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Question Number 4268    Answers: 3   Comments: 0

1+2=

$$\mathrm{1}+\mathrm{2}= \\ $$

Question Number 4219    Answers: 1   Comments: 0

Question Number 4225    Answers: 0   Comments: 0

^• A kite is a quadrilateral having two pairs of adjacent sides equal. Draw a semi-circle inside it touching all the sides using Eucledian tools. ^• Can we show that the above semi-circle is of the laregest possible area inside the kite?

$$\:^{\bullet} \mathrm{A}\:\mathrm{kite}\:\mathrm{is}\:\mathrm{a}\:\mathrm{quadrilateral}\:\mathrm{having}\:\mathrm{two} \\ $$$$\mathrm{pairs}\:\mathrm{of}\:\mathrm{adjacent}\:\mathrm{sides}\:\mathrm{equal}. \\ $$$$\mathrm{Draw}\:\mathrm{a}\:\mathrm{semi}-\mathrm{circle}\:\mathrm{inside}\:\mathrm{it}\:\mathrm{touching} \\ $$$$\mathrm{all}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{using}\:\mathrm{Eucledian}\:\mathrm{tools}. \\ $$$$\:^{\bullet} \:\mathrm{Can}\:\mathrm{we}\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\:\mathrm{above}\:\mathrm{semi}-\mathrm{circle} \\ $$$$\mathrm{is}\:\mathrm{of}\:\mathrm{the}\:\mathrm{laregest}\:\mathrm{possible}\:\mathrm{area}\:\mathrm{inside}\:\mathrm{the} \\ $$$$\mathrm{kite}? \\ $$

Question Number 4080    Answers: 2   Comments: 1

Prove that inside a square, a semi-circle, which touches all the four sides of the square, is possible with ruler and compass.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{inside}\:\mathrm{a}\:\mathrm{square},\:\mathrm{a}\:\mathrm{semi}-\mathrm{circle}, \\ $$$$\mathrm{which}\:\mathrm{touches}\:\mathrm{all}\:\mathrm{the}\:\mathrm{four}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{square},\:\mathrm{is}\:\mathrm{possible}\:\mathrm{with}\:\boldsymbol{\mathrm{ruler}}\:\mathrm{and}\:\boldsymbol{\mathrm{compass}}. \\ $$

Question Number 4077    Answers: 0   Comments: 4

For two co-planer circles to be tangent necessary and sufficient condition is, I think the distance between the centers of circles must be equal to r_1 +r_2 or ∣r_1 −r_2 ∣ , where r_1 and r_2 are the radii of the circles. If circles belong to different planes, what is necessary and sufficient condition for being tangent?

$$\mathrm{For}\:\mathrm{two}\:\boldsymbol{\mathrm{co}}-\boldsymbol{\mathrm{planer}}\:\mathrm{circles}\:\mathrm{to}\:\mathrm{be}\:\mathrm{tangent} \\ $$$$\:\mathrm{necessary}\:\mathrm{and}\:\mathrm{sufficient}\:\mathrm{condition}\:\mathrm{is}, \\ $$$$\mathrm{I}\:\mathrm{think} \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{centers}\:\mathrm{of}\: \\ $$$$\:\:\:\:\mathrm{circles}\:\:\mathrm{must}\:\mathrm{be}\:\mathrm{equal}\:\mathrm{to}\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} +\boldsymbol{\mathrm{r}}_{\mathrm{2}} \:\mathrm{or}\:\mid\boldsymbol{\mathrm{r}}_{\mathrm{1}} −\boldsymbol{\mathrm{r}}_{\mathrm{2}} \mid\:, \\ $$$$\:\:\:\:\:\mathrm{where}\:\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} \mathrm{and}\:\boldsymbol{\mathrm{r}}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{radii}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circles}. \\ $$$$\mathrm{If}\:\mathrm{circles}\:\mathrm{belong}\:\mathrm{to}\:\:\mathrm{different}\:\mathrm{planes},\:\mathrm{what} \\ $$$$\mathrm{is}\:\mathrm{necessary}\:\mathrm{and}\:\mathrm{sufficient}\:\mathrm{condition}\:\mathrm{for} \\ $$$$\mathrm{being}\:\mathrm{tangent}? \\ $$

Question Number 4033    Answers: 0   Comments: 15

One circle in a plane can produce one closed region at most(It produces one closed region at least).Two circles in a plane can produce at most three regions(They produce at least two regions).Three circles can produce seven closed regions at most(They produce three closed regions at least). How many distinct closed regions could be produced by 4, 5 and n circles at most? If n circles can produce f(n) closed regions at most(Of course they produce n closed regions at least), can they produce any number of closed regions between n and f(n)?

