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Question-78937




Question Number 78937 by mr W last updated on 21/Jan/20
Commented by mr W last updated on 21/Jan/20
As shown, the areas of three parts of  a rectangle are given. Find the area  of the fourth part.
$${As}\:{shown},\:{the}\:{areas}\:{of}\:{three}\:{parts}\:{of} \\ $$$${a}\:{rectangle}\:{are}\:{given}.\:{Find}\:{the}\:{area} \\ $$$${of}\:{the}\:{fourth}\:{part}. \\ $$
Commented by john santu last updated on 22/Jan/20
the Area is required =   (√((A+B+C)^2 −4AC))
$${the}\:{Area}\:{is}\:{required}\:=\: \\ $$$$\sqrt{\left({A}+{B}+{C}\right)^{\mathrm{2}} −\mathrm{4}{AC}} \\ $$
Commented by john santu last updated on 22/Jan/20
that right sir?
$${that}\:{right}\:{sir}? \\ $$
Commented by mr W last updated on 22/Jan/20
no sir!  correct result see solution below from  ajfour sir. it is (√((A+B+C)^2 −4AB)).
$${no}\:{sir}! \\ $$$${correct}\:{result}\:{see}\:{solution}\:{below}\:{from} \\ $$$${ajfour}\:{sir}.\:{it}\:{is}\:\sqrt{\left({A}+{B}+{C}\right)^{\mathrm{2}} −\mathrm{4}{AB}}. \\ $$
Commented by john santu last updated on 22/Jan/20
oo yes. i′m wrong to calculate
$${oo}\:{yes}.\:{i}'{m}\:{wrong}\:{to}\:{calculate} \\ $$
Answered by ajfour last updated on 21/Jan/20
let rectangle length=a+c  height=b+d  2B=(a+c)b  2A=(b+d)a  2C=cd  let  ?=Q  Q=(a+c)(b+d)−(A+B+C)     = ((4AB)/(ab))−(A+B+C)  (2B−ab)(2A−ab)=2Cab  (ab)^2 −2(A+B+C)(ab)+4AB=0  ab=(A+B+C)−(√((A+B+C)^2 −4AB))  hence  Q=((4AB)/((A+B+C)−(√((A+B+C)^2 −4AB))))−(A+B+C)
$${let}\:{rectangle}\:{length}={a}+{c} \\ $$$${height}={b}+{d} \\ $$$$\mathrm{2}{B}=\left({a}+{c}\right){b} \\ $$$$\mathrm{2}{A}=\left({b}+{d}\right){a} \\ $$$$\mathrm{2}{C}={cd} \\ $$$${let}\:\:?={Q} \\ $$$${Q}=\left({a}+{c}\right)\left({b}+{d}\right)−\left({A}+{B}+{C}\right) \\ $$$$\:\:\:=\:\frac{\mathrm{4}{AB}}{{ab}}−\left({A}+{B}+{C}\right) \\ $$$$\left(\mathrm{2}{B}−{ab}\right)\left(\mathrm{2}{A}−{ab}\right)=\mathrm{2}{Cab} \\ $$$$\left({ab}\right)^{\mathrm{2}} −\mathrm{2}\left({A}+{B}+{C}\right)\left({ab}\right)+\mathrm{4}{AB}=\mathrm{0} \\ $$$${ab}=\left({A}+{B}+{C}\right)−\sqrt{\left({A}+{B}+{C}\right)^{\mathrm{2}} −\mathrm{4}{AB}} \\ $$$${hence} \\ $$$${Q}=\frac{\mathrm{4}{AB}}{\left({A}+{B}+{C}\right)−\sqrt{\left({A}+{B}+{C}\right)^{\mathrm{2}} −\mathrm{4}{AB}}}−\left({A}+{B}+{C}\right) \\ $$
Commented by mr W last updated on 22/Jan/20
thank you sir! perfect!  the last line can be simplified to  Q=(√((A+B+C)^2 −4AB))
$${thank}\:{you}\:{sir}!\:{perfect}! \\ $$$${the}\:{last}\:{line}\:{can}\:{be}\:{simplified}\:{to} \\ $$$${Q}=\sqrt{\left({A}+{B}+{C}\right)^{\mathrm{2}} −\mathrm{4}{AB}} \\ $$
Commented by mr W last updated on 22/Jan/20
+ sign is suitable, since ab>A+B+C  ab=(A+B+C)+(√((A+B+C)^2 −4AB))
$$+\:{sign}\:{is}\:{suitable},\:{since}\:{ab}>{A}+{B}+{C} \\ $$$${ab}=\left({A}+{B}+{C}\right)+\sqrt{\left({A}+{B}+{C}\right)^{\mathrm{2}} −\mathrm{4}{AB}} \\ $$

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