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Question-23537

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Question Number 23537 by math solver last updated on 01/Nov/17
Commented by math solver last updated on 01/Nov/17
q.7 ?
$$\mathrm{q}.\mathrm{7}\:? \\ $$
Commented by ajfour last updated on 01/Nov/17
yes 24 , a=2, h=2 i have nearly solved, but some proof is still wanting..
$${yes}\:\mathrm{24}\:,\:{a}=\mathrm{2},\:{h}=\mathrm{2} \\ $$$${i}\:{have}\:{nearly}\:{solved},\:{but}\:{some} \\ $$$${proof}\:{is}\:{still}\:{wanting}.. \\ $$
Commented by math solver last updated on 01/Nov/17
sir it is not even in options !
$$\mathrm{sir}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{even}\:\mathrm{in}\:\mathrm{options}\:! \\ $$
Commented by math solver last updated on 01/Nov/17
the answer is 24. but i don′t know how ?
$$\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{24}. \\ $$$$\mathrm{but}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{how}\:? \\ $$
Commented by math solver last updated on 01/Nov/17
did you get the required proof?
$$\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{the}\:\mathrm{required}\:\mathrm{proof}? \\ $$
Commented by Rasheed.Sindhi last updated on 02/Nov/17
2a^2 +4ah=8a+4h a^2 +2ah−4a=2h a(a+2h−4)=2h ((2h)/a)=a+2h−4 ((2h)/a)∈Z a∣2 or a∣h a∣2⇒a=1,2 For a=1 2a^2 +4ah=8a+4h ⇒2+4h=8+4h (impossible) ⇒a=2 2a^2 +4ah=8a+4h 8+8h=16+4h⇒h=2 One solution is a=h=2 Hence sum of edges=24 For a∣h Let h=ka for some integer k. 2a^2 +4ah=8a+4h ⇒2a^2 +4a.ka=8a+4.ka ⇒2a^2 +4ka^2 =8a+4ka ⇒a+2ka=4+2k ⇒2k(a−1)=4−a ((4−a)/(2(a−1))) =k k=1⇒4−a=2a−2⇒a=2 Again a=h=2 and sum of edges=24 k=2⇒4−a=4(a−1)⇒a=(8/5) ∉N Continue
$$\mathrm{2a}^{\mathrm{2}} +\mathrm{4ah}=\mathrm{8a}+\mathrm{4h} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{2ah}−\mathrm{4a}=\mathrm{2h} \\ $$$$\mathrm{a}\left(\mathrm{a}+\mathrm{2h}−\mathrm{4}\right)=\mathrm{2h} \\ $$$$\frac{\mathrm{2h}}{\mathrm{a}}=\mathrm{a}+\mathrm{2h}−\mathrm{4} \\ $$$$\frac{\mathrm{2h}}{\mathrm{a}}\in\mathbb{Z} \\ $$$$\mathrm{a}\mid\mathrm{2}\:\:\:\mathrm{or}\:\:\:\mathrm{a}\mid\mathrm{h} \\ $$$$\mathrm{a}\mid\mathrm{2}\Rightarrow\mathrm{a}=\mathrm{1},\mathrm{2} \\ $$$$\mathrm{For}\:\mathrm{a}=\mathrm{1}\: \\ $$$$\:\:\:\:\mathrm{2a}^{\mathrm{2}} +\mathrm{4ah}=\mathrm{8a}+\mathrm{4h} \\ $$$$\:\:\:\:\:\Rightarrow\mathrm{2}+\mathrm{4h}=\mathrm{8}+\mathrm{4h}\:\left({impossible}\right) \\ $$$$\:\:\:\:\:\Rightarrow\mathrm{a}=\mathrm{2} \\ $$$$\:\:\:\:\mathrm{2a}^{\mathrm{2}} +\mathrm{4ah}=\mathrm{8a}+\mathrm{4h} \\ $$$$\:\:\:\:\:\:\mathrm{8}+\mathrm{8h}=\mathrm{16}+\mathrm{4h}\Rightarrow\mathrm{h}=\mathrm{2} \\ $$$$\mathrm{One}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{a}=\mathrm{h}=\mathrm{2} \\ $$$$\mathrm{Hence}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{edges}=\mathrm{24} \\ $$$$\mathrm{For}\:\mathrm{a}\mid\mathrm{h} \\ $$$$\:\:\:\mathrm{Let}\:\mathrm{h}=\mathrm{ka}\:\mathrm{for}\:\mathrm{some}\:\mathrm{integer}\:\mathrm{k}. \\ $$$$\:\:\:\:\mathrm{2a}^{\mathrm{2}} +\mathrm{4ah}=\mathrm{8a}+\mathrm{4h} \\ $$$$\:\:\:\:\Rightarrow\mathrm{2a}^{\mathrm{2}} +\mathrm{4a}.\mathrm{ka}=\mathrm{8a}+\mathrm{4}.\mathrm{ka} \\ $$$$\:\:\:\:\Rightarrow\mathrm{2a}^{\mathrm{2}} +\mathrm{4ka}^{\mathrm{2}} =\mathrm{8a}+\mathrm{4ka} \\ $$$$\:\:\:\:\Rightarrow\mathrm{a}+\mathrm{2ka}=\mathrm{4}+\mathrm{2k} \\ $$$$\:\:\:\:\Rightarrow\mathrm{2k}\left(\mathrm{a}−\mathrm{1}\right)=\mathrm{4}−\mathrm{a} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{4}−\mathrm{a}}{\mathrm{2}\left(\mathrm{a}−\mathrm{1}\right)}\:=\mathrm{k} \\ $$$$\:\:\mathrm{k}=\mathrm{1}\Rightarrow\mathrm{4}−\mathrm{a}=\mathrm{2a}−\mathrm{2}\Rightarrow\mathrm{a}=\mathrm{2} \\ $$$$\:\:\mathrm{Again}\:\mathrm{a}=\mathrm{h}=\mathrm{2}\:\mathrm{and}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{edges}=\mathrm{24} \\ $$$$\:\:\mathrm{k}=\mathrm{2}\Rightarrow\mathrm{4}−\mathrm{a}=\mathrm{4}\left(\mathrm{a}−\mathrm{1}\right)\Rightarrow\mathrm{a}=\frac{\mathrm{8}}{\mathrm{5}}\:\notin\mathbb{N}\:\:\: \\ $$$$\mathrm{Continue} \\ $$
Commented by ajfour last updated on 01/Nov/17
yes, just now.
