Part 1 Write your own real-world scenario where the Pythagorean Theorem can be applied to find a missing piece. You may choose t
o write a problem that is two- or three-dimensional in nature. Be sure that you will be able to draw a diagram of your scenario. Write out your problem and submit it for Part 1. Be sure to end your scenario with a question. Part 2 Draw a diagram of the scenario you created in Part 1. You may draw by hand and scan and upload your drawing or create a computer-generated drawing for submission. Be sure to label all parts and dimensions of the drawing. Part 3 Solve the question that you posed in Part 1. Show all of your steps in answering the question. For this option, you will need to submit all three parts for full credit—your real-world problem and question, the diagram that you created, and your work solving the problem, showing all steps. *Please note that your instructor is looking for your own original idea. While it is acceptable to use the Internet for research and inspiration, academic integrity policies apply.
1 answer:
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Answer:
∠EFD ≅ ∠EGD ⇒ A
Arc ED ≅ arc FD ⇒ C
m arc FD = 120° ⇒ E
Step-by-step explanation:
Let us revise some facts
- Equal chords subtended equal arcs
- The measure of an inscribed angle is one-half the measure of the central angle which subtended by the same arc
- The measure of a central angle is equal to the measure of its subtended arc
- If one angle of an isosceles triangle measure 60° then the triangle is equilateral
- The sum of the measures of the interior angles of any quadrilateral is 360°
In the quadrilateral CDGE
∵ m∠G = 60°
∵ m∠GDC = m∠GEC = 90°
- By using the 5th rule above
∴ m∠G + m∠GDC + m∠DCE + m∠GEC = 360°
∴ 60 + 90 + m∠DCE + 90 = 360
∴ 240 + m∠DCE = 360
- Subtract 240 from both sides
∵ m∠DCE = 120°
In circle C
∵ ∠DCE is a central angle subtended by arc DE
∵ ∠DFE is an inscribed angle subtended by arc DE
- By using the 2nd rule above
∴ m∠DFE = m∠∠DCE
∵ m∠DCE = 120°
∴ m∠DFE = (120)
∴ m∠DFE = 60°
- That means ∠EFD ≅ ∠EGD because their measure is 60°
∴ ∠EFD ≅ ∠EGD
In Δ EFD
∵ EF = FD
∵ m∠DFE = 60°
- By using the 4th rule above
∴ Δ EFD is an equilateral triangle
∴ ED = FD = FE
In circle C
∵ Side ED subtended by arc ED
∵ Side FD subtended by FD
∵ Side ED ≅ side FD ⇒ proved
- By using the 1st rule above
∴ Arc ED ≅ arc FD
∵ m∠ECD = 120°
∵ ∠ECD is a central angle subtended by arc ED
- By using the 3rd rule above
∴ m∠ECD = m arc ED
∴ m of arc ED = 120°
∵ Arc ED ≅ arc FD
∴ m arc ED = m arc FD
∴ m arc FD = 120°