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Question Number 40822 by MrW3 last updated on 28/Jul/18

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18

Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18

Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18

Commented by MrW3 last updated on 28/Jul/18

answer is correct sir! thanks!

$${answer}\:{is}\:{correct}\:{sir}! \\ $$$${thanks}! \\ $$

Answered by MrW3 last updated on 28/Jul/18

It can be seen that the sum of the diagonal parts must be equal. 20+?=16+32 ⇒?=48−20=28

$${It}\:{can}\:{be}\:{seen}\:{that}\:{the}\:{sum}\:{of}\:{the} \\ $$$${diagonal}\:{parts}\:{must}\:{be}\:{equal}. \\ $$$$\mathrm{20}+?=\mathrm{16}+\mathrm{32} \\ $$$$\Rightarrow?=\mathrm{48}−\mathrm{20}=\mathrm{28} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18

pls say whether my answer is right or wrong method may be diffdrent...

$${pls}\:{say}\:{whether}\:{my}\:{answer}\:{is}\:{right}\:{or}\:{wrong} \\ $$$${method}\:{may}\:{be}\:{diffdrent}... \\ $$

Commented by MrW3 last updated on 28/Jul/18

Commented by MrW3 last updated on 28/Jul/18

Proof: A=(1/2)×(a/2)×x+(1/2)×(a/2)×y=((ax)/4)+((ay)/4) B=(1/2)×(a/2)×(a−x)+(1/2)×(a/2)×y=(a^2 /4)−((ax)/4)+((ay)/4) C=(1/2)×(a/2)×(a−x)+(1/2)×(a/2)×(a−y)=(a^2 /2)−((ax)/4)−((ay)/4) D=(1/2)×(a/2)×x+(1/2)×(a/2)×(a−y)=(a^2 /4)+((ax)/4)−((ay)/4) A+C=(a^2 /2) B+D=(a^2 /2) ⇒A+C=B+D=(a^2 /2)=half of the square area

$${Proof}: \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{{a}}{\mathrm{2}}×{x}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{{a}}{\mathrm{2}}×{y}=\frac{{ax}}{\mathrm{4}}+\frac{{ay}}{\mathrm{4}} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{{a}}{\mathrm{2}}×\left({a}−{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}×\frac{{a}}{\mathrm{2}}×{y}=\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−\frac{{ax}}{\mathrm{4}}+\frac{{ay}}{\mathrm{4}} \\ $$$${C}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{{a}}{\mathrm{2}}×\left({a}−{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}×\frac{{a}}{\mathrm{2}}×\left({a}−{y}\right)=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}−\frac{{ax}}{\mathrm{4}}−\frac{{ay}}{\mathrm{4}} \\ $$$${D}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{{a}}{\mathrm{2}}×{x}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{{a}}{\mathrm{2}}×\left({a}−{y}\right)=\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\frac{{ax}}{\mathrm{4}}−\frac{{ay}}{\mathrm{4}} \\ $$$${A}+{C}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${B}+{D}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\boldsymbol{{A}}+\boldsymbol{{C}}=\boldsymbol{{B}}+\boldsymbol{{D}}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}={half}\:{of}\:{the}\:{square}\:{area} \\ $$

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