Tag: terms related to polygons

Questions Related to terms related to polygons

The straight line AB is divided at C so that $\bar{AC} = 3\bar{CB}$. Circles are described on AC and CB as diameters and a common tangent meets AB produced at D. Then $\bar{BD}$ equals.

maths fundamental concepts - geometry terms related to polygons curves open and closed figures

  1. the diameter of the smaller circle

  2. the radius of the smaller circle

  3. the radius of the larger circle

  4. $\bar{CB} \sqrt{3}$

  5. the difference of the two radii

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Correct Option: B
Explanation:

Let $ x=\overline{BD} $ and let $ r$ be the radius of the small circle. 


Draw the line from the center of each of the circles to the point of contact of the tangent of the circle. 

By similar triangles, 

$ \dfrac{x+r}{r}=\dfrac{x+5r}{3r} \implies x=r$.

$ \overline{BD} $ equals the radius of the smaller circle.

All chords of the curve $x^{2}+y^{2}-10x-4y+4=0$  which make a right angle at $(8,2)$ pass through

maths fundamental concepts - geometry terms related to polygons curves open and closed figures

  1. $(2,5)$

  2. $(-2,-5)$

  3. $(-5,-2)$

  4. $(5,2)$

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Correct Option: D

Let $m$ be the slope of tangent to the curve $e^{2y}=1+x^{2}$ then set of all values  of $m$ is :

maths fundamental concepts - geometry terms related to polygons curves open and closed figures

  1. $\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$

  2. $\left[-\infty, -\dfrac{1}{2}\right]\cup \left[\dfrac{1}{2}, 0\right]$

  3. $\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]-\left{0\right}$

  4. $[-2,2]-\left{0\right}$

Show answer
Correct Option: A
Explanation:

$m$ be the slope of tangent.

Given equation of curve is

${{e}^{2y}}=1+{{x}^{2}}$


Taking log both side and we get,

$ \log {{e}^{2y}}=\log \left( 1+{{x}^{2}} \right) $

$ 2y=\log \left( 1+{{x}^{2}} \right) $


On differentiating and we get,

$ 2\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\dfrac{d}{dx}\left( 1+{{x}^{2}} \right) $

$ \Rightarrow 2\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\dfrac{d}{dx}\left( 1+{{x}^{2}} \right) $

$ \Rightarrow 2\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\dfrac{d}{dx}2x $

$ \Rightarrow \dfrac{dy}{dx}=\dfrac{x}{1+{{x}^{2}}} $

$m=\dfrac{dy}{dx}=\dfrac{x}{1+{{x}^{2}}}$


On put $x=\left( -1,1 \right)$

So,

$ m=\dfrac{1}{1+1}=\dfrac{1}{2} $

$ m=\dfrac{-1}{1+1}=\dfrac{-1}{2} $

Hence, the value of $m$ is $\left[-\dfrac{1}{2},\dfrac{1}{2} \right]$


Hence, this is the answer.

The radius of the locus by the point represented by $z$, when $arg\dfrac {z-1}{z+1} =\dfrac {\pi}{4}$, is

maths fundamental concepts - geometry terms related to polygons curves open and closed figures

  1. $\sqrt {2}$

  2. $\sqrt {2}\pi$

  3. $\dfrac {\pi}{\sqrt {2}}$

  4. $none\ of\ these$

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Correct Option: A
Explanation:
Arg$\left[\cfrac{z-1}{z+1}\right]=\dfrac{\pi}{4}$

$\Rightarrow$Arg$\left[{z-1}\times \cfrac{1}{z+1}\right]=\dfrac{\pi}{4}$

$\Rightarrow$Arg$\left[{z-1}\right] \times$Arg$\left[\cfrac{1}{z+1}\right]=\dfrac{\pi}{4}$

$\Rightarrow$ Arg${\left[z-1\right]}-$Arg${\left[z+1\right]}=\dfrac{\pi}{4}$

Let $z=x+iy$ 

$\therefore$Arg$\left[\left(x-1\right)+iy\right]-$Arg$\left[\left(x+1\right)+iy\right]=\dfrac{\pi}{4}$

$\Rightarrow {\tan}^{-1}\left(\dfrac{y}{x-1}\right)-{\tan}^{-1}\left(\dfrac{y}{x+1}\right)=\dfrac{\pi}{4}$

$\Rightarrow {\tan}^{-1}{\left(\dfrac{\dfrac{y}{x-1}-\dfrac{y}{x+1}}{1+\dfrac{{y}^{2}}{{x}^{2}-1}}\right)}=\dfrac{\pi}{4}$

$\Rightarrow \dfrac{2y}{{x}^{2}-1+{y}^{2}}=1$

$\Rightarrow {x}^{2}-1+{y}^{2}=2y$

$\Rightarrow {x}^{2}-2y+{y}^{2}-1=0$

This represents a circle with center at $\left(0,1\right)$ and radius  $=\sqrt{0+1-\left(-1\right)}=\sqrt{2}$

A curve which begins and ends at the same point is called a:

maths fundamental concepts - geometry terms related to polygons curves open and closed figures

  1. closed curve

  2. open curve

  3. normal curve

  4. definite curve

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Correct Option: A
Explanation:

A close curve is made up of a closed boundary. It initialised by a fixed point and end with the same point.

So, a curve which begins and ends at the same point is called a closed curve.
Hence, the answer is a closed curve.

..................... are examples of simple closed curve.

maths fundamental concepts - geometry terms related to polygons curves open and closed figures

  1. Circle

  2. Rectangle

  3. Square

  4. All of the above

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Correct Option: D
Explanation:

A curve that does not cross itself and ends at the same point where it begins.
Therefore, D is the correct answer.

A point is moving along the curve ${ y }^{ 3 }=27x$. Find the interval of valued of $x$ in which the ordinate changes faster then abscissa is:

maths fundamental concepts - geometry terms related to polygons curves open and closed figures

  1. $x\in \left( -1,1 \right)$

  2. $x\in \left( -1,-1 \right) -\left{ 0 \right}$

  3. $x\in \left[ -1,1 \right] -\left{ 0 \right}$

  4. $x\in \left( -1,0 \right) $

Show answer
Correct Option: A
Explanation:

$y^3=27x$

Abcissa changes at slower rate than ordinate.
$\dfrac{dx}{dt}<\dfrac{dy}{dt}.................(1)$
$y^3=27x$
$3y^2\dfrac{dy}{dt}=27\dfrac{dx}{dt}.............(2)$
Putting $\dfrac{dx}{dt}$ in eq $(1)$
 $\dfrac{3y^2}{27}<\dfrac{dy}{dy}$
 $\dfrac{dy}{dt}\left(\dfrac{3y^2}{27}<1\right)<0$
By eq$(2)$ wecan say that $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ will be '+ve' or '-ve'.
So, $\dfrac{3y^2}{27}-1<0\Rightarrow{-3}<y<3$ and $-1<x<1$.

The curves $y = 2{\left( {x - a} \right)^2}andy = {e^{2x}}$ touches each other, then'a' is less than- 

maths fundamental concepts - geometry terms related to polygons curves open and closed figures

  1. $-1$

  2. $0$

  3. $1$

  4. $2$

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Correct Option: C

What is a curve?

maths fundamental concepts - geometry terms related to polygons curves open and closed figures

  1. A line which is not straight and does not any sharp edges.

  2. It is a polygon

  3. It is a quadrilateral

  4. A line with sharp edges.

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Correct Option: A
Explanation:

In mathematics, a straight line also is a curve with no bends.
Therefore, A is the correct answer.