What is the length of a diagonal of a cube with a side length of 10 cm? 200 cm 210cm 300 cm 320cm
2 answers:
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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<u>EXPLANATION</u><u>:</u>
Given that
sin θ = 1/2
We know that
sin 3θ = 3 sin θ - 4 sin³ θ
⇛sin 3θ = 3(1/2)-4(1/2)³
⇛sin 3θ = (3/2)-4(1/8)
⇛sin 3θ = (3/2)-(4/8)
⇛sin 3θ = (3/2)-(1/2)
⇛sin 3θ = (3-1)/2
⇛sin 3θ = 2/2
⇛sin 3θ = 1
and
cos 2θ = cos² θ - sin² θ
⇛cos 2θ = 1 - sin² θ - sin² θ
⇛cos 2θ = 1 - 2 sin² θ
Now,
cos 2θ = 1-2(1/2)²
⇛cos 2θ = 1-2(1/4)
⇛cos 2θ = 1-(2/4)
⇛cos 2θ = 1-(1/2)
⇛cos 2θ = (2-1)/2
⇛cos 2θ = 1/2
Now,
The value of sin 3θ /(1+cos 2θ
⇛1/{1+(1/2)}
⇛1/{(2+1)/2}
⇛1/(3/2)
⇛1×(2/3)
⇛(1×2)/3
⇛2/3
<u>Answer</u> : Hence, the req value of sin 3θ /(1+cos 2θ) is 2/3.
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