$$\:\:\:\:\mathrm{One}\:\mathrm{circle}\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane}\:\mathrm{can}\:\mathrm{produce}\:\mathrm{one}\:\:\mathrm{closed} \\ $$$$\mathrm{region}\:\mathrm{at}\:\mathrm{most}\left(\mathrm{It}\:\mathrm{produces}\:\mathrm{one}\:\mathrm{closed}\:\mathrm{region}\:\right. \\ $$$$\left.\mathrm{at}\:\mathrm{least}\right).\mathrm{Two}\:\:\mathrm{circles}\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane}\:\:\mathrm{can}\:\mathrm{produce} \\ $$$$\mathrm{at}\:\mathrm{most}\:\mathrm{three}\:\:\mathrm{regions}\left(\mathrm{They}\:\mathrm{produce}\:\mathrm{at}\:\mathrm{least}\right. \\ $$$$\left.\mathrm{two}\:\:\mathrm{regions}\right).\mathrm{Three}\:\mathrm{circles}\:\mathrm{can}\:\mathrm{produce}\:\mathrm{seven} \\ $$$$\mathrm{closed}\:\mathrm{regions}\:\mathrm{at}\:\mathrm{most}\left(\mathrm{They}\:\mathrm{produce}\:\mathrm{three}\right. \\ $$$$\left.\mathrm{closed}\:\mathrm{regions}\:\mathrm{at}\:\mathrm{least}\right). \\ $$$$ \\ $$$$\:\:\:\:\mathrm{How}\:\mathrm{many}\:\boldsymbol{\mathrm{distinct}}\:\mathrm{closed}\:\mathrm{regions}\:\mathrm{could}\:\mathrm{be}\:\mathrm{produced} \\ $$$$\mathrm{by}\:\mathrm{4},\:\mathrm{5}\:\mathrm{and}\:\mathrm{n}\:\:\mathrm{circles}\:\mathrm{at}\:\mathrm{most}? \\ $$$$\:\:\:\:\mathrm{If}\:\mathrm{n}\:\mathrm{circles}\:\mathrm{can}\:\mathrm{produce}\:\mathrm{f}\left(\mathrm{n}\right)\:\mathrm{closed}\:\mathrm{regions} \\ $$$$\mathrm{at}\:\mathrm{most}\left(\mathrm{Of}\:\mathrm{course}\:\mathrm{they}\:\mathrm{produce}\:\mathrm{n}\:\mathrm{closed}\right. \\ $$$$\left.\mathrm{regions}\:\mathrm{at}\:\mathrm{least}\right),\:\mathrm{can}\:\mathrm{they}\:\mathrm{produce}\:\mathrm{any} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{closed}\:\mathrm{regions}\:\mathrm{between}\:\mathrm{n}\:\mathrm{and}\:\mathrm{f}\left(\mathrm{n}\right)? \\ $$

Question Number 3950    Answers: 1   Comments: 0

We can make a cyllinder from a rectangle by connecting its opposite edges. Suppose we have two copies of a non−square rectangle.From one copy we make a long cyllinder by connecting long edges of it whereas from other copy by connecting short edges a short cyllinder is made. Compare the volumes of these two cyllinders.

$$\mathcal{W}{e}\:{can}\:{make}\:{a}\:\boldsymbol{{cyllinder}}\:{from}\:{a} \\ $$$$\boldsymbol{{rectangle}}\:{by}\:{connecting}\:{its}\:{opposite} \\ $$$$\boldsymbol{{edges}}. \\ $$$${Suppose}\:{we}\:{have}\:{two}\:{copies}\:{of}\:{a} \\ $$$$\boldsymbol{{non}}−\boldsymbol{{square}}\:\boldsymbol{{rectangle}}.{From} \\ $$$${one}\:{copy}\:{we}\:{make}\:{a}\:{long}\:{cyllinder} \\ $$$${by}\:{connecting}\:{long}\:{edges}\:{of}\:{it}\: \\ $$$${whereas}\:{from}\:{other}\:{copy}\:{by}\:{connecting} \\ $$$${short}\:{edges}\:{a}\:{short}\:{cyllinder}\:{is}\:{made}. \\ $$$${Compare}\:{the}\:\boldsymbol{{volumes}}\:{of}\:{these}\:{two} \\ $$$${cyllinders}. \\ $$

