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If all three side lengths are INTEGERS and form PYTHAGOREAN TRIPLE then we know decimals are not allowed
if 24 and 51 are both legs then
x = 3177
and squre root of that gives us 56.364.... NOT an integer
So that means one of them is not a leg and since hypotenuse is longer than both legs, 51 must be the hypotenuse so to get the third side
![51^{2} - 24^{2} = x^{2}[tex] [tex] x^{2}](https://tex.z-dn.net/?f=%2051%5E%7B2%7D%20-%20%2024%5E%7B2%7D%20%20%3D%20%20x%5E%7B2%7D%5Btex%5D%20%5Btex%5D%20x%5E%7B2%7D%20)
=2025
x=45, leg B
if 24 and 51 are both legs then
x = 3177
and squre root of that gives us 56.364.... NOT an integer
So that means one of them is not a leg and since hypotenuse is longer than both legs, 51 must be the hypotenuse so to get the third side
x=45, leg B
See the attached figure to better understand the problem
let
L-----> length side of the cuboid
W----> width side of the cuboid
H----> height of the cuboid
we know that
One edge of the cuboid has length 2 cm-----> <span>I'll assume it's L
so
L=2 cm
[volume of a cuboid]=L*W*H-----> 2*W*H
40=2*W*H------> 20=W*H-------> H=20/W------> equation 1
[surface area of a cuboid]=2*[L*W+L*H+W*H]----->2*[2*W+2*H+W*H]
100=</span>2*[2*W+2*H+W*H]---> 50=2*W+2*H+W*H-----> equation 2
substitute 1 in 2
50=2*W+2*[20/W]+W*[20/W]----> 50=2w+(40/W)+20
multiply by W all expresion
50W=2W²+40+20W------> 2W²-30W+40=0
using a graph tool------> to resolve the second order equation
see the attached figure
the solutions are
13.52 cm x 1.48 cm
so the dimensions of the cuboid are
2 cm x 13.52 cm x 1.48 cm
or
2 cm x 1.48 cm x 13.52 cm
<span>Find the length of a diagonal of the cuboid
</span>diagonal=√[(W²+L²+H²)]------> √[(1.48²+2²+13.52²)]-----> 13.75 cm
the answer is
the length of a diagonal of the cuboid is 13.75 cm
let
L-----> length side of the cuboid
W----> width side of the cuboid
H----> height of the cuboid
we know that
One edge of the cuboid has length 2 cm-----> <span>I'll assume it's L
so
L=2 cm
[volume of a cuboid]=L*W*H-----> 2*W*H
40=2*W*H------> 20=W*H-------> H=20/W------> equation 1
[surface area of a cuboid]=2*[L*W+L*H+W*H]----->2*[2*W+2*H+W*H]
100=</span>2*[2*W+2*H+W*H]---> 50=2*W+2*H+W*H-----> equation 2
substitute 1 in 2
50=2*W+2*[20/W]+W*[20/W]----> 50=2w+(40/W)+20
multiply by W all expresion
50W=2W²+40+20W------> 2W²-30W+40=0
using a graph tool------> to resolve the second order equation
see the attached figure
the solutions are
13.52 cm x 1.48 cm
so the dimensions of the cuboid are
2 cm x 13.52 cm x 1.48 cm
or
2 cm x 1.48 cm x 13.52 cm
<span>Find the length of a diagonal of the cuboid
</span>diagonal=√[(W²+L²+H²)]------> √[(1.48²+2²+13.52²)]-----> 13.75 cm
the answer is
the length of a diagonal of the cuboid is 13.75 cm