Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 4036 by Rasheed Soomro last updated on 27/Dec/15

Twin of Q#3943 Three circles are drawn in a plane in such a way that a closed region is produced which is not included in any of the circles. Determine the area of this region. Circles ∣ Radii ∣ Centers _(−) A r_1 C_1 B r_2 C_2 C r_3 C_3 Conditions: r_1 +r_2 ≥C_1 C_2 r_2 +r_3 ≥C_2 C_3 r_3 +r_1 ≥C_3 C_1

$$\mathrm{Twin}\:\mathrm{of}\:\mathrm{Q}#\mathrm{3943} \\ $$$$\mathrm{Three}\:\mathrm{circles}\:\mathrm{are}\:\mathrm{drawn}\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane} \\ $$$$\mathrm{in}\:\mathrm{such}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{region} \\ $$$$\mathrm{is}\:\mathrm{produced}\:\mathrm{which}\:\mathrm{is}\:\mathrm{not}\:\mathrm{included} \\ $$$$\mathrm{in}\:\mathrm{any}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circles}.\:\mathrm{Determine}\:\mathrm{the} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{this}\:\mathrm{region}. \\ $$$$\underset{−} {\mathrm{Circles}\:\mid\:\mathrm{Radii}\:\mid\:\mathrm{Centers}\:} \\ $$$$\:\:\:\:\:\:\mathrm{A}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} \:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{C}_{\mathrm{1}} \:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\mathrm{B}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{C}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\mathrm{C}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{3}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{C}_{\mathrm{3}} \\ $$$$\mathrm{Conditions}: \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} +\boldsymbol{\mathrm{r}}_{\mathrm{2}} \geqslant\mathrm{C}_{\mathrm{1}} \mathrm{C}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{2}} +\boldsymbol{\mathrm{r}}_{\mathrm{3}} \geqslant\mathrm{C}_{\mathrm{2}} \mathrm{C}_{\mathrm{3}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{3}} +\boldsymbol{\mathrm{r}}_{\mathrm{1}} \geqslant\mathrm{C}_{\mathrm{3}} \mathrm{C}_{\mathrm{1}} \\ $$

Commented by Yozzii last updated on 27/Dec/15

Commented by Yozzii last updated on 27/Dec/15

Area of region GHI is required.

$${Area}\:{of}\:{region}\:{GHI}\:{is}\:{required}. \\ $$

Commented by Rasheed Soomro last updated on 27/Dec/15

Yes. Exactly! In other cases two or all the three circles are tangent to one another.

$$\mathrm{Yes}.\:\mathrm{Exactly}! \\ $$$$\mathrm{In}\:\mathrm{other}\:\mathrm{cases}\:\mathrm{two}\:\mathrm{or}\:\:\mathrm{all}\:\mathrm{the}\:\mathrm{three} \\ $$$$\mathrm{circles}\:\mathrm{are}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{one}\:\mathrm{another}. \\ $$

Commented by Yozzii last updated on 27/Dec/15

Commented by Yozzii last updated on 27/Dec/15

Disregard the explicit measurements of angles given in the diagram. I only required the image to delineate the important angles for the calculation.

$${Disregard}\:{the}\:{explicit}\:{measurements} \\ $$$${of}\:{angles}\:{given}\:{in}\:{the}\:{diagram}.\: \\ $$$${I}\:{only}\:{required}\:{the}\:{image}\:{to}\:{delineate} \\ $$$${the}\:{important}\:{angles}\:{for}\:{the}\:{calculation}. \\ $$

Commented by Yozzii last updated on 27/Dec/15

Let A_1 be the area of △AEC. By the question we know the lengths AE,EC & CA. So, Heron′s formula could be used to find A_1 . A_1 =(√(s(s−CA)(s−AE)(s−EC))) s=((AE+EC+CA)/2). Let θ_1 =∠ACE,θ_2 =∠CEA and θ_3 =∠EAC. By cosine rule AE^2 =EC^2 +CA^2 −2(EC)(CA)cosθ_1 ⇒θ_1 =cos^(−1) {((EC^2 +CA^2 −AE^2 )/(2(EC)(CA)))}. In like fashion, θ_2 can be found and θ_3 =π−θ_1 −θ_2 . Let r_1 ,r_2 ,r_3 be the radii of the circles with centres C,E and A respectively. The area A_2 of the sector MCL (from diagram) is given by A_2 =(1/2)r_1 ^2 θ_1 . Similarly, the area A_3 of the sector OEN is A_3 =(1/2)r_2 ^2 θ_2 and the area A_4 of the sector KAJ is A_4 =(1/2)r_3 ^2 θ_3 . Now, the value of s=Σ_(i=2) ^4 A_i includes duplication of pairwise overlapping of circles. To find the correct value of s to subtract from A_1 to give the required area, work s by parts. Starting with A_2 , A_2 +A_3 considers the overlap of circles C and E two times. ⇒suitable area, s_1 =A_2 +A_3 −Area(LTN). Next, A_2 +A_3 +A_4 considers the overlap between circles A and E, and also A and C,twice. So, truly s=Σ_(r=2) ^4 A_r −Area(LTN)−Area(OKR)−Area(MBJ). Mr. Jain′s answer to question 3877 can help you find Areas LTN,OKR andMBJ, once you half each result for area found using his method directly. Afterwards, Area(RBT)=A_1 −s.

