If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution.
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
023. ``q Query 23
Question: `q Query 9.4.6 ABC, DEF transversed by EOB at rt angles; OB = EO; show triangles ABO and FEO congruent.
**** Explain the argument you used to show that the triangles were congruent.
Your solution:
Confidence Assessment:
Given Solution:
`a SAS: Angle AOB and Angle FOE are equal because they are vertical angles, so we have 2 sides and the included angle of triangle AOB equal, respectively, to 2 sides and the included angle of triangle FOE. Thus, the Side-Angle-Side property holds that triangle AOB is congruent to triangle FOE.
Self-critique (if necessary):
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Question: `q Query 9.4.18 ACB and QPR similar triangles, C and P rt angles, A=42 deg **** List the measures of the three angles of each triangle and explain how you obtained each.
Your solution:
Confidence Assessment:
Given Solution:
`a It is given that Angle A = 42 deg. and Angle C = 90 deg. Since all three angles must add up to equal 180 then Angle B = 48 deg.
In the second triangle, Angle P must equal 90 deg. since it is a right angle.
To find Angle R,
90(48) = 90R sp
4320 = 90R and
48 = R Angle R = 48 deg.
To find Angle Q,
90/90 = Q/42
Q = 42
Angle Q = 42 deg.
Self-critique (if necessary):
Self-critique Rating:
Question: `q Query 9.4.24 similar triangles, corresp sides a, b, 75; 10, 20, 25 **** What are the lengths of sides a and b and how did you obtain each?
Your solution:
Confidence Assessment:
Given Solution:
`a To find a,
75 (10) = 25a
750 = 25a
a= 30
To find b,
75/25 = b/20
1500/25 = 25b/25 so
b = 60.
a = 30, b = 60 and c = 75.
These values are triple the values of the similar triangle.
Self-critique (if necessary):
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Question: `q Query 9.4.42 rt triangle a = 7, c = 25, find b
**** What is the length of side b and how did you obtain it?
**** What does the Pythagorean Theorem say about the triangle as given and how did you use this Theorem to find the length of b?
Your solution:
Confidence Assessment:
Given Solution:
`a By the Pythagorean Theorem a^2 + b^2 = c^2. So we have
49 + b^2 = 625 Subtract 49 from both sides to get
b^2 = 576. Take the square root of both sides to get
b = 24.
Self-critique (if necessary):
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Question: `q Query 9.4.60 m, (m^2 +- 1) / 2 gives Pythagorean Triple **** What Pythagorean Triple is given by m = 5?
Your solution:
Confidence Assessment:
Given Solution:
`a ** If m = 5 then
(m^2 + 1) / 2 = (5^2 + 1 ) / 2 = 26 / 2 = 13
(m^2 - 1) / 2 = (5^2 - 1 ) / 2 = 24 / 2 = 12
So the Pythagorean triple is 5, 12, 13.
We can verify this:
5^2 + 12^2 should equal 13^2.
5^2 + 12^2 = 25 + 144 = 169.
13^2 = 169.
The two expressions are equal so this is indeed a Pythagorean triple. **
**** How did you verify that your result is indeed a Pythagorean Triple?
Student Answer: The numbers checked out when substituted into the Pythagorean Theorem.
Self-critique (if necessary):
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Question: `q Query 9.4.75 10 ft bamboo broken, upper end touches ground 3 ft from stem.
**** How high is the break, and how did you obtain your result?
Your solution:
Confidence Assessment:
Given Solution:
`a ** If the break is at height x then the hypotenuse, consisting of the broken part, is at height 10 - x.
The triangle formed by the vertical side, the break and the ground therefore has legs x and 3 and hypotenuse 10-x.
So we have
x^2 + 3^2 = (10-x)^2. Squaring the 3 and the right-hand side:
x^2 + 9 = 100 - 20 x + x^2. Subtracting x^2 from both sides
9 = 100 - 20 x so that
-20 x = -91 and
x = 4.55.
The break occurs at height 4.55 ft and the broken part has length 10 - 4.55 = 5.45, or 5.45 feet. **
Self-critique (if necessary):
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Question: `q Query 9.4.84 isosceles triangle perimeter 128 alt 48 **** What is the area of the triangle and how did you find
it?
Your solution:
Confidence Assessment:
Given Solution:
`a ** This problem is algebraically demanding. Your text might have a slicker way to do this, but the following works:
If the equal sides are x then the base is 128 - 2 x.
The altitude forms a right triangle with half the base and one of the equal sides. The sides of this right triangle are therefore 48, 1/2 (128 - 2x) = 64 - x, and x.
The right angle is formed between base and altitude so x is the hypotenuse.
We therefore have
48^2 + (64 - x)^2 = x^2 so that
48^2 + (64 - x) ( 64 - x) = x^2 or
48^2 + 64 ( 64-x) - x(64 - x) = x^2 or
48^2 + 64^2 - 64 x - 64 x + x^2 = x^2 or
48^2 + 64^2 - 128 x + x^2 = x^2. Subtracting x^2 from both sides we get
48^2 + 64^2 - 128 x = 0. Adding 128 x to both sides we get
48^2 + 64^2 = 128 x. Multiplying both sides by 1/128 get have
(48^2 + 64^2) / 128 = x. Evaluating this expression we end up with x = 50.
The base of the triangle is therefore 128 - 2x = 128 - 2 * 50 = 128 - 100 = 28.
So its area is 1/2 b h = 1/2 * 28 * 48 = 672. **
DRV
Self-critique (if necessary):
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Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment.