For the cuboid shown in figure,the lengths of AB, BC, CG are 12, 8, 6 units respectively. Two points P and Q are chosen such that they are the mid-points of the edges GH and CD respectively. Two cuts are then made using a knife so as to cut the cuboid in to three pieces. The first cut passes through PQ to meet EA. The second cut is made on other side of PQ to meet FB. The difference in the total surface area (in square units) of the biggest of the three new pieces and the total surface area of initial cuboid is :

Explanation

PH=6,PG=6

GC=HD=FB=EA=6

1 AND 2 Part is symmetric.

SA of 1=2(area of ADQ)+area of PQAE+area of PHDG+area of HDEA

=`2*1/2*8*6+sqrt(8^2+6^2)*6+6*6+8*6`

=192`unit^2`

SA of 2=2(area of AQB)+2(area of PQAE)+area of (EABF)

`=2*1/2*12*8+2*60+12*6`

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