Expert: Sherman D. - 5/9/2004

Question
"1. Determine whether EFGH is a parallelogram, rectangle, rhombus, or square

E(2,-3) F(-3,1) G(1,6) H(6,2)


2. The geometric mean between 4 and 6.
i think it is 4.9 just not sure


3. there is a triangle labeled C in the bottom left corner and A at the top vertex and D at the bottom right corner:

There is a line that splits the top vertex in the middle equally down to the middle of the other side at point B. Point B should be between CD.

The the left side (AC) is labeled 8 and the right side (AD) is labeled 6

bottom left (CB) is labeled 2x+8 and the left bottom (DB) is labeled 3x-10

the question is solve for x and AB is an angle bisector... i hope that wasnt too confusing!!


4. find the sum of the interior angle of a decagon.

5. find the measure of an interior and an exterior angle of a regular hexagon.

6. a parallelogram had an area of 36m (squared) and a base of 9m (squared). find its height

7. A square has a diagonal with a length of 10. What is the perimeter of the square?

A. 40
B. 5 square roots of 2
C. 4 square roots of 5
D. 20 sqaure roots of 2


SORRY SOO MANY QUESTIONS AGAIN!

Answer
I will try my best to solve all of them.

"1. Determine whether EFGH is a parallelogram, rectangle, rhombus, or square

E(2,-3) F(-3,1) G(1,6) H(6,2)

EF = sqrt((-3 - 2)^2 + (1 - (-3))^2)
EF = sqrt((-3 + (-2))^2 + (1 + 3)^2)
EF = sqrt((-5)^2 + (4)^2)
EF = sqrt(25 + 16)
EF = sqrt(41)

FG = sqrt((1 - (-3))^2 + (6 - 1)^2)
FG = sqrt((1 + 3)^2 + (5)^2)
FG = sqrt((4)^2 + 25)
FG = sqrt(16 + 25)
FG = sqrt(41)

GH = sqrt((6 - 1)^2 + (2 - 6)^2)
GH = sqrt((5)^2 + (-4)^2)
GH = sqrt(25 + 16)
GH = sqrt(41)

EH = sqrt((6 - 2)^2 + (2 - (-3))^2)
EH = sqrt((4)^2 + (2 + 3)^2)
EH = sqrt(16 + (5)^2)
EH = sqrt(16 + 25)
EH = sqrt(41)

So this is either a Rhombus or a Square.
Since none of the points line up, this can't be a square, so your answer is RHOMBUS.

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2.) The geometric mean between 4 and 6.
i think it is 4.9 just not sure
sqrt(4 * 6) = sqrt(24) = about 4.9 so yes you are correct.

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3.)the question is solve for x and AB is an angle bisector... i hope that wasnt too confusing!

a^2 + b^2 = c^2
c = a^2 + b^2

Since hieght will be the same for both

8^2 + (2x + 8)^2 = 6^2 + (3x - 10)^2

64 + ((2x + 8)(2x + 8)) = 36 + ((3x - 10)(3x - 10))^2

64 + (4x^2 + 16x + 16x + 64) = 36 + (9x^2 - 30x - 30x + 100)

64 + 4x^2 + 32x + 64 = 36 + 9x^2 - 60x + 100

4x^2 + 32x + 128 = 9x^2 - 60x + 136

5x^2 - 92x + 8 = 0

Using pathagoream thereom

x = (-b ± sqrt(b^2 - 4ac))/2a
x = (-(-92) ± ((-92)^2 - 4(5)(8)))/(2(5))
x = (92 ± sqrt(8464 - 160))/10
x = (92 ± sqrt(8304))/10
x = (92 ± 4sqrt(519))/10
x = (46 ± 2sqrt(519))/5

If i am wrong on that, then the answer is as simple as this

Since this is an angle bisector, it divides CD in half, so

CB = BD
2x + 8 = 3x - 10
-x = -18
x = 18

Not certain which is correct, but CB = BD, seems better.

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4.) find the sum of the interior angle of a decagon.
beta = 180°(n-2)/n
(180(10 - 2))/10
(180(8))/10
(1440)/10
Beta = 144

Since there are 10 sides equal, the sum is 1440°

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5.) find the measure of an interior and an exterior angle of a regular hexagon.

Internal angles beta = 2/3 radians = 120 degrees

Exterior angle would problem be 60°, since 120 + 60 = 180.

Not certain.

Found at
http://mathforum.org/dr.math/faq/formulas/faq.regpoly.html

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6.) a parallelogram had an area of 36m (squared) and a base of 9m (squared). find its height

How is it possible for a base to be squared, unless the object you are working with is 3D.

Not totally certain about this one, but here goes anyway

Info found at http://mathforum.org/dr.math/faq/formulas/faq.quad.html#parallelogram

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7.) A square has a diagonal with a length of 10. What is the perimeter of the square?

B.) 5 square roots of 2

a^2 + b^2 = c^2
Since this is a square, a = b, so

a^2 + a^2 = c^2
2a^2 = c^2
c = 10
2a^2 = 10^2
2a^2 = 100
a^2 = 50
a = sqrt(50)
a = sqrt(25 * 2)
a = 5sqrt(2)