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Question Number 210122 by SonGoku last updated on 31/Jul/24
Commented by SonGoku last updated on 31/Jul/24
How do I find the painted area?
$$\mathrm{How}\:\mathrm{do}\:\mathrm{I}\:\mathrm{find}\:\mathrm{the}\:\mathrm{painted}\:\mathrm{area}? \\ $$
Answered by mr W last updated on 01/Aug/24
Commented by mr W last updated on 31/Jul/24
z_0 =area of big triangle z_1 =area of circle z_2 =z_0 −z_1 =shaded area
$${z}_{\mathrm{0}} ={area}\:{of}\:{big}\:{triangle} \\ $$$${z}_{\mathrm{1}} ={area}\:{of}\:{circle} \\ $$$${z}_{\mathrm{2}} ={z}_{\mathrm{0}} −{z}_{\mathrm{1}} ={shaded}\:{area} \\ $$
Commented by mr W last updated on 01/Aug/24
excircle of ΔDEC is the incircle of ΔABC. s=((3+4+b)/2)=((7+b)/2) R=(√((s(s−3)(s−4))/(s−b))) t=((3+20+26+a+4)/2)=((53+a)/2) R=(√(((t−23)(t−26)(t−4−a))/t)) (((t−23)(t−26)(t−4−a))/t)=((s(s−3)(s−4))/(s−b)) ⇒(((7+a)(1+a)(45−a))/(45+a))=(((7+b)(1+b)(b−1))/(7−b)) ...(i) cos C=((3^2 +4^2 −b^2 )/(2×3×4))=((23^2 +(4+a)^2 −26^2 )/(2×23×(4+a))) ((25−b^2 )/(3×4))=(((4+a)^2 −147)/(23×(4+a))) ⇒b=(√(25−((12)/(23))(4+a−((147)/(4+a))))) ...(ii) from (i) and (ii) we get: a≈10.744363 b≈4.7444 R≈5.2913 shaded area =tR−πR^2 =(((53+a)R)/2)−πR^2 ≈80.6875 m^2 ✓
$${excircle}\:{of}\:\Delta{DEC}\:{is}\:{the}\:{incircle}\:{of} \\ $$$$\Delta{ABC}. \\ $$$${s}=\frac{\mathrm{3}+\mathrm{4}+{b}}{\mathrm{2}}=\frac{\mathrm{7}+{b}}{\mathrm{2}} \\ $$$${R}=\sqrt{\frac{{s}\left({s}−\mathrm{3}\right)\left({s}−\mathrm{4}\right)}{{s}−{b}}} \\ $$$${t}=\frac{\mathrm{3}+\mathrm{20}+\mathrm{26}+{a}+\mathrm{4}}{\mathrm{2}}=\frac{\mathrm{53}+{a}}{\mathrm{2}} \\ $$$${R}=\sqrt{\frac{\left({t}−\mathrm{23}\right)\left({t}−\mathrm{26}\right)\left({t}−\mathrm{4}−{a}\right)}{{t}}} \\ $$$$\frac{\left({t}−\mathrm{23}\right)\left({t}−\mathrm{26}\right)\left({t}−\mathrm{4}−{a}\right)}{{t}}=\frac{{s}\left({s}−\mathrm{3}\right)\left({s}−\mathrm{4}\right)}{{s}−{b}} \\ $$$$\Rightarrow\frac{\left(\mathrm{7}+{a}\right)\left(\mathrm{1}+{a}\right)\left(\mathrm{45}−{a}\right)}{\mathrm{45}+{a}}=\frac{\left(\mathrm{7}+{b}\right)\left(\mathrm{1}+{b}\right)\left({b}−\mathrm{1}\right)}{\mathrm{7}−{b}}\:\:\:…\left({i}\right) \\ $$$$\mathrm{cos}\:{C}=\frac{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}×\mathrm{4}}=\frac{\mathrm{23}^{\mathrm{2}} +\left(\mathrm{4}+{a}\right)^{\mathrm{2}} −\mathrm{26}^{\mathrm{2}} }{\mathrm{2}×\mathrm{23}×\left(\mathrm{4}+{a}\right)} \\ $$$$\frac{\mathrm{25}−{b}^{\mathrm{2}} }{\mathrm{3}×\mathrm{4}}=\frac{\left(\mathrm{4}+{a}\right)^{\mathrm{2}} −\mathrm{147}}{\mathrm{23}×\left(\mathrm{4}+{a}\right)} \\ $$$$\Rightarrow{b}=\sqrt{\mathrm{25}−\frac{\mathrm{12}}{\mathrm{23}}\left(\mathrm{4}+{a}−\frac{\mathrm{147}}{\mathrm{4}+{a}}\right)}\:\:\:…\left({ii}\right) \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get}: \\ $$$${a}\approx\mathrm{10}.\mathrm{744363} \\ $$$${b}\approx\mathrm{4}.\mathrm{7444} \\ $$$${R}\approx\mathrm{5}.\mathrm{2913} \\ $$$${shaded}\:{area}\:={tR}−\pi{R}^{\mathrm{2}} \\ $$$$\:\:\:\:=\frac{\left(\mathrm{53}+{a}\right){R}}{\mathrm{2}}−\pi{R}^{\mathrm{2}} \approx\mathrm{80}.\mathrm{6875}\:{m}^{\mathrm{2}} \:\checkmark \\ $$
Commented by mr W last updated on 31/Jul/24
Commented by SonGoku last updated on 01/Aug/24
I did as follows: I found the area of the triangle to be 291.8 (Approximately) I found the radius to be 7.32 (Approximately) I found the area of the circle to be 168.33 (Approximately) Therefore, the painteda area is Ap ≈ 291.8 − 168.33 Ap ≈ 123.47m^2
$$\mathrm{I}\:\mathrm{did}\:\mathrm{as}\:\mathrm{follows}: \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{to}\:\mathrm{be}\:\mathrm{291}.\mathrm{8}\:\left(\mathrm{Approximately}\right)\: \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{to}\:\mathrm{be}\:\mathrm{7}.\mathrm{32}\:\left(\mathrm{Approximately}\right) \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{to}\:\mathrm{be}\:\mathrm{168}.\mathrm{33}\:\left(\mathrm{Approximately}\right) \\ $$$$\mathrm{Therefore},\:\mathrm{the}\:\mathrm{painteda}\:\mathrm{area}\:\mathrm{is}\: \\ $$$$\mathrm{Ap}\:\approx\:\mathrm{291}.\mathrm{8}\:−\:\mathrm{168}.\mathrm{33} \\ $$$$\mathrm{Ap}\:\approx\:\mathrm{123}.\mathrm{47m}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 01/Aug/24
what are the side lengthes you got for the triangle? try to draw the triangle and the circle, then you′ll see that you are wrong. the figure which i showed above is the only one correct solution.