$${yes},\:{just}\:{now}. \\ $$
Commented by math solver last updated on 01/Nov/17
from here how you get a=h=2
$$\mathrm{from}\:\mathrm{here}\:\mathrm{how}\:\mathrm{you}\:\mathrm{get}\:\mathrm{a}=\mathrm{h}=\mathrm{2} \\ $$$$ \\ $$
Answered by ajfour last updated on 01/Nov/17
let the square base of cuboid be of edge length x. Let the height be y. Then, sum of the length of edges = 4x+4y+4x and area of all the faces is =x^2 +4xy+x^2 Given they are numerixally equal ⇒ 2x^2 +4xy = 8x+4y or x^2 +2xy = 4x+2y x^2 −4x+4 − 2y+2xy= 4 (x−2)^2 +2y(x−1)=4 ⇒ (x−2)^2 +2y(x−2)+y^2 =4−2y+y^2 (x−2+y)^2 = (y−1)^2 +3 ⇒ (x−2+y)^2 −(y−1)^2 =3 (x−2+y+y−1)(x−2+y−y+1)=3 (x+2y−3)(x−1)=3×1 =1×3 As x, y are positive integers, ⇒ both x−1=1 and x+2y−3=3 So x=2 and y=2 Sum of length of edges =24 . otherwise if x−1=3 ⇒ x=4 and x+2y−3 = 1 with x=4 yields y=0 ; this is true even but not a proper cuboid.
$${let}\:{the}\:{square}\:{base}\:{of}\:{cuboid}\:{be} \\ $$$${of}\:{edge}\:{length}\:\boldsymbol{{x}}.\:{Let}\:{the}\:{height} \\ $$$${be}\:\boldsymbol{{y}}.\:{Then},\:\:{sum}\:{of}\:{the}\:{length} \\ $$$${of}\:{edges}\:=\:\mathrm{4}{x}+\mathrm{4}{y}+\mathrm{4}{x} \\ $$$${and}\:{area}\:{of}\:{all}\:{the}\:{faces}\:{is} \\ $$$$\:={x}^{\mathrm{2}} +\mathrm{4}{xy}+{x}^{\mathrm{2}} \\ $$$${Given}\:{they}\:{are}\:{numerixally} \\ $$$${equal}\:\Rightarrow \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{xy}\:=\:\mathrm{8}{x}+\mathrm{4}{y} \\ $$$${or}\:\:\:{x}^{\mathrm{2}} +\mathrm{2}{xy}\:=\:\mathrm{4}{x}+\mathrm{2}{y} \\ $$$$\:\:\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}\:−\:\mathrm{2}{y}+\mathrm{2}{xy}=\:\mathrm{4} \\ $$$$\:\:\:\:\:\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{2}{y}\left({x}−\mathrm{1}\right)=\mathrm{4} \\ $$$$\Rightarrow\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}{y}\left({x}−\mathrm{2}\right)+{y}^{\mathrm{2}} =\mathrm{4}−\mathrm{2}{y}+{y}^{\mathrm{2}} \\ $$$$\:\left({x}−\mathrm{2}+{y}\right)^{\mathrm{2}} =\:\left({y}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3} \\ $$$$\Rightarrow\:\:\left({x}−\mathrm{2}+{y}\right)^{\mathrm{2}} −\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{3} \\ $$$$\left({x}−\mathrm{2}+{y}+{y}−\mathrm{1}\right)\left({x}−\mathrm{2}+{y}−{y}+\mathrm{1}\right)=\mathrm{3} \\ $$$$\left({x}+\mathrm{2}{y}−\mathrm{3}\right)\left({x}−\mathrm{1}\right)=\mathrm{3}×\mathrm{1}\:=\mathrm{1}×\mathrm{3} \\ $$$${As}\:{x},\:{y}\:{are}\:{positive}\:{integers}, \\ $$$$\Rightarrow\:{both}\:\:\:\:\:\:\:{x}−\mathrm{1}=\mathrm{1} \\ $$$${and}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}+\mathrm{2}{y}−\mathrm{3}=\mathrm{3} \\ $$$$\:{So}\:\:\:\:\:\:\:\:\:{x}=\mathrm{2}\:\:{and}\:\:\:{y}=\mathrm{2}\: \\ $$$$\:\:\:{Sum}\:{of}\:{length}\:{of}\:{edges}\:=\mathrm{24}\:. \\ $$$$\boldsymbol{{otherwise}}\:{if} \\ $$$$\:{x}−\mathrm{1}=\mathrm{3}\:\Rightarrow\:{x}=\mathrm{4} \\ $$$${and}\:\:{x}+\mathrm{2}{y}−\mathrm{3}\:=\:\mathrm{1}\:\:{with}\:{x}=\mathrm{4} \\ $$$${yields}\:\:{y}=\mathrm{0}\:;\:{this}\:{is}\:{true}\:{even} \\ $$$${but}\:{not}\:{a}\:{proper}\:{cuboid}. \\ $$

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