Tag: maths

Questions Related to maths

Relation between circumradius and number of sides is given by-

maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

  1. $Area=\dfrac{r^2n\sin(\dfrac{360}{n})}{3}$

  2. $Area=\dfrac{r^2n\sin(\dfrac{360}{n})}{2}$

  3. $Area=\dfrac{r^2n\cos(\dfrac{360}{n})}{2}$

  4. None of the above

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Correct Option: B

The sum of the radii of inscribed and circumscribed circles of an n sided regular polygon of side $'a'$ is

maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

  1. $=\frac{a}{2} \left ( \frac{1}{\sin \pi/2x} + \cot \frac{\pi}{x} \right )$

  2. $=\frac{a}{2} \left ( \frac{1}{\sin \pi/x} + \cot \frac{\pi}{2x} \right )$

  3. $=\frac{a}{2} \left ( \frac{1}{\sin \pi/x} + \cot \frac{\pi}{x} \right )$

  4. None of these

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Correct Option: C
Explanation:

$R\sin \theta  = \frac{a}{2}$
$R = \frac{a}{2\sin \theta }$
$\tan \theta = \frac{a/2}{r}$
$r = \frac{a}{2\tan \theta }                                   \theta = \frac{2\pi}{n} \times\frac{1}{2}$
$R+r = \frac{a}{2} \left ( \frac{1}{\sin \theta}+\frac{\sin \theta}{\cos \theta } \right )             = \frac{\pi}{x}$
    $= \frac{a}{2} \left ( \frac{1}{\sin \pi/x} + \cot \frac{\pi}{x} \right )$

In $\Delta ABC$, there are 35 lines drawn parallel to the base BC such that each line divides the other side into, equal parts. 
If BC =1.8 m find the length of $P _7 Q _7$.

maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

  1. 1.8 m

  2. 3.5 m

  3. 0.35 m

  4. 0.18 m

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Correct Option: C

State true or false.
A line from vortex C of  $\Delta ABC$ bisects the median from A. It divides the side AB in 1: 2.

maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

  1. True

  2. False

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Correct Option: A

For a regular hexagon with apothem $5m$, the side length is about $5.77m$. The area of the regular hexagon is (in $m^2$).

maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

  1. $75.5$

  2. $85.5$

  3. $76.5$

  4. $86.5$

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Correct Option: D
Explanation:

For a regular hexagon with apothem $5 m$ the side length is about $5.77m$.
Use the formula
$A = \frac{1}{2}pa$
to find the area of the hexagon.
The perimeter of the hexagon is about $6(5.77)$ or $34.62m$.
Now substitute the values.
$A = \frac{1}{2} (34.62)(5)$
Simplify.
$A = \frac{1}{2}(173.1)$
= $86.5$
$A = \frac{1}{2}(173.1)$
= $86.5$
Therefore, the area of the regular hexagon is about $86.5 m^2$.

If $D$ is the midpoint of side $BC$ of a triangle $ABC$ and $AD$ is perpendicular to $AC$ then

maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

  1. $3{a}^{2}={b}^{2}-3{c}^{2}$

  2. $3{b}^{2}={a}^{2}-{c}^{2}$

  3. ${b}^{2}={a}^{2}-{c}^{2}$

  4. ${a}^{2}+{b}^{2}=5{c}^{2}$

Show answer
Correct Option: B
Explanation:

In $\triangle ACD, \cos{C}=\dfrac{b}{\left(\dfrac{a}{2}\right)}$
$\Rightarrow \dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}=\dfrac{2b}{a}$
$\Rightarrow \dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2b}=2b$
$\Rightarrow {a}^{2}+{b}^{2}-{c}^{2}=4{b}^{2}$
$\Rightarrow {a}^{2}-{c}^{2}=3{b}^{2}$

If the angles of a triangle are in the ratio $2:3:7,$ then the sides opposite to these angles are in the ratio

maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

  1. $\sqrt{2}:2:\left(\sqrt{3}+1\right)$

  2. $2:\sqrt{2}:\left(\sqrt{3}+1\right)$

  3. $1:\sqrt{2}:\dfrac{\sqrt{2}}{\left(\sqrt{3}-1\right)}$

  4. $\dfrac{1}{\sqrt{2}}:1:\left(\dfrac{\sqrt{3}+1}{2}\right)$

Show answer
Correct Option: A,C,D
Explanation:

Let $A=2\alpha, B=3\alpha, C=7\alpha$
$\therefore A+B+C={180}^{0}$
$\Rightarrow 2\alpha+3\alpha+7\alpha={180}^{0}$
$\Rightarrow 12\alpha={180}^{0}$
$\Rightarrow \alpha={15}^{0}$
$\therefore A=2\alpha=2\times{15}^{0}={30}^{0}$
$B=3\alpha=3\times{15}^{0}={45}^{0}$
$C=7\alpha=7\times{15}^{0}={105}^{0}$
$a:b:c=\sin{{30}^{0}}:\sin{{45}^{0}}:\sin{{105}^{0}}$
       $=\dfrac{1}{2}:\dfrac{1}{\sqrt{2}}:\dfrac{\sqrt{3}+1}{2\sqrt{2}}$
$a:b:c=\sqrt{2}:2:\sqrt{3}+1$
        $=\dfrac{1}{\sqrt{2}}:1:\dfrac{\sqrt{3}+1}{2}$
or $1:\sqrt{2}:\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}$
or $1:\sqrt{2}:\dfrac{2\sqrt{2}}{2\left(\sqrt{3}-1\right)}$
$\therefore a:b:c= 1:\sqrt{2}:\dfrac{\sqrt{2}}{\sqrt{3}-1}$

In a triangle $ABC, \cos{A}+\cos{B}+\cos{C}=\dfrac{3}{2}$ then the triangle is

maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

  1. isosceles

  2. right-angled

  3. equilateral

  4. none of these.

