Let the length along the front be x
let the width be y
let the height be z
x^2+ z^2 = 25^2 = 625 , #1
y^2 + z^2 = 20^2 = 400 , #2
subtract them
x^2 - y^2 = 225 , #3
also x^2 + y^2 + z^2 = 28^2 = 784 , #4
add it to #3
2x^2 + z^2 = 1009, but x^2 + z^2 = 625 from #1
so
2x^2 + z^2 = 1009
x^2 + z^2 = 625
x^2 = 384
then from #1
z^2= 241
from #2
y^2 = 159
check:
#1 : x^2 + z^2 = 384+241 =625 , check
#2 : y^2 + z^2 = 159+241 = 400 , check
#4 : x^2 + y^2 + z^2 = 384+159+241=784 , check
so we need the distance D
D^2 = x^2 + y^2 = 384+159=543
D = √543 = appr 23.3
The length of the diagonal across the front of a rectangular box is 25 inches, and the length of the diagonal across the side of the box is 20 inches. The length of a 3 dimensional diagonal of the box is 28 inches. What is length of the diagonal across the top of box?
A. 15.5
B. 23.3
C. 45.9
D. 46.6
answer is B , but I can't come up with that answer, I come up with something else.
1 answer
1 answer