$${Let}\:{A}_{\mathrm{1}} \:{be}\:{the}\:{area}\:{of}\:\bigtriangleup{AEC}.\:{By}\:{the} \\ $$$${question}\:{we}\:{know}\:{the}\:{lengths}\:{AE},{EC} \\ $$$$\&\:{CA}.\:{So},\:{Heron}'{s}\:{formula}\:{could} \\ $$$${be}\:{used}\:{to}\:{find}\:{A}_{\mathrm{1}} .\: \\ $$$${A}_{\mathrm{1}} =\sqrt{{s}\left({s}−{CA}\right)\left({s}−{AE}\right)\left({s}−{EC}\right)} \\ $$$${s}=\frac{{AE}+{EC}+{CA}}{\mathrm{2}}. \\ $$$${Let}\:\theta_{\mathrm{1}} =\angle{ACE},\theta_{\mathrm{2}} =\angle{CEA}\:{and}\:\theta_{\mathrm{3}} =\angle{EAC}. \\ $$$${By}\:{cosine}\:{rule}\: \\ $$$${AE}^{\mathrm{2}} ={EC}^{\mathrm{2}} +{CA}^{\mathrm{2}} −\mathrm{2}\left({EC}\right)\left({CA}\right){cos}\theta_{\mathrm{1}} \\ $$$$\Rightarrow\theta_{\mathrm{1}} ={cos}^{−\mathrm{1}} \left\{\frac{{EC}^{\mathrm{2}} +{CA}^{\mathrm{2}} −{AE}^{\mathrm{2}} }{\mathrm{2}\left({EC}\right)\left({CA}\right)}\right\}. \\ $$$${In}\:{like}\:{fashion},\:\theta_{\mathrm{2}} \:{can}\:{be}\:{found} \\ $$$${and}\:\theta_{\mathrm{3}} =\pi−\theta_{\mathrm{1}} −\theta_{\mathrm{2}} . \\ $$$${Let}\:{r}_{\mathrm{1}} ,{r}_{\mathrm{2}} ,{r}_{\mathrm{3}} \:{be}\:{the}\:{radii}\:{of}\:{the}\:{circles} \\ $$$${with}\:{centres}\:{C},{E}\:{and}\:{A}\:{respectively}. \\ $$$${The}\:{area}\:{A}_{\mathrm{2}} \:{of}\:{the}\:{sector}\:{MCL}\:\left({from}\:{diagram}\right) \\ $$$${is}\:{given}\:{by}\:{A}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{r}_{\mathrm{1}} ^{\mathrm{2}} \theta_{\mathrm{1}} .\:{Similarly}, \\ $$$${the}\:{area}\:{A}_{\mathrm{3}} \:{of}\:{the}\:{sector}\:{OEN}\:{is}\:{A}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}{r}_{\mathrm{2}} ^{\mathrm{2}} \theta_{\mathrm{2}} \\ $$$${and}\:{the}\:{area}\:{A}_{\mathrm{4}} \:{of}\:{the}\:{sector}\:{KAJ}\:{is} \\ $$$${A}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{2}}{r}_{\mathrm{3}} ^{\mathrm{2}} \theta_{\mathrm{3}} .\:{Now},\:{the}\:{value}\:{of}\: \\ $$$${s}=\underset{{i}=\mathrm{2}} {\overset{\mathrm{4}} {\sum}}{A}_{{i}} \:{includes}\:{duplication}\:{of}\: \\ $$$${pairwise}\:{overlapping}\:{of}\:{circles}.\:{To} \\ $$$${find}\:{the}\:{correct}\:{value}\:{of}\:{s}\:{to}\:{subtract} \\ $$$${from}\:{A}_{\mathrm{1}} \:{to}\:{give}\:{the}\:{required}\:{area},\:{work} \\ $$$${s}\:{by}\:{parts}.\:{Starting}\:{with}\:{A}_{\mathrm{2}} ,\:{A}_{\mathrm{2}} +{A}_{\mathrm{3}} \\ $$$${considers}\:{the}\:{overlap}\:{of}\:{circles}\:{C}\:{and}\:{E} \\ $$$${two}\:{times}.\:\Rightarrow{suitable}\:{area}, \\ $$$${s}_{\mathrm{1}} ={A}_{\mathrm{2}} +{A}_{\mathrm{3}} −{Area}\left({LTN}\right). \\ $$$${Next},\:{A}_{\mathrm{2}} +{A}_{\mathrm{3}} +{A}_{\mathrm{4}} \:{considers}\:{the}\: \\ $$$${overlap}\:{between}\:{circles}\:{A}\:{and}\:{E},\:{and}\:{also}\:{A}\:{and}\:{C},{twice}. \\ $$$${So},\:{truly}\:{s}=\underset{{r}=\mathrm{2}} {\overset{\mathrm{4}} {\sum}}{A}_{{r}} −{Area}\left({LTN}\right)−{Area}\left({OKR}\right)−{Area}\left({MBJ}\right). \\ $$$${Mr}.\:{Jain}'{s}\:{answer}\:{to}\:{question}\:\mathrm{3877}\:{can}\:{help} \\ $$$${you}\:{find}\:{Areas}\:{LTN},{OKR}\:{andMBJ}, \\ $$$${once}\:{you}\:{half}\:{each}\:{result}\:{for}\:{area}\: \\ $$$${found}\:{using}\:{his}\:{method}\:{directly}. \\ $$$$ \\ $$$${Afterwards},\: \\ $$$${Area}\left({RBT}\right)={A}_{\mathrm{1}} −{s}. \\ $$

Commented by Rasheed Soomro last updated on 27/Dec/15

GREAT ! YOU HAVE DONE SUCCESSFULLY!

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathbb{G}\mathcal{REA}\mathbb{T}\:! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathbb{Y}\mathcal{OU}\:\:\:\mathbb{H}\mathcal{AVE}\:\mathbb{D}\mathcal{ONE}\:\:\mathbb{S}\mathcal{UCCESSFULLY}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com