Query 23

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course Mth 152

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:

confidence rating #$&*: 3

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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): ok

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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:

Angle P = 90 deg.

Angle R = 48 deg.

Angle Q = 42 deg.

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): ok

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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:

a = 30, b = 60

Triple the size

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): ok

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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:

a^2 + b^2 = c^2

49 + b^2 = 625

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): ok

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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:

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): ok

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Self-critique Rating: 3

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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:

x^2 + 3^2 = (10-x)^2

x = 4.55 ft

confidence rating #$&*: 3

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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): ok

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Self-critique Rating: 3

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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:

48^2 + (64 - x)^2 = x^2

48^2 + 64 ( 64-x) - x(64 - x) = x^2

48^2 + 64^2 - 128 x = 0

(48^2 + 64^2) / 128 = x

x = 50.

128 - 2x = 128 - 2 * 50 = 128 - 100 = 28.

= 672

confidence rating #$&*: 3

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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): ok

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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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