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 = [tex] \sqrt{(side) ^{2}+(side) ^{2} } [/tex]
Diagonal = [tex] \sqrt{(10) ^{2}+(10) ^{2} } [/tex]
Diagonal = [tex] \sqrt{100 + 100} [/tex]
Diagonal = [tex] \sqrt{200} [/tex] = [tex] \sqrt{(100)(2)} [/tex]
Diagonal = [tex]10 \sqrt{2} [/tex] inches
Step 2: Find the distance between the vertices farthest from each other,
Edge = 10 inches
Diagonal of the lateral face/square = [tex]10 \sqrt{2} [/tex] inches
Distance =[tex] \sqrt{(10) ^{2}+(10 \sqrt{2}) ^{2} } [/tex]
Distance = [tex] \sqrt{100 +(100)( \sqrt{4}) } [/tex]
Distance = [tex] \sqrt{100 + 100(2)} [/tex]
Distance = [tex] \sqrt{100 + 200} [/tex]
Distance = [tex] \sqrt{300} [/tex]
Distance = [tex] \sqrt{(100)(3)} [/tex]
Distance = [tex]10 \sqrt{3} [/tex] inches
Distance ≈ (10) (1.732) inches
Distance ≈ 17. 32 inches
ANSWER: The distance between the vertices farthest from each other in a cube is [tex]10 \sqrt{3} [/tex] inches or approx. 17.32 inches.
Please click image below for my illustration with solution.
