Answer :
Kyle should calculare diagonal (3D)of that box because its the longest segment in the box.
First we want to find diagonal of the base, so
diagonal of base is from Pythagorean theorem, so
[tex]8^2+8^2=d^2[/tex]
[tex]d^2=128[/tex]
[tex]d=8 \sqrt{2} [/tex]
Now we can use again P theorem but with the triangle with sides, diagonal of the base, height it box and 3D diagonal.
So we have
[tex](8 \sqrt{2} )^2+8^2=(d3D)^2[/tex]
[tex](d3D)^2=192 d3D=8 \sqrt{3} d3D=13.85[/tex] so its more that paintbrush
First we want to find diagonal of the base, so
diagonal of base is from Pythagorean theorem, so
[tex]8^2+8^2=d^2[/tex]
[tex]d^2=128[/tex]
[tex]d=8 \sqrt{2} [/tex]
Now we can use again P theorem but with the triangle with sides, diagonal of the base, height it box and 3D diagonal.
So we have
[tex](8 \sqrt{2} )^2+8^2=(d3D)^2[/tex]
[tex](d3D)^2=192 d3D=8 \sqrt{3} d3D=13.85[/tex] so its more that paintbrush