Question Number 3944    Answers: 0   Comments: 0

Prove that, inside a given square, a semicircle of the largest possible area, can be constructed using ruler and compass.

$$\mathcal{P}{rove}\:{that},\:{inside}\:\:{a}\:{given}\:{square},\:{a}\:{semicircle}\: \\ $$$${of}\:{the}\:{largest}\:{possible}\:{area},\:{can}\:{be}\:{constructed}\: \\ $$$${using}\:{ruler}\:{and}\:{compass}. \\ $$

Question Number 3943    Answers: 0   Comments: 11

What is the area of overlapping region of three circles of radii r_1 , r_2 , r_3 with their respective centres C_1 , C_2 and C_3 when r_1 +r_2 > C_1 C_2 , r_2 +r_3 > C_2 C_3 and r_3 +r_1 > C_3 C_1 . Note that C_i C_j is the distance between centres C_(i ) and C_j .

$$\mathcal{W}{hat}\:{is}\:{the}\:{area}\:\:{of}\:\:{overlapping}\:{region} \\ $$$${of}\:{three}\:{circles}\:{of}\:{radii}\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} \:,\:\boldsymbol{\mathrm{r}}_{\mathrm{2}} \:,\:\boldsymbol{\mathrm{r}}_{\mathrm{3}} \:{with}\:{their} \\ $$$${respective}\:{centres}\:\boldsymbol{\mathrm{C}}_{\mathrm{1}} \:,\:\boldsymbol{\mathrm{C}}_{\mathrm{2}} \:{and}\:\boldsymbol{\mathrm{C}}_{\mathrm{3}} \:{when} \\ $$$$\boldsymbol{\mathrm{r}}_{\mathrm{1}} +\boldsymbol{\mathrm{r}}_{\mathrm{2}} >\:\boldsymbol{\mathrm{C}}_{\mathrm{1}} \boldsymbol{\mathrm{C}}_{\mathrm{2}} \:\:,\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{2}} +\boldsymbol{\mathrm{r}}_{\mathrm{3}} >\:\boldsymbol{\mathrm{C}}_{\mathrm{2}} \boldsymbol{\mathrm{C}}_{\mathrm{3}} \:\:{and} \\ $$$$\boldsymbol{\mathrm{r}}_{\mathrm{3}} +\boldsymbol{\mathrm{r}}_{\mathrm{1}} >\:\boldsymbol{\mathrm{C}}_{\mathrm{3}} \boldsymbol{\mathrm{C}}_{\mathrm{1}} . \\ $$$${Note}\:{that}\:\boldsymbol{\mathrm{C}}_{\boldsymbol{\mathrm{i}}} \boldsymbol{\mathrm{C}}_{\boldsymbol{\mathrm{j}}} \:{is}\:{the}\:{distance}\:{between}\:{centres} \\ $$$$\boldsymbol{\mathrm{C}}_{\boldsymbol{\mathrm{i}}\:} {and}\:\:\boldsymbol{\mathrm{C}}_{\boldsymbol{\mathrm{j}}} \:. \\ $$

Question Number 3930    Answers: 0   Comments: 3

y=((log_e ((x/m)−sa))/r^2 ) yr^2 =log_e ((x/m)−sa) e^(yr^2 ) =(x/m)−sa me^(rry) =x−mas Merry christmas everyone! Let our venture for knowledge continue through to the new year!

$${y}=\frac{\mathrm{log}_{{e}} \left(\frac{{x}}{{m}}−{sa}\right)}{{r}^{\mathrm{2}} } \\ $$$${yr}^{\mathrm{2}} =\mathrm{log}_{\mathrm{e}} \left(\frac{{x}}{{m}}−{sa}\right) \\ $$$${e}^{{yr}^{\mathrm{2}} } =\frac{{x}}{{m}}−{sa} \\ $$$${me}^{{rry}} ={x}−{mas} \\ $$$$ \\ $$$$\mathrm{Merry}\:\mathrm{christmas}\:\mathrm{everyone}! \\ $$$$\mathrm{Let}\:\mathrm{our}\:\mathrm{venture}\:\mathrm{for}\:\mathrm{knowledge}\:\mathrm{continue} \\ $$$$\mathrm{through}\:\mathrm{to}\:\mathrm{the}\:\mathrm{new}\:\mathrm{year}! \\ $$

