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Question-75902




Question Number 75902 by Mr. K last updated on 20/Dec/19
Commented by Mr. K last updated on 20/Dec/19
find the area of the square
$${find}\:{the}\:{area}\:{of}\:{the}\:{square} \\ $$
Answered by mr W last updated on 20/Dec/19
let a=side length of square  OP=(a/( (√2)))  PM=(√2)a  ((a/( (√2))))^2 +((√2)a)^2 =r^2   (5/2)a^2 =r^2   a^2 =((2r^2 )/5)=((2×10)/5)=4=area of square
$${let}\:{a}={side}\:{length}\:{of}\:{square} \\ $$$${OP}=\frac{{a}}{\:\sqrt{\mathrm{2}}} \\ $$$${PM}=\sqrt{\mathrm{2}}{a} \\ $$$$\left(\frac{{a}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}{a}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\frac{\mathrm{5}}{\mathrm{2}}{a}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{2}{r}^{\mathrm{2}} }{\mathrm{5}}=\frac{\mathrm{2}×\mathrm{10}}{\mathrm{5}}=\mathrm{4}={area}\:{of}\:{square} \\ $$
Commented by TawaTawa last updated on 22/Dec/19
Great sir.  Please how did you know that  ∠OPM  =  90°
$$\mathrm{Great}\:\mathrm{sir}. \\ $$$$\mathrm{Please}\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{know}\:\mathrm{that}\:\:\angle\mathrm{OPM}\:\:=\:\:\mathrm{90}° \\ $$
Commented by mr W last updated on 22/Dec/19
due to symmetry you can see that  OP=OQ ⇒PM//OQ.
$${due}\:{to}\:{symmetry}\:{you}\:{can}\:{see}\:{that} \\ $$$${OP}={OQ}\:\Rightarrow{PM}//{OQ}. \\ $$
Commented by TawaTawa last updated on 22/Dec/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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