NEED HELP PLZ Each edge of the cube measures 4 inches in length
What is the distance, in inches, from vertex B to vertex H? Round your answer to the nearest tenth of an inch

The distance from vertex B to vertex H is 6.9 inches.

Step-by-step explanation:

The length of the line BH can be thought of as the hypotenuse of a triangle rectangle where the catheti are the lines HD (whit a measure of 4in) and line DB.

The length of the line DB can be thought of as the hypotenuse of a triangle rectangle where the catheti are lines DA and AB (both are 4in long)

Then if we use the Pythagorean's theorem, the length of line DB is:

(DB)^2 = (DA)^2 + (AB)^2

(DB)^2 = (4in)^2 + (4in)^2 = 32in^2

(DB) = √(32in^2) = 5.66 in

Whit this, we can find the length of line HB as:

(HB)^2 = (HD)^2 + (DB)^2

(HB)^2 = (4in)^2 + (DB)^2 = 16in^2 + 32in^2 = 48in^2

HB = √(48in^2) = 6.93 in

If we round to the nearest thent, we get:

HB = 6.9 in

The distance from vertex B to vertex H is 6.9 inches.

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351.23

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