# PYTHAGOREAN THEOREM IN THREE DIMENSIONS

## The Pythagorean Theorem In a right triangle, the sum of the squares of the lengths of the  legs is equal to the square of the length of the hypotenuse.

If a and b are legs and c is the hypotenuse, then

a2 + b2 = c2

Example 1 :

A box used for shipping narrow silver sticks measures 6 inches by 6 inches by 20 inches. What is the length of the longest stick that will fit in the box, given that the length of the tube must be a whole number of inches ?

Solution :

Step 1 :

Draw an appropriate diagram for the given information. From the diagram given above, the box has the following dimensions.

Length (l)  =  20 in.

Width (w)  =  6 in.

Height (h)  =  6 in.

Step 2 :

We want to find d, the length from a bottom corner to the opposite top corner. First, find s, the length of the diagonal across the bottom of the box.

w2 + l2  =  s2

Step 3 :

Substitute the given measures.

62 + 202  =  s2

Simplify.

36 + 400  =  s2

436  =  s2

Step 3 :

Use the expression for s to find d.

h2 + s2  =  d2

Step 4 :

Plug h = 6 and s2 = 436.

62 + 436  =  d2

Simplify.

36 + 436  =  d2

472  =  d2

Take square root on both sides.

√472  =  √d2

21.7  ≈  d

Hence, the length of the longest stick that will fit in the box is 21 inches.

Example 2 :

Lily ordered a replacement part for her desk. It was shipped in a box that measures 4 in. by 4 in. by 14 in. What is the greatest length in whole inches that the part could have been ?

Solution :

Step 1 :

Draw an appropriate diagram for the given information. From the diagram given above, the box has the following dimensions.

Length (l)  =  14 in.

Width (w)  =  4 in.

Height (h)  =  4 in.

Step 2 :

We want to find d, the length from a bottom corner to the opposite top corner. First, find s, the length of the diagonal across the bottom of the box.

w2 + l2  =  s2

Step 3 :

Substitute the given measures.

42 + 142  =  s2

Simplify.

16 + 196  =  s2

212  =  s2

Step 3 :

Use the expression for s to find d.

h2 + s2  =  d2

Step 4 :

Plug h = 4 and s2 = 212.

42 + 212  =  d2

Simplify.

16 + 212  =  d2

228  =  d2

Take square root on both sides.

√228  =  √d2

15.1  ≈  d

Hence, the greatest length that the part could have been 15 inches.

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