Question Number 3900    Answers: 1   Comments: 0

I asked this question a while ago, but I forgot how to solve it: S=∫_0 ^( n) ⌊x⌋dx

$$\mathrm{I}\:\mathrm{asked}\:\mathrm{this}\:\mathrm{question}\:\mathrm{a}\:\mathrm{while}\:\mathrm{ago}, \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{forgot}\:\mathrm{how}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}: \\ $$$$ \\ $$$${S}=\int_{\mathrm{0}} ^{\:{n}} \lfloor{x}\rfloor{dx} \\ $$

Question Number 3877    Answers: 0   Comments: 3

What is the area of overlapping region of two circles having radii r_1 and r_2 when the distance between their centres is c, given that r_1 +r_2 >c.

$$\mathcal{W}{hat}\:{is}\:{the}\:{area}\:{of}\:\:{overlapping} \\ $$$${region}\:{of}\:{two}\:{circles}\:{having}\:{radii} \\ $$$$\boldsymbol{\mathrm{r}}_{\mathrm{1}} \:{and}\:\boldsymbol{\mathrm{r}}_{\mathrm{2}} \:{when}\:{the}\:{distance}\:{between} \\ $$$${their}\:{centres}\:{is}\:\:\boldsymbol{\mathrm{c}},\:{given}\:{that}\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} +\boldsymbol{\mathrm{r}}_{\mathrm{2}} >\boldsymbol{\mathrm{c}}. \\ $$

Question Number 3850    Answers: 0   Comments: 7

How many dimention/s does the point have?

$${How}\:{many}\:{dimention}/{s}\:{does}\:{the}\:{point}\:{have}? \\ $$

Question Number 3843    Answers: 1   Comments: 4

Let the side of the following mentioned figures is s: The area of square is s^2 , the 3D area(volume) of a cube is s^3 ,the 4D area/volume of 4D hypercube can be said s^4 and so on. Now if radius is r, the area of circle is 𝛑r^2 , the 3D area(volume) of a sphere is(4/3)𝛑r^3 , what will be the 4D area/volume of 4D sphere and 5D area/volume of 5D sphere?

$$\mathcal{L}{et}\:{the}\:{side}\:{of}\:{the}\:{following}\:{mentioned} \\ $$$${figures}\:{is}\:\boldsymbol{\mathrm{s}}: \\ $$$${The}\:{area}\:{of}\:{square}\:{is}\:\boldsymbol{\mathrm{s}}^{\mathrm{2}} ,\:{the}\:\mathrm{3}{D}\:{area}\left({volume}\right) \\ $$$${of}\:{a}\:{cube}\:{is}\:\boldsymbol{\mathrm{s}}^{\mathrm{3}} ,{the}\:\mathrm{4}{D}\:{area}/{volume}\:{of}\:\mathrm{4}{D} \\ $$$${hypercube}\:{can}\:{be}\:{said}\:\boldsymbol{\mathrm{s}}^{\mathrm{4}} \:{and}\:{so}\:{on}. \\ $$$$ \\ $$$${Now}\:{if}\:{radius}\:{is}\:\boldsymbol{\mathrm{r}},\:{the}\:{area}\:{of}\:{circle}\:{is}\:\boldsymbol{\pi\mathrm{r}}^{\mathrm{2}} , \\ $$$${the}\:\mathrm{3}{D}\:{area}\left({volume}\right)\:{of}\:{a}\:{sphere}\:{is}\frac{\mathrm{4}}{\mathrm{3}}\boldsymbol{\pi\mathrm{r}}^{\mathrm{3}} , \\ $$$${what}\:{will}\:{be}\:{the}\:\mathrm{4}{D}\:{area}/{volume}\:{of}\:\mathrm{4}{D}\: \\ $$$${sphere}\:{and}\:\mathrm{5}{D}\:{area}/{volume}\:{of}\:\mathrm{5}{D}\:{sphere}? \\ $$

Question Number 3840    Answers: 1   Comments: 0

A semicircle contains a square of possible largest area.If s is the measure of the side of the square,what is the radius of the semicircle?

$$\mathcal{A}\:\:{semicircle}\:\:{contains}\:{a}\:{square}\:\:{of}\:\: \\ $$$${possible}\:{largest}\:{area}.{If}\:\:{s}\:\:{is}\:{the}\:{measure} \\ $$$${of}\:{the}\:{side}\:{of}\:{the}\:{square},{what}\:{is}\:{the} \\ $$$${radius}\:{of}\:{the}\:{semicircle}? \\ $$

