1.2.F - Quiz_ Introduction to the Coordinate Plane.pdf - 1.2.5 Week 2 Quiz Geometry(Geometry 1.2.5 Week 2 Quiz Score for this quiz 90 out of 100

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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 ...
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  • Fall '19
  • Melissa Jeffers
  • Math, Geometry

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