```Question 482509
a trough is formed by nailing together, edge to edge, two boards 11 ft. in length, so that the right section is a right triangle.
If 15 gal. of water are poured into the trough ad if the trough is held level so that a right section of the water is an isosceles right triangle, how deep is the water? (231 cu. in. = 1 gal.)
:
Find the vol of the water in cu/in
15 * 231 = 3465 cu/in
:
Change 11 ft:  11*12 = 132 inches
:
Find the triangular area of the end of the trough; 3465/132 = 26.25 sq/inches
:
Let x = width of the trough; A right triangle area =
{{{1/2}}}x*x = 26.25
x^2 = 2(26.25)
x = {{{sqrt(52.5)}}}
x = 7.2457 inches the width of each board
;
Find the width of the trough which is the the hypotenuse
h = {{{sqrt(x^2 + x^2)}}}
h = {{{sqrt(52.5 + 52.5)}}}
h = {{{sqrt(105)}}}
h = 10.247 inches wide, also the base of the triangle
:
Find the altitude of the triangle, which is the depth of the water; we know it's area
{{{1/2}}}*10.247*d = 26.25
10.247d = 2(26.25)
10.247d = 52.5
d = {{{52.5/10.247}}}
d = 5.12 inches is the depth of the water
:
:
Check this by finding the vol of water with these measurements
{{{1/2}}}*10.247*5.12*132 = 3462.366 cu/inches which is about 15 gal
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