Question Number 3838    Answers: 0   Comments: 4

Show that the construction of the rectangle of minimum perimeter when its area is ab where a=AB and b=CD are given is possible with ruler and compass.

$${Show}\:{that}\:{the}\:{construction}\:{of} \\ $$$$\:\:\:\:\:\:\mathrm{the}\:\mathrm{rectangle}\:\mathrm{of}\:\mathrm{minimum}\: \\ $$$$\:\:\:\:\:\:\mathrm{perimeter}\:\mathrm{when}\:\mathrm{its}\:\mathrm{area}\:\:\mathrm{is}\:\boldsymbol{\mathrm{ab}}\: \\ $$$$\:\:\:\:\:\:\mathrm{where}\:\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{AB}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{CD}}\:\mathrm{are}\:\:\mathrm{given} \\ $$$${is}\:{possible}\:{with}\:{ruler}\:{and}\:{compass}.\:\:\:\:\: \\ $$$$ \\ $$

Question Number 3836    Answers: 0   Comments: 5

A square,whose area is s^2 ,contains a semicircle of possible largest area. Determine radius of the semicircle.

$${A}\:{square},{whose}\:{area}\:{is}\:{s}^{\mathrm{2}} ,{contains}\: \\ $$$${a}\:{semicircle}\:{of}\:{possible}\:{largest}\:{area}. \\ $$$${Determine}\:{radius}\:{of}\:{the}\:{semicircle}. \\ $$

Question Number 3830    Answers: 0   Comments: 3

Draw a rectangle of maximum perimeter, by ruler and compass,when area is ab. (AB =a,CD=b are given.)

$$\mathcal{D}{raw}\:{a}\:{rectangle}\:{of}\:{maximum}\:{perimeter}, \\ $$$${by}\:{ruler}\:{and}\:{compass},{when}\:{area}\:{is}\:\boldsymbol{\mathrm{ab}}.\: \\ $$$$\left(\boldsymbol{\mathrm{AB}}\:=\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{CD}}=\boldsymbol{\mathrm{b}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{given}}.\right) \\ $$

Question Number 3823    Answers: 0   Comments: 2

Consider a triangle ABC. Let D and E are two points on AB and AC respectively such that DE ∥ BC. Now there are two parts of △ABC : △ADE and trapizoid DBCE. If these two regions have same area What will be the ratio of two distances : (i) distance of DE from point A and (ii) distance between BC and DE ?

$${Consider}\:{a}\:{triangle}\:\mathrm{ABC}.\:{Let}\:\mathrm{D}\:\:{and}\:\:\mathrm{E} \\ $$$${are}\:{two}\:{points}\:{on}\:\mathrm{AB}\:\:{and}\:\:\mathrm{AC}\:{respectively} \\ $$$${such}\:{that}\:\mathrm{DE}\:\parallel\:\mathrm{BC}.\:{Now}\:{there}\:{are}\:{two} \\ $$$${parts}\:{of}\:\bigtriangleup\mathrm{ABC}\::\:\bigtriangleup\mathrm{ADE}\:\:\:{and}\:\:{trapizoid} \\ $$$$\mathrm{DBCE}.\:{If}\:{these}\:{two}\:{regions}\:{have}\:{same}\:{area} \\ $$$${What}\:{will}\:{be}\:{the}\:{ratio}\:{of}\:{two}\:{distances}\:: \\ $$$$\left({i}\right)\:{distance}\:{of}\:\mathrm{DE}\:{from}\:{point}\:\mathrm{A}\:{and} \\ $$$$\left({ii}\right)\:{distance}\:{between}\:\mathrm{BC}\:{and}\:\mathrm{DE}\:\:? \\ $$

Question Number 3808    Answers: 2   Comments: 1

A chord divides the circle in two segments,having areas s_1 and s_2 . If diameter, perpendicular to this chord is cut into 1:3 by the chord ,what is s_1 :s_2 ?

$${A}\:{chord}\:{divides}\:\:{the}\:{circle}\:{in}\:{two} \\ $$$${segments},{having}\:{areas}\:{s}_{\mathrm{1}} \:{and}\:\:{s}_{\mathrm{2}} . \\ $$$${If}\:{diameter},\:{perpendicular}\:{to}\:{this} \\ $$$${chord}\:{is}\:{cut}\:{into}\:\mathrm{1}:\mathrm{3}\:{by}\:{the}\:{chord}\:,{what}\:{is}\:{s}_{\mathrm{1}} :{s}_{\mathrm{2}} \:? \\ $$$$ \\ $$

