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Illusion [34]
3 years ago
15

Inside a square with side length 10, two congruent equilateral triangles are drawn such that they share one side and each has on

e vertex on a vertex of the square. What is the side length of the largest square that can be inscribed in the space inside the square and outside of the triangles?

Mathematics
1 answer:
Sphinxa [80]3 years ago
4 0

Answer:

5 length

Step-by-step explanation:

The diagram attached shows two equilateral triangles ABC & CDE. Since both squares share  one side of the square  BDFH of length 10, then their lengths will be 5 each. To obtain the largest square inscribed inside the original square BDFH, it makes sense to draw two other equilateral triangles AGH & EFG at the upper part of BDFH with length equal to 5.

So, the largest square that can be inscribe in the space outside the two equilateral triangles ABC & CDE and within BDFH is the square ACEG.

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Solve equations show all steps 1. 2x-10=44+8x 2. 7x-4=20=3x 3. 2(x-3)=-20 4. 15-4x+5=32
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Answer:

2x - 10 = 44 + 8x

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Step-by-step explanation:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                    2*x-10-(44+8*x)=0

Pull out like factors :

  -6x - 54  =   -6 • (x + 9)

  -6   =  0

 Solve  :    x+9 = 0

Subtract  9  from both sides of the equation :

                     x = -9

x = -9

Move all terms containing

x

to the left side of the equation.

4

x

−

4

=

20

Move all terms not containing

x

to the right side of the equation.

4

x

=

24

divide each term by 4

x = 6

2(x−3)=−20

Step 1: Simplify both sides of the equation.

2(x−3)=−20

2x−6=−20

Step 2: Add 6 to both sides.

2x−6+6=−20+6

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Step 3: Divide both sides by 2.

2x

2

=

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x=−7

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Step 2: Subtract 20 from both sides.

−4x+20−20=32−20

−4x=12

Step 3: Divide both sides by -4.

−4x

−4

=

12

−4

x=−3

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