![McDougal Littell Jurgensen Geometry: Student Edition Geometry](https://www.bartleby.com/isbn_cover_images/9780395977279/9780395977279_largeCoverImage.gif)
Concept explainers
a
To sketch: a line which shows a set of points equidistant from F and T as shown in the figure below.
a
![Check Mark](/static/check-mark.png)
Explanation of Solution
Given information: Points F,T and P are shown as below in the diagram and a sketch has to be made of a line that specifies a set of points which are equidistant from F and T .
Sketch: Take a centimeter ruler and find a point midway between F and T and mark it. Now draw a line through that point perpendicular to the line joining F and T, which is the required line.
Interpretation: The required line is equidistant from points F and T which can be checked using a centimeter ruler.
b
To sketch: a
b
![Check Mark](/static/check-mark.png)
Explanation of Solution
Given information: Points F, T and P are shown as below in the diagram and a sketch has to be made of a circle whose radius is 2 cm and the center is at P as shown below.
Sketch: Use a ruler to set the distance from the point of the compass needle to the pencil's lead at 2 cm. Place the compass needle at P and rotate it about 360 degrees. The required object in the plane is a circle of radius 2 cm.
Interpretation: The circle so formed has a radius of 2 cm which can be checked by a centimeter ruler.
c
To sketch: the points 2 cm from P which are equidistant from F and T as shown in the figure below,
c
![Check Mark](/static/check-mark.png)
Explanation of Solution
Given information: Points F,T and P are shown as below in the diagram; it is to be found that how many points are equidistant from F and T and are also 2 cm away from P.
Graph: Take a centimeter ruler and find a point midway between F and T and mark it. Now draw a line through that point perpendicular to the line joining F and T. Use a ruler to set the distance from the point of the compass needle to the pencil's lead at 2 cm. Place the compass needle at P and rotate it about 360 degrees. The required object in the plane is a circle of radius 2 cm. But it does not cross the line joining F and T even once.
So there are no point of intersection obtained as a result.
Interpretation: There are no points which are equidistant from F and T and 2 cm away from P.
Chapter 1 Solutions
McDougal Littell Jurgensen Geometry: Student Edition Geometry
Additional Math Textbook Solutions
Elementary Statistics: Picturing the World (7th Edition)
Thinking Mathematically (6th Edition)
Precalculus
Elementary Statistics (13th Edition)
Elementary Statistics
College Algebra with Modeling & Visualization (5th Edition)
- Please help me answer this question!. Please handwrite it. I don't require AI answers. Thanks for your time!.arrow_forward1 What is the area of triangle ABC? 12 60° 60° A D B A 6√√3 square units B 18√3 square units 36√3 square units D 72√3 square unitsarrow_forwardPar quel quadrilatère est-elle représentée sur ce besoin en perspective cavalièrearrow_forward
- -10 M 10 y 5 P -5 R 5 -5 Ο 10 N -10 Οarrow_forwardDescribe enlargement on map gridarrow_forward◆ Switch To Light Mode HOMEWORK: 18, 19, 24, 27, 29 ***Please refer to the HOMEWORK sheet from Thursday, 9/14, for the problems ****Please text or email me if you have any questions 18. Figure 5-35 is a map of downtown Royalton, showing the Royalton River running through the downtown area and the three islands (A, B, and C) connected to each other and both banks by eight bridges. The Down- town Athletic Club wants to design the route for a marathon through the downtown area. Draw a graph that models the layout of Royalton. FIGURE 5-35 North Royalton Royalton River South Royption 19. A night watchman must walk the streets of the Green Hills subdivision shown in Fig. 5-36. The night watch- man needs to walk only once along each block. Draw a graph that models this situation.arrow_forward
- Solve this question and check if my answer provided is correctarrow_forwardProof: LN⎯⎯⎯⎯⎯LN¯ divides quadrilateral KLMN into two triangles. The sum of the angle measures in each triangle is ˚, so the sum of the angle measures for both triangles is ˚. So, m∠K+m∠L+m∠M+m∠N=m∠K+m∠L+m∠M+m∠N=˚. Because ∠K≅∠M∠K≅∠M and ∠N≅∠L, m∠K=m∠M∠N≅∠L, m∠K=m∠M and m∠N=m∠Lm∠N=m∠L by the definition of congruence. By the Substitution Property of Equality, m∠K+m∠L+m∠K+m∠L=m∠K+m∠L+m∠K+m∠L=°,°, so (m∠K)+ m∠K+ (m∠L)= m∠L= ˚. Dividing each side by gives m∠K+m∠L=m∠K+m∠L= °.°. The consecutive angles are supplementary, so KN⎯⎯⎯⎯⎯⎯∥LM⎯⎯⎯⎯⎯⎯KN¯∥LM¯ by the Converse of the Consecutive Interior Angles Theorem. Likewise, (m∠K)+m∠K+ (m∠N)=m∠N= ˚, or m∠K+m∠N=m∠K+m∠N= ˚. So these consecutive angles are supplementary and KL⎯⎯⎯⎯⎯∥NM⎯⎯⎯⎯⎯⎯KL¯∥NM¯ by the Converse of the Consecutive Interior Angles Theorem. Opposite sides are parallel, so quadrilateral KLMN is a parallelogram.arrow_forwardQuadrilateral BCDE is similar to quadrilateral FGHI. Find the measure of side FG. Round your answer to the nearest tenth if necessary. BCDEFGHI2737.55arrow_forward
- An angle measures 70.6° more than the measure of its supplementary angle. What is the measure of each angle?arrow_forwardName: Date: Per: Unit 7: Geometry Homework 4: Parallel Lines & Transversals **This is a 2-page document! ** Directions: Classify each angle pair and indicate whether they are congruent or supplementary. 1 1.23 and 25 2. 24 and 28 3. 22 and 25 4. 22 and 28 5. 21 and 27 6. 22 and 26 Directions: Find each angle measure. 7. Given: wvm25-149 m21- 8. Given: mn: m1=74 mz2- m22- m.23- m23- mz4= V mz4= m25= m26- m26= m27- m27 m28- m48= 9. Given: a || b: m28 125 m2- 10. Given: xy: m22-22 m21- = mz2- m43- m3- mZA m24-> m. 5- m25- m26- m.26=> m2]=> m27= m28- 11. Given: rm2-29: m15-65 m2=> m29-> m3- m. 10- mc4= m25= m212- m.46- m213- mat- m214- m28- & Gina when (N) Things ALICE 2017arrow_forwardMatch each statement to the set of shapes that best describes them. 1. Similar triangles by SSS 2. Similar triangles by SAS 3. Similar triangles by AA 4. The triangles are not similar > U E 35° 89° S F 89° J 35° 94° G 52° 90° E K 52° Iarrow_forward
- Elementary Geometry For College Students, 7eGeometryISBN:9781337614085Author:Alexander, Daniel C.; Koeberlein, Geralyn M.Publisher:Cengage,Elementary Geometry for College StudentsGeometryISBN:9781285195698Author:Daniel C. Alexander, Geralyn M. KoeberleinPublisher:Cengage Learning
![Text book image](https://www.bartleby.com/isbn_cover_images/9781337614085/9781337614085_smallCoverImage.jpg)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781285195698/9781285195698_smallCoverImage.gif)