the distance formula. 5/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry)
2 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
RP = √‾
(6
− 2) + (0 − (−1)) ‾ Simplify.
2 2 ‾‾‾‾‾‾‾‾
RP = √‾
(4)
+ (1)‾ Take the squares.
RP = √‾
16
+ 1
‾‾‾‾
‾ Simplify.
RP = √17
‾
‾‾ Use a calculator to approximate.
RP ≈ 4. 1 Use the formula for the perimeter of a triangle.
P = a + b + c Substitute 3. 6 for a , 2. 8 for b , and 4. 1 for c .
P = 3. 6 + 2. 8 + 4. 1 Simplify.
P = 10. 5 Therefore, the perimeter of △ is approximately 10. 5 units. PQR 10 / 10 Question banks: edit ( ?
iframe_request=true#&iframe_request=true) Show quiz names ( ) (?iframe_request=true) Question 5: 10 pts
Find the coordinates of the image of Q (3, 4) after the translation (x, y) → (x + 3, y − 5) . (−2, 7)
0% of points
(6, −1)
100% of points
(−1, 6)
0% of points
(0, −1)
0% of points The translation is deﬁned by the rule (x, y) → (x + 3, y − 5) . Apply the translation rule to the preimage.
(x, y) → (x + 3, y − 5) Q (3, 4) → Q ′ (3 + 3, 4 − 5) Simplify the coordinates of Q .
′ 6/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry) Q (3, 4) → Q 6 ′ (6, − 1) y 5
Q(3, 4)
4 3 3
2 5 1
1 2 3 4 -1 5 6 x Q'(6, -1) Therefore, the coordinates of the image are (6, − 1) .
10 / 10 Question banks: edit ( ?
iframe_request=true#&iframe_request=true) Show quiz names ( ) (?iframe_request=true) Question 6: 10 pts
T ⎯⎯⎯⎯⎯ is the midpoint of K L . K has coordinates (2, −6) , and T has coordinates (−4, 2) . Identify the coordinates of L . (−1, −2)
0% of points
(−2, −4)
0% of points
(−2, 8)
0% of points
(−10, 10)
100% of points Given: K (2, −6), T (−4, 2).
Let the coordinates of L equal (x, y) . 7/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry) y
y L 2 T x 2x −4 −6
Use the Midpoint Formula: M ( K
x1 +x2
2 y +y , 1 2 2 ) Substitute the coordinates into the midpoint formula where (x 1, 2+x ( , 2 −6+y
2 ) y1 ) = (2, −6) and (x 2, y2 ) = (x, y) . = (−4, 2) Find the x -coordinate.
Set the coordinates equal.
2+x = −4 2 Multiply both sides by 2 .
2( 2+x
2 ) = 2 (−4) Simplify.
2 + x = −8 Subtract 2 from both sides.
x = −10 Find the y -coordinate.
Set the coordinates equal.
−6+y
2 = 2 Multiply both sides by 2 .
2 −6+y ( 2 ) = 2 (2) Simplify.
−6 + y = 4 8/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry) Add 6 to both sides.
y = 10 y
10 L 2 T −10 2x −4 −6 K Therefore, the coordinates of L are (−10, 10).
10 / 10 Question banks: edit ( ?
iframe_request=true#&iframe_request=true) Show quiz names ( ) (?iframe_request=true) Question 7: 10 pts
⎯⎯⎯⎯⎯ Identify the coordinates of the midpoint of AD , which has endpoints A (3, −2) and D (−5, 4) .
(−1, 1)
100% of points
(1, −1)
0% of points
(−15, −8)
0% of points
(−2, 1)
0% of points First identify the values of x , x , y , and y .
1 2 1 2 9/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry) y
4 D M y
M xM −5 −2 A Here, A (3, −2) is (x 1, So, x 1 x 3 , y1 ) and D (−5, 4) is (x 2, , , and y = 3 x2 = −5 y1 = −2 Use the Midpoint Formula: M ( 2 x1 +x2
2 = 4 y2 ) . . y +y , 1 2 2 ) Substitute the coordinates into the midpoint formula.
M 3+(−5) ( 2 , −2+4
2 ) Simplify.
M ( −2
2 , 2
2 ) Simplify.
M (−1, 1) y
4 D
M −5 1 −1
−2 x 3
A ⎯⎯⎯⎯⎯ Therefore, the coordinates of the midpoint of AD are (−1, 1).
