Question

IB + CL = 28 cm - 20 cm = 8 cm IB = CL = 4 cm Hence, IB = BJ = CK = CL = DM = DN = AO = AP = 4 cm Area of section BEFC = Area of section DGHA Area of section ABEH = Area of section CDGF Area of section ABEH = Area of section CDGF = [12(16+24)(4)]=80 Exercise 11.3 : Solutions of Questions on Page Number : 186 Q1 : There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make Answer : We know that, Total surface area of the cuboid = 2 (lh + bh + lb) Total surface area of the cube = 6 (l) 2 Total surface area of cuboid (a) = [2{(60) (40) + (40) (50) + (50) (60)}] cm2 = [2(2400 + 2000 + 3000)] cm2 = (2 x 7400) cm2 = 14800 cm2



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Answer 1

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answer-IB + CL = 28 cm - 20 cm = 8 cm IB = CL = 4 cm Hence, IB = BJ = CK = CL = DM = DN = AO = AP = 4 cm Area of section BEFC = Area of section DGHA Area of section ABEH = Area of section CDGF Area of section ABEH = Area of section CDGF = [12(16+24)(4)]=80

Q1 : There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make

Answer : We know that, Total surface area of the cuboid = 2 (lh + bh + lb) Total surface area of the cube = 6 (l) 2 Total surface area of cuboid (a) = [2{(60) (40) + (40) (50) + (50) (60)}] cm2 = [2(2400 + 2000 + 3000)] cm2 = (2 x 7400) cm2 = 14800^2 cm.

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Answer 2

Answer:

Total surface area of the cube = 6 (l) 2 Total surface area of cuboid (a) = [2{(60) (40) + (40) (50) + (50) (60)}] cm2 = [2(2400 +