Question-222482

Question Number 222482 by fantastic last updated on 28/Jun/25

Commented by fantastic last updated on 28/Jun/25

$${it}\:{is}\:{not}\:{the}\:{length} \\ $$$${its}\:{is}\:{representing}\:{areas} \\ $$

Answered by MathematicalUser2357 last updated on 28/Jun/25

FORGET ABOUT THE CIRCLE!!! Let x=EF^(−) (DE^(−) )^2 +(EF^(−) )^2 =(a+b)^2 20^2 +x^2 =(a+b)^2 x^2 =(a+b)^2 −20^2 =(a+b)^2 −400 Since x>0, x=(√((a+b)^2 −400)) In terms of a, x=(√((2a−47)^2 −400)) In terms of b, x=(√((2b+47)^2 −400))

$$\mathrm{FORGET}\:\mathrm{ABOUT}\:\mathrm{THE}\:\mathrm{CIRCLE}!!! \\ $$$$\mathrm{Let}\:{x}=\overline {\mathrm{EF}} \\ $$$$\left(\overline {\mathrm{DE}}\right)^{\mathrm{2}} +\left(\overline {\mathrm{EF}}\right)^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\mathrm{20}^{\mathrm{2}} +{x}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{20}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{400} \\ $$$$\mathrm{Since}\:{x}>\mathrm{0},\:{x}=\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{400}} \\ $$$$\mathrm{In}\:\mathrm{terms}\:\mathrm{of}\:{a},\:{x}=\sqrt{\left(\mathrm{2}{a}−\mathrm{47}\right)^{\mathrm{2}} −\mathrm{400}} \\ $$$$\mathrm{In}\:\mathrm{terms}\:\mathrm{of}\:{b},\:{x}=\sqrt{\left(\mathrm{2}{b}+\mathrm{47}\right)^{\mathrm{2}} −\mathrm{400}} \\ $$

Commented by A5T last updated on 28/Jun/25

From the circle, we get a(a+b)=DE^2 =400 So, a and b are unique which we can get from the additional equation a=b+47. We get negative b, so the question is vague. Are a and b actually side lengths?

$$\mathrm{From}\:\mathrm{the}\:\mathrm{circle},\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\left(\mathrm{a}+\mathrm{b}\right)=\mathrm{DE}^{\mathrm{2}} =\mathrm{400} \\ $$$$\mathrm{So},\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{are}\:\mathrm{unique}\:\mathrm{which}\:\mathrm{we}\:\mathrm{can}\:\mathrm{get}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{additional}\:\mathrm{equation}\:\mathrm{a}=\mathrm{b}+\mathrm{47}. \\ $$$$\mathrm{We}\:\mathrm{get}\:\mathrm{negative}\:\mathrm{b},\:\mathrm{so}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{vague}. \\ $$$$\mathrm{Are}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{actually}\:\mathrm{side}\:\mathrm{lengths}? \\ $$

Commented by fantastic last updated on 28/Jun/25

$${it}\:{is}\:{not}\:{the}\:{length} \\ $$$${its}\:{is}\:{representing}\:{areas} \\ $$

Answered by mr W last updated on 28/Jun/25

Commented by fantastic last updated on 28/Jun/25

$${wow} \\ $$

Commented by mr W last updated on 28/Jun/25

r=((DE)/2)=10 EF=h a=((πr^2 )/2)−c b=((2rh)/2)−c ((πr^2 )/2)−c=((2rh)/2)−c+47 ⇒h=((πr)/2)−((47)/r)=5π−((47)/(10))≈11

$${r}=\frac{{DE}}{\mathrm{2}}=\mathrm{10} \\ $$$${EF}={h} \\ $$$${a}=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}−{c} \\ $$$${b}=\frac{\mathrm{2}{rh}}{\mathrm{2}}−{c} \\ $$$$\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}−{c}=\frac{\mathrm{2}{rh}}{\mathrm{2}}−{c}+\mathrm{47} \\ $$$$\Rightarrow{h}=\frac{\pi{r}}{\mathrm{2}}−\frac{\mathrm{47}}{{r}}=\mathrm{5}\pi−\frac{\mathrm{47}}{\mathrm{10}}\approx\mathrm{11} \\ $$

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