Answers

  • Brainly User
2015-12-31T00:47:18+08:00
The cube has 6 lateral square faces.  The corners are all right angles.  Therefore, use Pythagorean Theorem to solve for hypotenuse/ diagonals/distance between the vertices farthest from each other.

Step 1:  Find the hypotenuse/diagonal of the square (lateral face) given the edge which measures 10 inches.

Diagonal =  \sqrt{(side) ^{2}+(side) ^{2}  }

Diagonal =  \sqrt{(10) ^{2}+(10) ^{2}  }

Diagonal =  \sqrt{100 + 100}

Diagonal =  \sqrt{200}   =   \sqrt{(100)(2)}

Diagonal = 10  \sqrt{2}   inches

Step 2:  Find the distance between the vertices farthest from each other,
Edge = 10 inches
Diagonal of the lateral face/square = 10 \sqrt{2}    inches

Distance =  \sqrt{(10) ^{2}+(10 \sqrt{2}) ^{2}   }

Distance =  \sqrt{100 +(100)( \sqrt{4}) }

Distance =  \sqrt{100 + 100(2)}

Distance =  \sqrt{100 + 200}

Distance =  \sqrt{300}

Distance =  \sqrt{(100)(3)}

Distance = 10 \sqrt{3}   inches

Distance ≈ (10) (1.732) inches

Distance ≈ 17. 32 inches


ANSWER:  The distance between the vertices farthest from each other in a cube is 10 \sqrt{3} inches or approx. 17.32 inches.

Please click image below for my illustration with solution.
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