how does the total surface area of a box change if
i) Each dimension is doubled ?
ii)Each dimension is tripled?
Express in words. Can you find the area if each dimension is raised to n times?
Answers
Answered by
214
let length =L
width=B
height=H
surface area of box=2{LB+BH+HL}
(a) all dimensions are double
so ,
new length =2L
width=2B
height =2H
now surface area=2{(2L)(2B)+(2B)(2H)+(2H)(2L)}
=4 x 2{LB+BH+HL}
compare we see 4times surface of box
(b) if all dimensions triples
then
surface area of box=2{(3L)(3B)+(3B)(3H)+(3H)(3L)}
=9 x 2 {LB+BH+HL}
we see surface area of box now, 9 times
yes " i find if dimensions raised n times
surface area of box then n^2 times
by above observation .
width=B
height=H
surface area of box=2{LB+BH+HL}
(a) all dimensions are double
so ,
new length =2L
width=2B
height =2H
now surface area=2{(2L)(2B)+(2B)(2H)+(2H)(2L)}
=4 x 2{LB+BH+HL}
compare we see 4times surface of box
(b) if all dimensions triples
then
surface area of box=2{(3L)(3B)+(3B)(3H)+(3H)(3L)}
=9 x 2 {LB+BH+HL}
we see surface area of box now, 9 times
yes " i find if dimensions raised n times
surface area of box then n^2 times
by above observation .
mysticd:
abhi , ur's also correct
no problem sir
you understand me this is enough for me
but last lines mension tsa = 4*2(LB+BH+HL) = 4times of FIRST TSA
by the shivam write good answer then me so he deserves this
okay i correct it
now
very good
Answered by
86
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