What is the diagonal of a box with dimensions 2 feet by 3 feet by 5 feet?

Answer 1

The first step to take is to solve the hypotenuse of one of the sides of the box.

Use Pythagoras to figure out the hypotenuse of the side with lengths 3 feet and 5 feet.

Then you can use that diagonal as one side of a triangle with the other side of the triangle being the 2 feet length. Use Pythagoras to solve for the hypotenuse of that triangle.

You should get
First hypotenuse = 5.83 ft
Second hypotenuse (your answer) = 6.16 ft

Answer 2

Diagonal of a box :

A^2 + B^2 + C^2 = D^2 .

= 2^2 + 3^2 + 5^2
= 4 + 9 + 25
= 38

D = sqrt.(38) = 6.164

Answer 3

It's a 2 triangle problem. Assume the base is 2x3 and the height is 5. First, draw a line from corner to corner across the bottom of the box. That is the hypotenuse of the triangles that make up the bottom of the box and its length is SQRT(2x2 + 3x3)=SQRT (13). This hypotenuse becomes the base of the second triangle, and the height of the box is the other side of that triangle. The diagonal of the box is the hypotenuse of this second triangle which is SQRT[SQRT(13)xSQRT(13) + 5x5)] = SQRT(13 + 25) = SQRT(38) = 6.164 feet.

Answer 4

d ² = 2 ² + 3 ² + 5 ²
d ² = 4 + 9 + 25
d ² = 38
d = √38 ft
d = 6.16 ft ( to 2 decimal places)

Answer 5

Diagonal of the Base=sq rt(2^2+3^2)
=sq rt (4+9)=sq rt(13)
Now diagonal of the Box=sq rt[5^2+{(sq rt13)}^2]
or =sq rt(25+13)
=sq rt(38) ans

Answer 6

d=sqrt[2^2+3^2+5^2]
=sqrt[38]
=6.16. ANS.

Answer 7

(2^2 + 3^2 + 5^2)^1/2 =(38)^1/2 = 6.164

the square root of the sum of the squares of the sides

Answer 8

sqrt(2^2+3^2+5^2) = 6.1644