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Question Number 72194 by mr W last updated on 26/Oct/19

Commented by mr W last updated on 26/Oct/19

find the sum of areas of two squares inside the semicircle with radius 10.

$${find}\:{the}\:{sum}\:{of}\:{areas}\:{of}\:{two}\:{squares} \\ $$$${inside}\:{the}\:{semicircle}\:{with}\:{radius}\:\mathrm{10}. \\ $$

Answered by MJS last updated on 27/Oct/19

I turned it around if the larger square has side length 1, which side lengths are possible for the smaller square? put the right upper vertice ((1),(1) ), the left upper one (((−s)),(s) ) both must be on a circle with center on the x−axis (x−p)^2 +y^2 =r^2 ⇒ upper semi circle: y=(√(2s−2(s−1)x−x^2 )) the upper vertices of the larger square must fit into the circle ⇒ y≥1 at x=0 ⇒ (1/2)≤s<1 now choose s ⇒ area=s^2 +1; r=(√(s^2 +1)) area within circle with radius 10 is ((s^2 +1)/(((√(s^2 +1)))^2 ))×100=100

$$\mathrm{I}\:\mathrm{turned}\:\mathrm{it}\:\mathrm{around} \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{larger}\:\mathrm{square}\:\mathrm{has}\:\mathrm{side}\:\mathrm{length}\:\mathrm{1},\:\mathrm{which} \\ $$$$\mathrm{side}\:\mathrm{lengths}\:\mathrm{are}\:\mathrm{possible}\:\mathrm{for}\:\mathrm{the}\:\mathrm{smaller}\:\mathrm{square}? \\ $$$$\mathrm{put}\:\mathrm{the}\:\mathrm{right}\:\mathrm{upper}\:\mathrm{vertice}\:\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix},\:\mathrm{the}\:\mathrm{left}\:\mathrm{upper} \\ $$$$\mathrm{one}\:\begin{pmatrix}{−{s}}\\{{s}}\end{pmatrix} \\ $$$$\mathrm{both}\:\mathrm{must}\:\mathrm{be}\:\mathrm{on}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{center}\:\mathrm{on}\:\mathrm{the} \\ $$$${x}−\mathrm{axis} \\ $$$$\left({x}−{p}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\mathrm{upper}\:\mathrm{semi}\:\mathrm{circle}:\:{y}=\sqrt{\mathrm{2}{s}−\mathrm{2}\left({s}−\mathrm{1}\right){x}−{x}^{\mathrm{2}} } \\ $$$$\mathrm{the}\:\mathrm{upper}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{the}\:\mathrm{larger}\:\mathrm{square}\:\mathrm{must} \\ $$$$\mathrm{fit}\:\mathrm{into}\:\mathrm{the}\:\mathrm{circle}\:\Rightarrow\:{y}\geqslant\mathrm{1}\:\mathrm{at}\:{x}=\mathrm{0} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant{s}<\mathrm{1} \\ $$$$\mathrm{now}\:\mathrm{choose}\:{s} \\ $$$$\Rightarrow\:\mathrm{area}={s}^{\mathrm{2}} +\mathrm{1};\:{r}=\sqrt{{s}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{area}\:\mathrm{within}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{10}\:\mathrm{is} \\ $$$$\frac{{s}^{\mathrm{2}} +\mathrm{1}}{\left(\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} }×\mathrm{100}=\mathrm{100} \\ $$

Commented by mr W last updated on 27/Oct/19

thanks sir!

$${thanks}\:{sir}! \\ $$

Answered by mr W last updated on 27/Oct/19

Commented by mr W last updated on 27/Oct/19

∠AOB=((2π)/4)=(π/2) AB^2 =(a+b)^2 +(b−a)^2 =2(a^2 +b^2 ) AB^2 =r^2 +r^2 =2r^2 ⇒a^2 +b^2 =r^2 =10^2 =100 i.e. the sum of both squares is contant.

$$\angle{AOB}=\frac{\mathrm{2}\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{2}} \\ $$$${AB}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} +\left({b}−{a}\right)^{\mathrm{2}} =\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right) \\ $$$${AB}^{\mathrm{2}} ={r}^{\mathrm{2}} +{r}^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={r}^{\mathrm{2}} =\mathrm{10}^{\mathrm{2}} =\mathrm{100} \\ $$$${i}.{e}.\:{the}\:{sum}\:{of}\:{both}\:{squares}\:{is}\:{contant}. \\ $$

Commented by ajfour last updated on 27/Oct/19

Great way Sir, i myself had thought to attempt it somewhat this way..

$${Great}\:{way}\:{Sir},\:{i}\:{myself}\:{had} \\ $$$${thought}\:{to}\:{attempt}\:{it}\:{somewhat} \\ $$$${this}\:{way}.. \\ $$

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