Question

an amazing formula from ancient mathematics was used by the egyptians to find the volume of the frustum of a square pyramid, as shown in the figure. its volume is given by (v=\frac{1}{3}h(a^{2}+ab + b^{2})), where b is the length of the base, a is the length of the top, and h is the height. a. when a pyramid was partially completed to a height h of 206 ft, b was 760 ft, and a was 318 ft. calculate its volume at this stage of construction. the volume is (square) ft³. (round to the nearest integer as needed.)

Explanation:

Step1: Substitute given values into formula

Given $h = 206$, $a=318$, $b = 760$, and $V=\frac{1}{3}h(a^{2}+ab + b^{2})$. Substitute these values: $V=\frac{1}{3}\times206\times(318^{2}+318\times760 + 760^{2})$.

Step2: Calculate individual - squared and product terms

First, calculate $318^{2}=318\times318 = 101124$, $318\times760=241680$, and $760^{2}=760\times760 = 577600$.

Step3: Calculate the sum inside the parentheses

$318^{2}+318\times760 + 760^{2}=101124+241680 + 577600=920404$.

Step4: Calculate the volume

$V=\frac{1}{3}\times206\times920404=\frac{206\times920404}{3}=\frac{189603224}{3}\approx63201075$.

Answer:

$63201075$