10 / 10 Question banks: edit ( ?
iframe_request=true#&iframe_request=true) Show quiz names ( ) (?iframe_request=true) Question 8: 10 pts
Given points U (−4, 4) and V ⎯⎯⎯⎯⎯⎯ , identify U V and its reﬂection across the x -axis. (5, 1) 10/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry) 0% of points 0% of points 0% of points 11/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry) 100% of points
⎯⎯⎯⎯⎯⎯
UV is the line segment with endpoints at U and V . The coordinates of point U are given to be (−4, 4) and the coordinates of point V are given to be (5, 1) .
⎯⎯⎯⎯⎯⎯
Therefore, U V is the line segment with endpoints at U (−4, 4) and V (5, 1). A reﬂection is a transformation across a line, called the line of reﬂection, where each point and its image are the
⎯⎯⎯⎯⎯⎯
⎯⎯⎯⎯⎯⎯⎯⎯⎯
same distance from the line of reﬂection. So, the reﬂection of U V (denoted U V ) across the x -axis, is a line
′ ⎯⎯⎯⎯⎯⎯⎯⎯⎯
′
′
U V segment where each point on ⎯⎯⎯⎯⎯⎯
corresponting point on U V .
⎯⎯⎯⎯⎯⎯⎯⎯⎯
′
′
V To sketch U
U (−4, 4)
V (5, 1) ′ is the same distance from the x -axis (the line of reﬂection) as the
⎯⎯⎯⎯⎯⎯ , ﬁrst reﬂect the endpoints of U V across the x -axis. is 4 units above the x -axis, so U is 4 units below the x -axis, at (−4, −4) .
′ is 1 unit above the x -axis, so V is 1 unit below the x -axis, at (5, −1).
′ 12/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry) ⎯⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯
′
′
V Connect endpoints U and V to sketch the reﬂection of U V across the x -axis, U
′ ′ . 10 / 10 Question banks: edit ( ?
iframe_request=true#&iframe_request=true) Show quiz names ( ) (?iframe_request=true) Question 9: 10 pts
Find the circumference and area of a circle with diameter 12 ft. Express your answer in terms of π .
12π ft; 64π ft2
0% of points
8π ft; 36π ft2
0% of points
6π ft; 38π ft2
0% of points
12π ft; 36π ft2
13/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry) 100% of points 12 ft Use the formula for the radius of a circle.
r = 1
2 d Substitute 12 for d .
r = 1 · 12 2 1 Multiply and 12.
2 r = 6 ft Therefore, the radius of the circle is 6 ft . 6 ft Use the formula for the circumference of a circle.
C = 2π r Substitute 6 for r .
C = 2π (6) Multiply 2 and 6 . C = 12π ft Therefore, the circumference of the circle is 12π ft .
Use the formula for the area of a circle.
A = πr 2 Substitute 6 for r .
A = π(6) 2 Simplify. 2 A = 36π ft 14/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry)
2 Therefore, the area of the circle is 36π ft .
10 / 10 Question banks: edit ( ?
iframe_request=true#&iframe_request=true) Show quiz names ( ) (?iframe_request=true) Question 10: 10 pts
A ﬁgure has vertices at A (−5, 4), B (−1, 4), C (−1, 5) , and D (−5, 5) . After a transformation the image of the ﬁgure
has vertices at A ′ (1, − 5) , B ′ (5, − 5) , C ′ (5, − 4), and D′ (1, − 4). Identify the preimage, the image, and the
transformation.
rotation 0% of points
translation 0% of points 15/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry) translation 100% of points
reﬂection 0% of points Begin by plotting the points to identify the vertices. Then, draw lines to connect the vertices. These are the sides
of the ﬁgures. The original ﬁgure in a transformation is called the preimage. Therefore, the preimage is ABCD.
The resulting ﬁgure in a transformation is called the image. Therefore, the image is A
Now determine if the transformation of ABCD ′ ′ ′ → A B C D ′ ′ ′ ′ B C D ′ . is a reﬂection, translation, or rotation. 16/17 10/17/2019 1.2.5 Week 2 Quiz: Geometry (Geometry) Notice that all the points of the rectangle ABCD move the same distance in the same direction.
Transformation in which all the points of a ﬁgure move the same distance in the same direction is a translation. Therefore, the transformation of ABCD ′ ′ ′ → A B C D ′ is a translation. 0 / 10 Question banks: edit ( ?
iframe_request=true#&iframe_request=true) Show quiz names ( ) - Quiz Score: 90 out of 100 17/17 ...

View
Full Document

- Fall '19
- Melissa Jeffers
- Math, Geometry