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a) The blue shaded region A₁ of the unit circle is a sector with central angle y. The tan shaded region A₂ is a right triangle with base measuring x and acute angles a, ß. The elementary geometric area formulas give 1 A₁ = r2 and b) Since B,y are [ CIRCLE ONE: vertical / alternate / corresponding] angles, we conclude that B = 2 c) We know that so that we can write sin ß = y = y(x) = A₂ = 2 h. d) The vertex of the triangle corresponding with ß is B(x, h). Use the fact that B lies on the unit circle centered at the origin to write h = h(x) = e) With the above we are able to write the shaded areas as functions of x: and A₁ = A₁(x) = A₂ = A₂(x) = f) Write the sum of A₁ and A₂ as a single definite integral ·b(x) f(t) dt. A₁ + A₂ =

a) The blue shaded region A₁ of the unit circle is a sector with central angle y. The tan shaded region A₂ is a right triangle with base measuring x and acute angles a, ß. The elementary geometric area formulas give 1 A₁ = r2 and b) Since B,y are [ CIRCLE ONE: vertical / alternate / corresponding] angles, we conclude that B = 2 c) We know that so that we can write sin ß = y = y(x) = A₂ = 2 h. d) The vertex of the triangle corresponding with ß is B(x, h). Use the fact that B lies on the unit circle centered at the origin to write h = h(x) = e) With the above we are able to write the shaded areas as functions of x: and A₁ = A₁(x) = A₂ = A₂(x) = f) Write the sum of A₁ and A₂ as a single definite integral ·b(x) f(t) dt. A₁ + A₂ =

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter3: Triangles
Section3.1: Congruent Triangles
Problem 7E: In ABC, the midpoints of the sides are joined. a what does intuition suggest regarding the...
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The second picture is the graph and the x changest from 0 to 1
~ = B (₁,√1-2²) 0 SIS X 7 X = 0.5 ⒸVI-X²³XSYS√1-2² (0sxsx) √1-X² x{0≤x≤X} X 0 0≤x≤ (x,0) ✓Label: X X X 1 X X X X -0.2 1.2+ 0.8- 0.6+ 0.4+ 0.2+ 0 -0.2+ -0.4+ 0.2 0.4 X 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Transcribed Image Text:~ = B (₁,√1-2²) 0 SIS X 7 X = 0.5 ⒸVI-X²³XSYS√1-2² (0sxsx) √1-X² x{0≤x≤X} X 0 0≤x≤ (x,0) ✓Label: X X X 1 X X X X -0.2 1.2+ 0.8- 0.6+ 0.4+ 0.2+ 0 -0.2+ -0.4+ 0.2 0.4 X 0.6 0.8 1 1.2 1.4 1.6 1.8 2
A₁ Va a) The blue shaded region A₁ of the unit circle is a sector with central angle y. The tan shaded region A₂ is a right triangle with base measuring x and acute angles a, ß. The elementary geometric area formulas give = 2 √1-x² c) We know that 1 so that we can write X r2 and A₂ = b) Since B,y are [ CIRCLE ONE: vertical / alternate / corresponding] angles, we conclude that B = sin ß = =1/1² y = y(x) = h. d) The vertex of the triangle corresponding with ß is B(x, h). Use the fact that B lies on the unit circle centered at the origin to write h = h(x) = e) With the above we are able to write the shaded areas as functions of x: and A₁ = A₁(x) = A₂ = A₂(x) = f) Write the sum of A₁ and A₂ as a single definite integral pb(x) S f(t) dt. A₁ + A₂ =
Transcribed Image Text:A₁ Va a) The blue shaded region A₁ of the unit circle is a sector with central angle y. The tan shaded region A₂ is a right triangle with base measuring x and acute angles a, ß. The elementary geometric area formulas give = 2 √1-x² c) We know that 1 so that we can write X r2 and A₂ = b) Since B,y are [ CIRCLE ONE: vertical / alternate / corresponding] angles, we conclude that B = sin ß = =1/1² y = y(x) = h. d) The vertex of the triangle corresponding with ß is B(x, h). Use the fact that B lies on the unit circle centered at the origin to write h = h(x) = e) With the above we are able to write the shaded areas as functions of x: and A₁ = A₁(x) = A₂ = A₂(x) = f) Write the sum of A₁ and A₂ as a single definite integral pb(x) S f(t) dt. A₁ + A₂ =
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