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Question 3
The area of the top of a rectangular box is 252 in2, the area of the front of the box is 105 in2, and the surface area of the box is 834 in2. What is the volume of the box?
First solution:
Let a,b,c be the sides of the rectangular box.
Let T be the area of the top/bottom of the box.
Let F be the area of the front/back of the box
Let R be the area of the right/left side of the box
T = ab = 252 sq. in.
F = ac = 105 sq. in.
R = bc
Let S be the surface area of this box.
S = 834 sq. in
S = 2ab+ 2ac + 2bc
S = 2 (ab + ac + bc)
Substitute 252 for ab, 105 for ac, 834 for S to solve for bc
834 = 2 (252 + 105 + bc)
834 = 2 (357 + bc)
834 / 2 = (357 + bc)
417 = 357 + bc
417 - 357 = bc
bc = 60 sq in.
Let V be the volume
V=abc
ab = 252 sq. in.
b = 252/a
ac = 105 sq. in.
c = 105/a
bc = 60 sq. in.
Substitute 252/a for b, 105/a for c to solve for a
252/a(105/a) = 60 sq.in.
252 (105) / a^2 = 60
26460 = 60a^2
26460 / 60 = a^2
441 = a^2
a = square root of 441
a = 21 in.
Since bc=60 sq. in. and a=21 in. the volume of the box is
product of them.
V = abc
Substitute 21 for a, 60 for bc
V = 21(60)
V = 1260 cubic inches
Ans: The volume is 1260 cubic inches.
Second Solution:
Let: x be the length of the box
y be the breadth of the box
z be the height of the box
From the problem, we know that:
xy = 252 sq.in .
xz = 105 sq. in.
and
2(xy + yz + xz) = 834 sq. in.
xy + yz + xz = 417 sq. in.
xy + yz + xz - xy - xz = 417 - 252 - 105
yz = 60 sq. in.
V=xyz = square root of ( x^2 * y^2 * z^2 )
x^2(y^2)(z^2)= 60(105)(252)
xyz = 60(105)(252)
xyz = square root of 1587600
xyz = 1260 cubic inches
Ans: The volume is 1260 cubic inches.