Complete question:
The base of a pyramid covers an area of 14.0 acres (1 acre = 43, 560 ft² ) and has a height of 481 ft. If the volume of a pyramid is given by the expression V = bh/3, where b is the area of the base and h is the height. If the height of the pyramid were increased to 512 ft and the height to base area ratio of the pyramid were kept constant, by what percentage would the volume of the pyramid increase
Answer:
The percentage increase of the volume is 6.45 %
Step-by-step explanation:
The volume of pyramid is calculated as;
V = bh/3
The initial volume of the pyramid is calculated as follows;
[tex]V = \frac{bh}{3} \\\\V_0 = (\frac{1}{3} ) \times (14 \ acres \ \times \frac{43,560 \ ft^2}{1 \ acre} ) \times (481 \ ft)\\\\V_0 = 97,777,680 \ ft^3[/tex]
The final volume of the pyramid when the height is increased to 512 ft.
[tex]V_1 = (\frac{1}{3} ) \times (14 \ acres \ \times \frac{43,560 \ ft^2}{1 \ acre} ) \times (512 \ ft)\\\\V_1 = 104,079,360 \ ft^3[/tex]
The percentage increase of the volume;
[tex]= \frac{104,079,360 - 97,777,680}{97,777,680} \times 100\%\\\\= 6.45 \ \%[/tex]
The volume of the pyramid will increase by 11.79%
The volume of a Pyramid is calculated using:
[tex]V = \frac{Bh}3[/tex]
Where:
When the height is increased from 458 to 512, then the initial and the final volumes are:
[tex]V_1 = \frac{458}{3}b[/tex]
[tex]V_2 = \frac{512}{3}b[/tex]
The percentage change is then calculated as:
[tex]\% Change = \frac{V_2 -V_1}{V_1} \times 100\%[/tex]
This gives
[tex]\% Change = \frac{512b/3 -458b/3}{458b/3} \times 100\%[/tex]
Simplify the numerator
[tex]\% Change = \frac{54b/3}{458b/3} \times 100\%[/tex]
Evaluate the quotient
[tex]\% Change = 0.1179 \times 100\%[/tex]
Express as percentage
[tex]\% Change = 11.79 \%[/tex]
Hence, the volume of the pyramid will increase by 11.79%
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