Truncated pyramid

Find the volume and surface area of a regular quadrilateral truncated pyramid if base lengths a1 = 17 cm, a2 = 5 cm, height v = 8 cm.

Correct result:

V =  1064 cm3
S =  754 cm2

Solution:

a1=17 cm a2=5 cm v=8 cm  S1=a12=172=289 cm2 S2=a22=52=25 cm2  V=13⋅ v⋅ (S1+S1⋅ S2+S2)=13⋅ 8⋅ (289+289⋅ 25+25)=1064 cm3
h2=v2+((a1−a2)/2)2=82+((17−5)/2)2=10 cm S3=a1+a22⋅ h2=17+52⋅ 10=110 cm2 S=S1+S2+4⋅ S3=289+25+4⋅ 110=754 cm2

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