the height of a storage bin, in the form of a frustum of a rectangular pyramid, is 6 ft. its volume is 818cu.ft , and the upper base is 4 ft. by 16 ft. what are the dimensions of the lower base?
$V = \frac{1}{3}(A_1 + A_2 + \sqrt{A_1A_2})h$
$818 = \frac{1}{3}\left[ \, 4(16) + A_2 + \sqrt{4(16)A_2} \, \right](6)$
$345 - A_2 = \sqrt{64A_2}$
$(345 - A_2)^2 = 64A_2$
$119,025 - 690A_2 + {A_2}^2 = 64A_2$
${A_2}^2 - 754A_2 + 119,025 = 0$
$A_2 = 529 ~ \text{and} ~ 225$
If A2 = 529, V = 1554 (not okay) If A2 = 225, V = 818 (okay!)
Use A2 = 225 ft2
The lower base is proportional to the upper base
$a = 7.5 ~ \text{ft}$
$\dfrac{b^2}{16^2} = \dfrac{225}{64}$
$b = 30 ~ \text{ft}$
Dimensions = 7.5 ft × 30 ft answer
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$V = \frac{1}{3}(A_1 + A_2 + \sqrt{A_1A_2})h$
$818 = \frac{1}{3}\left[ \, 4(16) + A_2 + \sqrt{4(16)A_2} \, \right](6)$
$345 - A_2 = \sqrt{64A_2}$
$(345 - A_2)^2 = 64A_2$
$119,025 - 690A_2 + {A_2}^2 = 64A_2$
${A_2}^2 - 754A_2 + 119,025 = 0$
$A_2 = 529 ~ \text{and} ~ 225$
If A2 = 529, V = 1554 (not okay)
If A2 = 225, V = 818 (okay!)
Use A2 = 225 ft2
The lower base is proportional to the upper base
$a = 7.5 ~ \text{ft}$
$\dfrac{b^2}{16^2} = \dfrac{225}{64}$
$b = 30 ~ \text{ft}$
Dimensions = 7.5 ft × 30 ft answer
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