$${what}\:{are}\:{the}\:{side}\:{lengthes}\:{you}\:{got} \\ $$$${for}\:{the}\:{triangle}? \\ $$$${try}\:{to}\:{draw}\:{the}\:{triangle}\:{and}\:{the}\:{circle}, \\ $$$${then}\:{you}'{ll}\:{see}\:{that}\:{you}\:{are}\:{wrong}. \\ $$$${the}\:{figure}\:{which}\:{i}\:{showed}\:{above} \\ $$$${is}\:{the}\:{only}\:{one}\:{correct}\:{solution}. \\ $$
Commented by SonGoku last updated on 01/Aug/24
I applied triangle similarity: 3:23 = x:26 4:y = 4:(4+y) Then I used the Herons formula to determine the area of the triangle and then I determined the radiusof the circle: ((23R)/2) + ((30.67R)/2) + ((26R)/2) = 291.8
$$\mathrm{I}\:\mathrm{applied}\:\mathrm{triangle}\:\mathrm{similarity}: \\ $$$$\mathrm{3}:\mathrm{23}\:=\:\mathrm{x}:\mathrm{26}\: \\ $$$$\mathrm{4}:\mathrm{y}\:=\:\mathrm{4}:\left(\mathrm{4}+\mathrm{y}\right) \\ $$$$\mathrm{Then}\:\mathrm{I}\:\mathrm{used}\:\mathrm{the}\:\mathrm{Herons}\:\mathrm{formula}\:\mathrm{to}\:\mathrm{determine}\:\mathrm{the}\: \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{and}\:\mathrm{then}\:\mathrm{I}\:\mathrm{determined}\:\mathrm{the}\: \\ $$$$\mathrm{radiusof}\:\mathrm{the}\:\mathrm{circle}: \\ $$$$\frac{\mathrm{23R}}{\mathrm{2}}\:+\:\frac{\mathrm{30}.\mathrm{67R}}{\mathrm{2}}\:+\:\frac{\mathrm{26R}}{\mathrm{2}}\:=\:\mathrm{291}.\mathrm{8} \\ $$
Commented by mr W last updated on 01/Aug/24
this is wrong! the small triangle mustn′t be similar to the big triangle. the only important condition is that the circle must tangent the four sides. when you draw the triangle you got, you will see that the circle doesn′t tangent the fourth line.
$${this}\:{is}\:{wrong}! \\ $$$${the}\:{small}\:{triangle}\:{mustn}'{t}\:{be}\:{similar} \\ $$$${to}\:{the}\:{big}\:{triangle}.\:{the}\:{only}\: \\ $$$${important}\:{condition}\:{is}\:{that}\:{the} \\ $$$${circle}\:{must}\:{tangent}\:{the}\:{four}\:{sides}. \\ $$$${when}\:{you}\:{draw}\:{the}\:{triangle}\:{you}\:{got}, \\ $$$${you}\:{will}\:{see}\:{that}\:{the}\:{circle}\:{doesn}'{t} \\ $$$${tangent}\:{the}\:{fourth}\:{line}. \\ $$
Commented by SonGoku last updated on 01/Aug/24
I got it. If the fourth line were to touch the circle, would it be correct to solve it by applying triangle simlarity?
$$\mathrm{I}\:\mathrm{got}\:\mathrm{it}.\: \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{fourth}\:\mathrm{line}\:\mathrm{were}\:\mathrm{to}\:\mathrm{touch}\:\mathrm{the}\:\mathrm{circle}, \\ $$$$\mathrm{would}\:\mathrm{it}\:\mathrm{be}\:\mathrm{correct}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{by}\:\mathrm{applying}\:\mathrm{triangle} \\ $$$$\mathrm{simlarity}? \\ $$
Commented by mr W last updated on 01/Aug/24
no! the condition which must be fulfilled is that the excircle of the small triangle is also the incircle of the big triangle. both triangles may be but not must be similar.
$${no}!\:{the}\:{condition}\:{which}\:{must}\:{be} \\ $$$${fulfilled}\:{is}\:{that}\:{the}\:{excircle}\:{of}\:{the} \\ $$$${small}\:{triangle}\:{is}\:{also}\:{the}\:{incircle}\:{of} \\ $$$${the}\:{big}\:{triangle}.\:{both}\:{triangles}\:{may} \\ $$$${be}\:{but}\:{not}\:{must}\:{be}\:{similar}. \\ $$
Commented by mr W last updated on 01/Aug/24
Commented by SonGoku last updated on 01/Aug/24
Understood. Thank you very much for thei informaton and for your patience. I learn a lot herei wth you all.
$$\mathrm{Understood}. \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{for}\:\mathrm{thei} \\ $$$$\mathrm{informaton}\:\mathrm{and}\:\mathrm{for}\:\mathrm{your}\:\mathrm{patience}. \\ $$$$\mathrm{I}\:\mathrm{learn}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{herei}\:\mathrm{wth}\:\mathrm{you}\:\mathrm{all}. \\ $$
Commented by mr W last updated on 01/Aug/24
Commented by mr W last updated on 01/Aug/24

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