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kotykmax [81]
3 years ago
10

EFGH is a rhombus. Given EG = 16 and FH = 12, what is the length of one side of the rhombus?

Mathematics
2 answers:
natulia [17]3 years ago
7 0
We can calculate the side length with this formula:
<span>4 • Side² = Long Diagonal² + Short Diagonal²
</span><span>4 • Side² = 16² + 12²
</span><span>4 • Side² = 256 + 144
</span><span>4 • Side² = 400
</span><span>Side² = 100
</span>Side = 10

Source:
http://www.1728.org/quadltrl.htm



Tom [10]3 years ago
4 0

<u>Answer</u>

= 10


<u>Explanation</u>

EG is one of the diagonals of the rhombus and the other diagonal is FH.

EG = 16 and FH = 12

The diagonals of a rhombus meet at 90°

From this information we can form a right triangle with legs (16/2)=8 and (12/2)=6

The hypotenuse of this triangle is the side of the rhombus.

Using the pythagorean theorem;

L² = 6²+8²

Where L is the length of the rhombus.

L² = 6²+8²

L = √36+64

= √100

= 10

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olga55 [171]
For the first one,

you multiply 5x and -5 to get -25.

Then you move the constant to the right and change the sign to get 3x= -21 +25.

Calculate the sum of -21+25 to get 4.

Divide both sides of the equation by 3 to get x=4/3.

So then your final answer is x=4
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For the second one,

Distribute 7 through the parentheses to get 14x + 14+8=120

The you add 14+8 to get 22. Your equation should be 14x+22=120

Move the constant to the right side and change its sign to get 14x=120-22

Subtract 120-22 to get 98. Now your equation should be 14x=98

Divide both sides of the equation by 14 to get x=7.

Your finial answer for the second one should be x=7.
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I hope this made sense.
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Answer:

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