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Question-175709

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Question Number 175709 by Shrinava last updated on 05/Sep/22
Answered by mr W last updated on 05/Sep/22
r=radius of small circle R=radius of big semicircle πr^2 =4π ⇒r=2 R=((3+x)/2) (R−r)^2 =r^2 +(R−3)^2 2R(3−r)=9 R=(9/(2(3−r)))=(9/2) ((3+x)/2)=(9/2) ⇒x=6
$${r}={radius}\:{of}\:{small}\:{circle} \\ $$$${R}={radius}\:{of}\:{big}\:{semicircle} \\ $$$$\pi{r}^{\mathrm{2}} =\mathrm{4}\pi\:\Rightarrow{r}=\mathrm{2} \\ $$$${R}=\frac{\mathrm{3}+{x}}{\mathrm{2}} \\ $$$$\left({R}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} +\left({R}−\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{R}\left(\mathrm{3}−{r}\right)=\mathrm{9} \\ $$$${R}=\frac{\mathrm{9}}{\mathrm{2}\left(\mathrm{3}−{r}\right)}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\frac{\mathrm{3}+{x}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{6} \\ $$
Commented by Tawa11 last updated on 05/Sep/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by Shrinava last updated on 06/Sep/22
cool professor thank you
$$\mathrm{cool}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$
Answered by HeferH last updated on 05/Sep/22
Let R be the radius of the semicircle, Q the center of this semicircle and r the radius of the tangent circle: r^2 π = 4π r = 2 OC = 2 QC = R − 3 QO = R − 2 x = QC + R The triangle OCQ is a right triangle: (R− 2)^2 = (R− 3)^2 + 2^2 (R− 2)^2 − (R−3)^2 = 4 (R−2 + R − 3)(R−2−(R − 3)) = 4 (2R −5)(1) = 4 R = (9/2) x = (9/2) − 3 + (9/2) = 6
$$\: \\ $$$$\:{Let}\:{R}\:{be}\:{the}\:{radius}\:{of}\:{the}\:{semicircle},\:{Q}\:{the}\: \\ $$$$\:{center}\:{of}\:{this}\:{semicircle}\:{and}\:\:{r}\:{the}\:{radius} \\ $$$$\:{of}\:{the}\:{tangent}\:{circle}: \\ $$$$\:{r}^{\mathrm{2}} \pi\:=\:\mathrm{4}\pi \\ $$$$\:{r}\:=\:\mathrm{2} \\ $$$$\:{OC}\:=\:\mathrm{2} \\ $$$$\:{QC}\:=\:{R}\:−\:\mathrm{3} \\ $$$$\:{QO}\:=\:{R}\:−\:\mathrm{2} \\ $$$$\:{x}\:=\:{QC}\:+\:{R}\: \\ $$$$\: \\ $$$$\:{The}\:{triangle}\:{OCQ}\:{is}\:{a}\:{right}\:{triangle}: \\ $$$$\:\left({R}−\:\mathrm{2}\right)^{\mathrm{2}} \:=\:\left({R}−\:\mathrm{3}\right)^{\mathrm{2}} \:+\:\mathrm{2}^{\mathrm{2}} \\ $$$$\:\left({R}−\:\mathrm{2}\right)^{\mathrm{2}} \:−\:\left({R}−\mathrm{3}\right)^{\mathrm{2}} \:=\:\mathrm{4} \\ $$$$\:\left({R}−\mathrm{2}\:+\:{R}\:−\:\mathrm{3}\right)\left({R}−\mathrm{2}−\left({R}\:−\:\mathrm{3}\right)\right)\:=\:\mathrm{4} \\ $$$$\:\left(\mathrm{2}{R}\:−\mathrm{5}\right)\left(\mathrm{1}\right)\:=\:\mathrm{4} \\ $$$$\:{R}\:=\:\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\:{x}\:=\:\frac{\mathrm{9}}{\mathrm{2}}\:−\:\mathrm{3}\:+\:\frac{\mathrm{9}}{\mathrm{2}}\:=\:\mathrm{6} \\ $$$$\: \\ $$
Commented by Tawa11 last updated on 05/Sep/22
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Commented by Shrinava last updated on 06/Sep/22
cool professor thank you
$$\mathrm{cool}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$

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