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serious [3.7K]
3 years ago
12

What is the length of a diagonal of a cube with a side length of 10 cm? 200 cm 210cm 300 cm 320cm

Mathematics
2 answers:
Andrej [43]3 years ago
8 0

Answer:

its c on endeguity

Step-by-step explanation:

Scilla [17]3 years ago
7 0

Answer:

The length of the diagonal of the cube = √(3 × 10²) = √300 cm

Step-by-step explanation:

* Lets revise the properties of the cube

- It has six equal faces all of them are squares

- It has 12 vertices

- The diagonal of the cube is the line joining two vertices in opposite

 faces (look to the attached figure)

- To find the length of the diagonal do that:

# Find the diagonal of the base using Pythagoras theorem

∵ The length of the side of the cube is L

∵ The base is a square

∴ The length of the diagonal d = √(L² + L²) = √(2L²)

- Now use the diagonal of the base and a side of a side face to find the

 diagonal of the cube by Pythagoras theorem

∵ d = √(2L²)

∵ The length of the side of the square = L

∴ The length of the diagonal of the cube = √[d² + L²]

∵ d² = [√(2L²)]² = 2L² ⇒ power 2 canceled the square root

∴ The length of the diagonal of the cube = √[2L² + L²] = √(3L²)

* Now lets solve the problem

∵ The length of the side of the square = 10 cm

∴ The length of the diagonal of the cube = √(3 × 10²) = √300 cm

- Note: you can find the length of the diagonal of any cube using

 this rule Diagonal = √(3L²)

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Step-by-step explanation:

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Line l1 passes through (-2, 5), and (-1, -10)
zalisa [80]

The equations of the lines l1 with points (-2, 5), and (-1, -10) and the line l2 with points (5, 15), and (3, 8), gives the coordinate of the intersection between the lines as the point; (-45/37, -250/37)

<h3>Which method can be used to describe the lines to find the intersection point?</h3>

The slope, m1, of line 1 l1 is found as follows;

  • m1 = (5 - (-10))/(-2- (-1)) = -15

The equation of line l1 in point and slope form is therefore;

y1 - 5 = -15•(x - (-2))

Which gives;

y1 = -15•(x - (-2)) + 5 = -15•x - 25

  • y1 = -15•x - 25

The slope, m2, of line 2 l2 is found as follows;

m2 = (15 - 8)/(5 - 3) = 3.5

Equation of line l2 is therefore;

y2 - 15 = 3.5•(x - 5)

Which gives;

y2 = 3.5•(x - 5) + 15 = 3.5•x - 2.5

  • y2 = 3.5•x - 2.5

At the intersection point, we have;

y1 = y2

Therefore;

-15•x - 25 = 3.5•x - 2.5

18.5•x = -22.5

x = -22.5/18.5 = -45/37

y2 = 3.5•x - 2.5

At the intersection point, we have;

y = y2 = 3.5×(-22.5/18.5) - 2.5 = -250/37

y = -250/37

The coordinates of the intersection between the lines is therefore;

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