The edges of the lower and upper faces of a three-sided regular straight truncated pyramid are: a1=103 cm , a2=43 cm, and side edge is: s=37 cm. Calculate the volume and surface area of the truncated pyramid?
h1=a12-a122=a12-a124=4·a124-a124=3·a124
h1=3·a12
h1=3·10·32=3·102
h1=15
h2=3·a22
h2=3·4·32=3·42
h2=6
H2=372-62=9·7-36=63-36
H2=27=9·3
H=9·3
H=33
B1=a1·h12=103·152
B1=753
B2=a2·h22=43·62
B2=123
V=H3B1+B1·B2+B2
V=333·753+753·123+123
V=3·753+2700+123
V=3·753+900·3+123
V=3·753+303+123
V=3·1173
V=1053 cm3
h32=H2+32
h32=332+32
h32=27+9=36
h3=36
h3=6
A=B1+B2+M
The lateral area consists of three isosceles trapezoids.
M=3·a1+a22·h3
M=3·103+432·6
M=1263
A=B1+B2+M=753+123+1263
A=2133 cm2
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