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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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