PLEASE SOLVE QUESTION 10.
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Hola User________________
〽⭐Here is Your Answer..!!!
_________________________
↪Actually welcome to the concept of the MENSURATION..
↪Basically here as given ...
↪The Lenght of the side of a individual square = 4 cm
↪thus here after fusing it into a big cube ...
↪The EDGE LENGTH OF THE NEW BIG CUBE = 2 times the edge of the old small cube ..
↪there fore New edge length = 2 × 4 = 8 cm
since the Length will be L = 8 cm , height will be H= 8 cm but the breadth will be B = 4 cm
↪...Now since we know that
the total SURFACE AREA OF A CUBOID = 2 ( LB + BH + LH )
↪since L = 8 cm , B =4 cm and H = 8cm
↪thus T.S.A = 2 ( 8 (4) + 4 (8) + 8 (8) ) ===》 256 cm^2 ...
↪thus the T.S.A OF THE NEW CUBOID = 256 cm^2
↪NOW COST OF PAINTING IS Rs. 32 / cm^2
↪Thus cost of painting the whole cube = T.S.A × RATE
↪=====》 256 × 32== 》 Rs. 8192 /-
↪Thus the total expenditure on painting is Rs. 8192/-
_________________________
↪〽Hope it helps u....☺
〽⭐Here is Your Answer..!!!
_________________________
↪Actually welcome to the concept of the MENSURATION..
↪Basically here as given ...
↪The Lenght of the side of a individual square = 4 cm
↪thus here after fusing it into a big cube ...
↪The EDGE LENGTH OF THE NEW BIG CUBE = 2 times the edge of the old small cube ..
↪there fore New edge length = 2 × 4 = 8 cm
since the Length will be L = 8 cm , height will be H= 8 cm but the breadth will be B = 4 cm
↪...Now since we know that
the total SURFACE AREA OF A CUBOID = 2 ( LB + BH + LH )
↪since L = 8 cm , B =4 cm and H = 8cm
↪thus T.S.A = 2 ( 8 (4) + 4 (8) + 8 (8) ) ===》 256 cm^2 ...
↪thus the T.S.A OF THE NEW CUBOID = 256 cm^2
↪NOW COST OF PAINTING IS Rs. 32 / cm^2
↪Thus cost of painting the whole cube = T.S.A × RATE
↪=====》 256 × 32== 》 Rs. 8192 /-
↪Thus the total expenditure on painting is Rs. 8192/-
_________________________
↪〽Hope it helps u....☺
Cherrg01:
no let it be...
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