Let's assume that the pyramid is "SABCD" where "ABCD" is the base square and "O" is the center of the "ABCD" (i.e. "O=AC\\cap BD" ).
Let "M" be the middle of "AB" ("M\\in AB" ,"AM=MB=\\frac12 AB" ).
Given "OS=2" and "AB=1.2" we have to find "0.750\/2\\cdot (4S_{ABS})" .
Now we find:
1. "OM"
2. "MS"
3. "S_{ABS}=\\frac12 ah=\\frac12 AB\\cdot MS"
4. and produce the answer.
"OM=\\frac12 AB=0.6" ; from "\\Delta OSM" , "\\angle MOS=90^\\circ" , "MS=\\sqrt{OM^2+OS^2}=\\sqrt{4.36}"
"S_{ABS}=\\frac12 AB\\cdot MS=0.6\\sqrt{4.36}" .
The answer is "0.750\/2\\cdot (4S_{ABS})=0.750\/2\\cdot (4\\cdot 0.6\\sqrt{4.36})=0.9\\sqrt{4.36}\\approx 1.8792551716038992\\approx 1.88" so the right answer is B.
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