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 5825 by Rasheed Soomro last updated on 30/May/16

Commented by Rasheed Soomro last updated on 30/May/16

A is the center of arc BC. AB=AE=AG=AC AB means distance between A and B Similarly AE,AG,AC are also distances.

$$\mathrm{A}\:\mathrm{is}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{arc}\:\mathrm{BC}. \\ $$$$\mathrm{AB}=\mathrm{AE}=\mathrm{AG}=\mathrm{AC} \\ $$$$\mathrm{AB}\:{m}\mathrm{eans}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B} \\ $$$$\mathrm{Similarly}\:\mathrm{AE},\mathrm{AG},\mathrm{AC}\:\mathrm{are}\:\mathrm{also}\:\mathrm{distances}. \\ $$

Answered by Yozzii last updated on 30/May/16

The centre of the arc AE is B and ∣AB∣=∣AE∣>0. Since AE is an arc ⇒∣AB∣=∣BE∣ and ∣AE∣=∣AB∣θ where θ=∠ABE. ∴ θ=((∣AE∣)/(∣AB∣))=1 rad. C is the centre of the arc AG and ∣AG∣=∣AC∣. We similarly find ∠ACG=α=1 rad. We are given that A,B & C are collinear points. So ∠BAC=γ=π rad. A is the centre of the arc BC. We now employ the formula for the area of the sectors whose angles are θ,α and γ respectively. A_θ =(1/2)(∣AB∣)^2 θ=(((AB)^2 )/2) A_α =(1/2)(∣AC∣)^2 α=(((∣AC∣)^2 )/2)=(((∣AB∣)^2 )/2) ∵ ∣AB∣=∣AC∣ A_γ =(1/2)(∣AB∣)^2 γ=(((∣AB∣)^2 )/2)π. The total area of the closed figure is then A=A_γ +A_θ +A_α or A=(((∣AB∣)^2 )/2)(2+π)=(r^2 /2)(2+π) ∵∣AB∣=r.

$${The}\:{centre}\:{of}\:{the}\:{arc}\:{AE}\:{is}\:{B}\:{and}\: \\ $$$$\mid{AB}\mid=\mid{AE}\mid>\mathrm{0}.\:{Since}\:{AE}\:{is}\:{an}\:{arc} \\ $$$$\Rightarrow\mid{AB}\mid=\mid{BE}\mid\:{and}\:\mid{AE}\mid=\mid{AB}\mid\theta\: \\ $$$${where}\:\theta=\angle{ABE}.\:\therefore\:\theta=\frac{\mid{AE}\mid}{\mid{AB}\mid}=\mathrm{1}\:{rad}. \\ $$$${C}\:{is}\:{the}\:{centre}\:{of}\:{the}\:{arc}\:{AG}\:{and} \\ $$$$\mid{AG}\mid=\mid{AC}\mid.\:{We}\:{similarly}\:{find}\:\angle{ACG}=\alpha=\mathrm{1}\:{rad}. \\ $$$${We}\:{are}\:{given}\:{that}\:{A},{B}\:\&\:{C}\:{are}\: \\ $$$${collinear}\:{points}.\:{So}\:\angle{BAC}=\gamma=\pi\:{rad}. \\ $$$${A}\:{is}\:{the}\:{centre}\:{of}\:{the}\:{arc}\:{BC}. \\ $$$${We}\:{now}\:{employ}\:{the}\:{formula}\:{for} \\ $$$${the}\:{area}\:{of}\:{the}\:{sectors}\:{whose}\:{angles} \\ $$$${are}\:\theta,\alpha\:{and}\:\gamma\:{respectively}. \\ $$$${A}_{\theta} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mid{AB}\mid\right)^{\mathrm{2}} \theta=\frac{\left({AB}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${A}_{\alpha} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mid{AC}\mid\right)^{\mathrm{2}} \alpha=\frac{\left(\mid{AC}\mid\right)^{\mathrm{2}} }{\mathrm{2}}=\frac{\left(\mid{AB}\mid\right)^{\mathrm{2}} }{\mathrm{2}}\:\:\:\because\:\mid{AB}\mid=\mid{AC}\mid \\ $$$${A}_{\gamma} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mid{AB}\mid\right)^{\mathrm{2}} \gamma=\frac{\left(\mid{AB}\mid\right)^{\mathrm{2}} }{\mathrm{2}}\pi. \\ $$$${The}\:{total}\:{area}\:{of}\:{the}\:{closed}\:{figure} \\ $$$${is}\:{then}\:{A}={A}_{\gamma} +{A}_{\theta} +{A}_{\alpha} \:\:{or} \\ $$$${A}=\frac{\left(\mid{AB}\mid\right)^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{2}+\pi\right)=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{2}+\pi\right)\:\:\:\:\because\mid{AB}\mid={r}. \\ $$

Commented by Rasheed Soomro last updated on 30/May/16

Very sorry that my question was not clear. According to your understanding you are very right! But I didn′t mean that. By AE ,AB,AG,...I meant distances between points. Where I meant arc I wrote arc AG etc.

$$\mathrm{Very}\:\mathrm{sorry}\:\mathrm{that}\:\mathrm{my}\:\mathrm{question}\:\mathrm{was}\:\mathrm{not}\:\mathrm{clear}. \\ $$$$\mathrm{According}\:\mathrm{to}\:\mathrm{your}\:\mathrm{understanding}\:\mathrm{you}\:\mathrm{are} \\ $$$$\mathrm{very}\:\mathrm{right}! \\ $$$$\mathrm{But}\:\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{mean}\:\mathrm{that}. \\ $$$$\mathrm{By}\:\mathrm{AE}\:,\mathrm{AB},\mathrm{AG},...\mathrm{I}\:\mathrm{meant}\:\mathrm{distances}\:\mathrm{between} \\ $$$$\mathrm{points}.\:\mathrm{Where}\:\mathrm{I}\:\mathrm{meant}\:\mathrm{arc}\:\mathrm{I}\:\mathrm{wrote}\:\mathrm{arc}\:\mathrm{AG}\:{etc}. \\ $$

Commented by Yozzii last updated on 30/May/16

My calculation considers the symbolism you used initially as measures of distances between points. If you′d like, we can write for example the distance between points A and B as ∣AB∣.

$${My}\:{calculation}\:{considers}\:{the}\:{symbolism} \\ $$$${you}\:{used}\:{initially}\:{as}\:{measures}\:{of} \\ $$$${distances}\:{between}\:{points}.\:{If}\:{you}'{d} \\ $$$${like},\:{we}\:{can}\:{write}\:{for}\:{example}\:{the}\:{distance} \\ $$$${between}\:{points}\:{A}\:{and}\:{B}\:{as}\:\mid{AB}\mid. \\ $$$$ \\ $$

Commented by Rasheed Soomro last updated on 07/Jun/16

!SknaHT_(←)

$$\underset{\leftarrow} {!\mathcal{S}{kna}\mathcal{HT}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com