Query 23

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course Mth 152

8/4 11:00pm

If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://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

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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.

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Your solution:

With angles boa and eof being vertical which would make them vertical, then using the sas congruence property would make abo and feo congruent.

confidence rating #$&*: 2

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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.

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Self-critique (if necessary):

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Self-critique Rating: 3

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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.

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Your solution:

The question gives us that angle A=42 deg, angle C=90 deg, So all three angles should add to 180 deg. So if 42+B+90=180 then B+132=180 then B=48 so for triangle QPR, angle P would be 90 deg then 4320/90=Angle R=48, To find angle Q=90+48+Q=180 then angle Q=42

confidence rating #$&*: 3

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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.

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Self-critique (if necessary):

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Self-critique Rating: 3

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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?

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Your solution:

Finding A you take 75(10)=25a then 750=25a then a=30 for b 75(20)=25b, then 1500=25b then b=60 so a=30,b=60, c=75

confidence rating #$&*: 3

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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.

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Self-critique (if necessary):

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Self-critique Rating: 3

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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?

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Your solution:

The Pythagorean Theorem formula is a^2+b^2+c^2 so 7^2+b^2=25^2

49+b^2=625

B^2=576

Then square root of 576 would be b=24

confidence rating #$&*: 3

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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.

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Self-critique (if necessary):

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Self-critique Rating: 3

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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?

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Your solution:

Finding the Pythogorean Triple if m=5

5^2-1/2=24/2=12

5^2+1/2=26/2=13

The triple would be 5, 12, 13

confidence rating #$&*: 3

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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.

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Self-critique (if necessary):

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Self-critique Rating: OK

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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?

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Your solution:

With there being a break you want to take x^2 +3^2=(10-x)^2 then you would have x^2+9=100+x^2 if you take x^2=91

confidence rating #$&*: 2

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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. **

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Self-critique (if necessary): Looking at the given solution I started out right but did not come out with the right solution, and how was -20x achieved.

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Self-critique Rating: 2

@&

(a - b ) ^ 2 = a^2 - 2 a b + b^2, which is a common special product, but which can be obtained by the distributive law.

(10 - x) ^2 can be also obtained directly by the distributive law:

(10 - x) ( 10 - x) = 10 ( 10 - x) - x ( 10 - x ) = 100 - 10 x - 10 x + x^2 = 100 - 20 x + x^2.

*@

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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?

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Your solution:

Basing upon that equal sides are x then base would be 128-2x=1/2(128-2x)=64-x so then 48^2+(64-x)^2+x^2

48^2+64^2-128x+x^2=x^2 then 48^2 +64^2/128x, 2304+4096=6400/128x, x=50 so then 128-2(50)=128-100=28 so take 1/2bh=1/2*28*48=672

confidence rating #$&*: 2

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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

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Self-critique (if necessary):

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Self-critique Rating: 3

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Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment.

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Self-critique (if necessary):

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Self-critique rating:

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Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment.

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Self-critique (if necessary):

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Self-critique rating:

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