if the radius of the circular region is 5 yards what was the total length of the border to the nearest tenth of a yard
1 answer:
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Explanation:
a. The line joining the midpoints of the parallel bases is perpendicular to both of them. It is the line of symmetry for the trapezoid. This means the angles and sides on one side of that line of symmetry are congruent to the corresponding angles and sides on the other side of the line. The diagonals are the same length.
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b. We observe that adjacent pairs of points have the same x-coordinate, so are on vertical lines, which have undefined slope. KN is a segment of the line x=1; LM is a segment of the line x=3. If the trapezoid is isosceles, the midpoints of these segments will be on a horizontal line. The midpoint of KN is at y=(3-2)/2 = 1/2. The midpoint of LM is at y=(1+0)/2 = 1/2. These points are on the same horizontal line, so the trapezoid <em>is isosceles</em>.
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c. We observed in part (b) that the parallel sides are KN and LM. The coordinate difference between K and L is (1, 3) -(3, 1) = (-2, 2). That is, segment KL is the hypotenuse of an isosceles right triangle with side lengths 2, so the lengths of KL and MN are both 2√2.
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For part (c), we used the shortcut that the hypotenuse of an isosceles right triangle is √2 times the leg length.
Proof of the Law of Sines
The Law of Sines states that for any triangle ABC, with sides a,b,c (see below)
a
sin A
=
b
sin B
=
c
sin C
For more see Law of Sines.
Acute triangles
Draw the altitude h from the vertex A of the triangle
From the definition of the sine function
sin B =
h
c
a n d sin C =
h
b
or
h = c sin B a n d h = b sin C
Since they are both equal to h
c sin B = b sin C
Dividing through by sinB and then sinC
c
sin C
=
b
sin B
Repeat the above, this time with the altitude drawn from point B
Using a similar method it can be shown that in this case
c
sin C
=
a
sin A
Combining (4) and (5) :
a
sin A
=
b
sin B
=
c
sin C
- Q.E.D
Obtuse Triangles
The proof above requires that we draw two altitudes of the triangle. In the case of obtuse triangles, two of the altitudes are outside the triangle, so we need a slightly different proof. It uses one interior altitude as above, but also one exterior altitude.
First the interior altitude. This is the same as the proof for acute triangles above.
Draw the altitude h from the vertex A of the triangle
sin B =
h
c
a n d sin C =
h
b
or
h = c sin B a n d h = b sin C
Since they are both equal to h
c sin B = b sin C
Dividing through by sinB and then sinC
c
sin C
=
b
sin B
Draw the second altitude h from B. This requires extending the side b:
The angles BAC and BAK are supplementary, so the sine of both are the same.
(see Supplementary angles trig identities)
Angle A is BAC, so
sin A =
h
c
or
h = c sin A
In the larger triangle CBK
sin C =
h
a
or
h = a sin C
From (6) and (7) since they are both equal to h
c sin A = a sin C
Dividing through by sinA then sinC:
a
sin A
=
c
sin C
Combining (4) and (9):
a
sin A
=
b
sin B
=
c
sin C