Question Number 3714    Answers: 1   Comments: 0

Derive a formula of volume of right circular cone when the formula of volume of cyllinder is given.

$$\mathcal{D}\mathrm{erive}\:\mathrm{a}\:\mathrm{formula}\:\mathrm{of}\:\mathrm{volume}\:\mathrm{of}\:\boldsymbol{\mathrm{right}}\:\boldsymbol{\mathrm{circular}}\:\boldsymbol{\mathrm{cone}} \\ $$$$\mathrm{when}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{of}\:\mathrm{volume}\:\mathrm{of}\:\boldsymbol{\mathrm{cyllinder}}\:\:\mathrm{is}\:\mathrm{given}. \\ $$

Question Number 3695    Answers: 0   Comments: 12

Can we say that A line is a circle whose radius is ∞ Or A circle with ∞ radius is a line ?

$$\mathcal{C}{an}\:{we}\:{say}\:{that} \\ $$$$\mathcal{A}\:{line}\:{is}\:{a}\:{circle}\:{whose}\:{radius}\:{is}\:\infty \\ $$$$\mathcal{O}{r} \\ $$$${A}\:{circle}\:{with}\:\infty\:{radius}\:{is}\:{a}\:{line}\:\:? \\ $$

Question Number 3662    Answers: 0   Comments: 11

Lets say we have an n−gon. All sides are equal. When n=3, interior angles θ=((180)/3) θ=60° n=4, θ=((360)/4)=90° ⋮ n=t, θ=((180(t−2))/t) For a circle (essentially an ∞−gon): n=∞ ∴θ=180lim_(t→∞) ((t−2)/t) θ=180°????

$$\mathrm{Lets}\:\mathrm{say}\:\mathrm{we}\:\mathrm{have}\:\mathrm{an}\:{n}−\mathrm{gon}. \\ $$$$\mathrm{All}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{equal}. \\ $$$$ \\ $$$$\mathrm{When}\:{n}=\mathrm{3},\:\mathrm{interior}\:\mathrm{angles}\:\theta=\frac{\mathrm{180}}{\mathrm{3}} \\ $$$$\theta=\mathrm{60}° \\ $$$$ \\ $$$${n}=\mathrm{4},\:\theta=\frac{\mathrm{360}}{\mathrm{4}}=\mathrm{90}° \\ $$$$\vdots \\ $$$${n}={t},\:\theta=\frac{\mathrm{180}\left({t}−\mathrm{2}\right)}{{t}} \\ $$$$ \\ $$$$\mathrm{For}\:\mathrm{a}\:\mathrm{circle}\:\left(\mathrm{essentially}\:\mathrm{an}\:\infty−{gon}\right): \\ $$$${n}=\infty \\ $$$$\therefore\theta=\mathrm{180}\underset{{t}\rightarrow\infty} {\mathrm{lim}}\frac{{t}−\mathrm{2}}{{t}} \\ $$$$\theta=\mathrm{180}°???? \\ $$

Question Number 3647    Answers: 0   Comments: 2

Draw a line segment equal to ab units when AB=a units and CD=b units are given. Only ruler and compass may be used.

$${Draw}\:{a}\:{line}\:{segment}\:{equal}\:{to} \\ $$$$\boldsymbol{{ab}}\:{units}\:{when}\:{AB}=\boldsymbol{{a}}\:{units}\:\:{and}\:{CD}=\boldsymbol{{b}}\:{units}\: \\ $$$${are}\:{given}.\:{Only}\:{ruler}\:{and}\:{compass}\:{may}\:{be} \\ $$$${used}. \\ $$

Question Number 3607    Answers: 1   Comments: 0

Construct a line segment of a^2 units using ruler and compass only, when a line segment of a units is given.

$${Construct}\:{a}\:{line}\:{segment}\:{of}\:\boldsymbol{{a}}^{\mathrm{2}} \:{units}\: \\ $$$${using}\:{ruler}\:{and}\:\:{compass}\:{only},\:{when} \\ $$$${a}\:{line}\:{segment}\:{of}\:\boldsymbol{{a}}\:{units}\:{is}\:{given}. \\ $$

Question Number 3348    Answers: 1   Comments: 3

Prove that the regular pentagon is possible with ruler and compass.

$${Prove}\:{that}\:{the}\:{regular}\:{pentagon} \\ $$$${is}\:{possible}\:{with}\:{ruler}\:{and}\:{compass}. \\ $$

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