Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 181085 by mr W last updated on 21/Nov/22

Commented by mr W last updated on 21/Nov/22

find the diameter of semicircle.

$${find}\:{the}\:{diameter}\:{of}\:{semicircle}. \\ $$

Commented by HeferH last updated on 21/Nov/22

R = 12.5?

$${R}\:=\:\mathrm{12}.\mathrm{5}? \\ $$

Commented by mr W last updated on 21/Nov/22

yes!

$${yes}! \\ $$

Answered by mr W last updated on 21/Nov/22

Commented by mr W last updated on 21/Nov/22

r_1 +r_2 =(√((R−r_1 )^2 −r_1 ^2 ))+(√((R−r_2 )^2 −r_2 ^2 )) (r_1 +r_2 )−(√(R(R−2r_1 )))=(√(R(R−2r_2 ))) (r_1 +r_2 )^2 −2R(r_1 −r_2 )=2(r_1 +r_2 )(√(R(R−2r_1 ))) (r_1 +r_2 )^4 −4R(r_1 −r_2 )(r_1 +r_2 )^2 +4R^2 (r_1 −r_2 )^2 =4(r_1 +r_2 )^2 R(R−2r_1 ) 4r_1 r_2 (2R)^2 −2(r_1 +r_2 )^3 (2R)−(r_1 +r_2 )^4 =0 ⇒2R=(((r_1 +r_2 )^2 (r_1 +r_2 +(√((r_1 +r_2 )^2 +4r_1 r_2 ))))/(4r_1 r_2 )) ⇒2R=((10^2 (10+(√(10^2 +4×24))))/(4×24))=25 ✓

$${r}_{\mathrm{1}} +{r}_{\mathrm{2}} =\sqrt{\left({R}−{r}_{\mathrm{1}} \right)^{\mathrm{2}} −{r}_{\mathrm{1}} ^{\mathrm{2}} }+\sqrt{\left({R}−{r}_{\mathrm{2}} \right)^{\mathrm{2}} −{r}_{\mathrm{2}} ^{\mathrm{2}} } \\ $$$$\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)−\sqrt{{R}\left({R}−\mathrm{2}{r}_{\mathrm{1}} \right)}=\sqrt{{R}\left({R}−\mathrm{2}{r}_{\mathrm{2}} \right)} \\ $$$$\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{R}\left({r}_{\mathrm{1}} −{r}_{\mathrm{2}} \right)=\mathrm{2}\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)\sqrt{{R}\left({R}−\mathrm{2}{r}_{\mathrm{1}} \right)} \\ $$$$\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{4}} −\mathrm{4}{R}\left({r}_{\mathrm{1}} −{r}_{\mathrm{2}} \right)\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{R}^{\mathrm{2}} \left({r}_{\mathrm{1}} −{r}_{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{4}\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} {R}\left({R}−\mathrm{2}{r}_{\mathrm{1}} \right) \\ $$$$\mathrm{4}{r}_{\mathrm{1}} {r}_{\mathrm{2}} \left(\mathrm{2}{R}\right)^{\mathrm{2}} −\mathrm{2}\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{3}} \left(\mathrm{2}{R}\right)−\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{4}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{R}=\frac{\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} \left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} +\sqrt{\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{r}_{\mathrm{1}} {r}_{\mathrm{2}} }\right)}{\mathrm{4}{r}_{\mathrm{1}} {r}_{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{2}{R}=\frac{\mathrm{10}^{\mathrm{2}} \left(\mathrm{10}+\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{4}×\mathrm{24}}\right)}{\mathrm{4}×\mathrm{24}}=\mathrm{25}\:\checkmark \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com