MR-838. problem

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