Submitted by Editor
Objective
To explore the changes in surface areas and volumes of cuboids with respect to each other.
Description
Case 1:
Took the cuboids having equal volumes and following dimensions:-
1) Length = 20cm, Breadth = 10cm, Height = 5cm.
2) Length = 10cm, Breadth = 10cm, Height = 10cm.
3) Length = 100cm, Breadth = 5cm, Height = 2cm.
Calculation:-
= 20 x 10 x 5 = 1000 cubic cm. Surface area of cuboid = 2(lb + bh + hl)
= 2(20 x 10 + 10 x 5 + 5 x 12) = 2(200 + 50 + 60) = 620 square cm.
2. Volume of cuboid =lbh
= 10 x 10 x 10 = 1000 cubic cm. Surface area of cuboid = 2(lb + bh + hl)
= 2(10 x 10 + 10 x 10 + 10 x 10) = 2(100 + 100 + 100) = 600 square cm.
= 100 x 5 x 2
= 1000 cubic cm. Surface area of cuboid = 2(lb + bh + hl)
= 2(100 x 5 + 5 x 2 + 2 x 100) = 2(500 + 10 + 200) = 1420 square cm.
Observation:-
All these cuboids have volume =1000 cubic cm that is volumes are equal. The surface areas are not equal.
The surface of cuboid which is a cube is minimum.
Case 2:
Took the cuboids having equal volumes and following dimensions:-
1) Length = 14cm, Breadth = 6cm, Height = 5.4cm.
2) Length = 8cm, Breadth = 8cm, Height = 8cm.
3) Length = 16cm, Breadth = 6.4cm, Height = 4cm.
= 14 x 6 x 5.4 = 453.6 cubic cm.
Surface area of cuboid = 2(lb + bh + hl)
= 2(14 x 6 + 6 x 5.4 + 5.4 x 14) = 2(84 + 32.4 + 75.6) = 384 square cm.
2) Volume of cuboid =lbh
= 8 x 8 x 8 = 512 cubic cm.
Surface area of cuboid = 2(lb + bh + hl)
= 2(8 x 8 + 8 x 8 + 8 x 8) = 2(64 + 64 + 64) = 384 square cm.
= 16 x 6.4 x 4 = 409.6 cubic cm. Surface area of cuboid = 2(lb + bh + hl)
= 2(16 x 6.4 + 6.4 x 4+ 4 x 16) = 2(102.4 + 25.6+ 64) = 384 square cm.
Observation:
All these cuboids have surface area = 384 square cm that is surface areas are equal. The volumes are not equal.
The cuboid which a cube has largest volume.
Final Conclusion:
1) Of all the cuboids with equal volumes, the cube has the minimum surface area.
2) Of all the cuboids with equal surface areas, the cube has the maximum volume.
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