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InterviewSolution
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Find the volume and whole surface of frustum of a square pyramid , thje sides of whosebase and top are 24 cm and 16 cm respectively and each of the edges of the frustum is 20cm. |
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Answer» Solution :Given that a =24, b cm, Edge =e=20 cm AB=(1)/(2)`xx`Diagonal `=(1)/(2)xxasqrt(2)=(b)(sqrt(2))` `DC=(1)/(2)xx`Diagonal `=(1)/(sqrt(2)=(a)/(sqrt(2))` DE=DC-EC=DC-AB=`(a)(sqrt(2))-(b)/(sqrt(2))` To find height h, we have for sqare FRUSTUM `e^(2)=h^(2)+(1)/(2)(24-16)^(2)` `400=h^(2)+(1)/(2)xx8xx=h^(2)+32` `h^(2)=400-32=368 rarrh=19.18 cm` volume `V=(1)/(3)h(A_(1)+sqrt(A_(1)A_(2))+A_(2)` `A_(1)=a^(2)=24^(2)=576, A_(2)=b^(2)=16^(2)=256` `V=(1)/(3)xx19.18(576+384+256)=777.29 cm^(2)` Total SURFACE area =slant surface + Area of the TOP and bottom faces `=(1)/(2)` (sum of the perimeters `xx` slant height +`a^(2)+b^(2)`  [AF=b,DG=a] [KM=b] But DK=MG DK=Mg=`(a-b)/(2)` `e^(2)=l^(2)+DK^(2)` `e^(2)=l^(2)+((a)/(2)-(b)/(2))^(2)` `(20)^(2)=l^(2)+((24)/(2)(16)/(2^(2)))rArr400=l^(2)+(4^(2))` `[l^(2)`=384 `rArr l=sqrt(384)`=19.6] Total surface area `=(1)/(20)xx4xx16xx4)xx(19.6)+24^(2)+16^(2)` `=(1)/(2)(96+64)xx19.6+576+256` `=(1)/(2)xx160xx19.6+576+256` =1568+576+256=2400 `cm^(2)` 
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