QUESTION IMAGE
Question
find the surface area and volume of the solid. round each measure to the nearest tenth, if necessary. surface area ft² volume ft³
Step1: Find base area
The base is a square with side length $s = 16$ ft. The area of the base $B$ is given by $B=s^{2}=16^{2}=256$ ft².
Step2: Find area of triangular faces
There are 4 triangular faces. Two pairs of congruent triangles.
For the triangles with height $h_1 = 15$ ft and base $b = 16$ ft, the area of one such triangle $A_1=\frac{1}{2}\times b\times h_1=\frac{1}{2}\times16\times15 = 120$ ft².
For the triangles with height $h_2 = 17$ ft and base $b = 16$ ft, the area of one such triangle $A_2=\frac{1}{2}\times b\times h_2=\frac{1}{2}\times16\times17 = 136$ ft².
The total lateral - surface area $L = 2A_1+2A_2=2\times120 + 2\times136=240+272 = 512$ ft².
Step3: Calculate surface area
The surface area $SA$ of the pyramid is the sum of the base area and the lateral - surface area. So $SA=B + L=256+512=768$ ft².
Step4: Find the volume
First, we need to find the height $h$ of the pyramid. We can use the Pythagorean theorem. Let's consider the right - triangle formed by half of the base side and the slant height. The base of the right - triangle for finding the height of the pyramid: $a = 8$ ft.
For the slant height $l = 15$ ft, using the Pythagorean theorem $h=\sqrt{15^{2}-8^{2}}=\sqrt{225 - 64}=\sqrt{161}\approx 12.7$ ft.
The volume $V$ of a pyramid is given by $V=\frac{1}{3}Bh$, where $B = 256$ ft² and $h\approx12.7$ ft. So $V=\frac{1}{3}\times256\times12.7=\frac{3251.2}{3}\approx1083.7$ ft³.
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surface area: $768.0$ ft²
volume: $1083.7$ ft³