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Correct Option: C
Explanation:

$\cos{A}+\cos{B}+\cos{C}=\dfrac{3}{2}$
Using transformation angle formula, we have
$\left(\cos{A}+\cos{B}\right)+\cos{C}=\dfrac{3}{2}$
$\Rightarrow 2\cos{\left(\dfrac{A+B}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}+\cos{C}=\dfrac{3}{2}$
Using sub-multiple angle formula to $\cos{C}=1-2{\sin}^{2}{\dfrac{C}{2}}$ we get
$\Rightarrow  2\cos{\left(\dfrac{A+B}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}+1-2{\sin}^{2}{\dfrac{C}{2}}=\dfrac{3}{2}$ 
Since $A+B+C=\pi\Rightarrow \dfrac{A+B}{2}=\dfrac{\pi}{2}-\dfrac{C}{2}$
$\Rightarrow  2\cos{\left(\dfrac{\pi}{2}-\dfrac{C}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}-2{\sin}^{2}{\dfrac{C}{2}}=\dfrac{3}{2}-1$
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}-2{\sin}^{2}{\dfrac{C}{2}}=\dfrac{1}{2}$
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\left[\cos{\left(\dfrac{A-B}{2}\right)}+{\sin\left(\dfrac{C}{2}\right)}\right]=\dfrac{1}{2}$
Again
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\left[\cos{\left(\dfrac{A-B}{2}\right)}+{\sin\left(\dfrac{\pi}{2}-\dfrac{A+B}{2}\right)}\right]=\dfrac{1}{2}$
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\left[\cos{\left(\dfrac{A-B}{2}\right)}+{\cos\left(\dfrac{A+B}{2}\right)}\right]=\dfrac{1}{2}$
Using transformation angle formula, we get
$\Rightarrow \sin{\left(\dfrac{C}{2}\right)}\left[2\sin{\left(\dfrac{B}{2}\right)}\sin{\left(\dfrac{A}{2}\right)}\right]=\dfrac{1}{4}$
$\Rightarrow \sin{\left(\dfrac{A}{2}\right)}\sin{\left(\dfrac{B}{2}\right)}\sin{\left(\dfrac{C}{2}\right)}=\dfrac{1}{8}$
$\therefore \sin{\left(\dfrac{A}{2}\right)}=\dfrac{1}{2}, \sin{\left(\dfrac{B}{2}\right)}=\dfrac{1}{2},\sin{\left(\dfrac{C}{2}\right)}=\dfrac{1}{2}$
$\Rightarrow \sin{\left(\dfrac{A}{2}\right)}=\sin{\dfrac{\pi}{6}}, \sin{\left(\dfrac{B}{2}\right)}=\sin{\dfrac{\pi}{6}},\sin{\left(\dfrac{C}{2}\right)}=\sin{\dfrac{\pi}{6}}$
$\therefore {\left(\dfrac{A}{2}\right)}={\left(\dfrac{B}{2}\right)}={\left(\dfrac{C}{2}\right)}=\dfrac{\pi}{6}$
$\Rightarrow \angle{A}=\angle{B}=\angle{C}=2\times\dfrac{\pi}{6}=\dfrac{\pi}{3}$
Hence, the triangle is equilateral.

Let ${A} _{0}{A} _{1}{A} _{2}{A} _{3}{A} _{4}{A} _{5}$ be a regular hexagon inscribed in a circle of unit radius.The product of the length of the line segments ${A} _{0}{A} _{1},{A} _{0}{A} _{2}$ and ${A} _{0}{A} _{4}$ is

maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

  1. $\dfrac{3}{4}$

  2. $3\sqrt{3}$

  3. $3$

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

Show answer
Correct Option: C
Explanation:

${A} _{0}{A} _{1}=2.1.\cos{\dfrac{\pi}{3}}=2.1.\dfrac{1}{2}=1={A} _{1}{A} _{2}$
$\cos{\dfrac{2\pi}{3}}=\dfrac{{\left({A} _{0}{A} _{1}\right)}^{2}+{\left({A} _{1}{A} _{2}\right)}^{2}-{\left({A} _{0}{A} _{2}\right)}^{2}}{2{A} _{0}{A} _{1}.{A} _{1}{A} _{2}}$
$\Rightarrow \cos{\left(\pi-\dfrac{\pi}{3}\right)}=\dfrac{1+1-{\left({A} _{0}{A} _{2}\right)}^{2}}{2.1.1}$
$\Rightarrow -\cos{\dfrac{\pi}{3}}=\dfrac{2-{\left({A} _{0}{A} _{2}\right)}^{2}}{2}$
$\Rightarrow \dfrac{-1}{2}=\dfrac{2-{\left({A} _{0}{A} _{2}\right)}^{2}}{2}$
$\Rightarrow -1=2-{\left({A} _{0}{A} _{2}\right)}^{2}$
$\Rightarrow {\left({A} _{0}{A} _{2}\right)}^{2}=3$
$\therefore {A} _{0}{A} _{2}=\sqrt{3}={A} _{0}{A} _{4}$
$\therefore {A} _{0}{A} _{1}\times {A} _{0}{A} _{2}\times{A} _{0}{A} _{4}=1\times\sqrt{3}\times\sqrt{3}=3$

The ratio of the areas of two regular octagons which are respectively inscribed and circumscribed to a circle of radius $r$ is

maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

  1. $\cos{\dfrac{\pi}{8}}$

  2. ${\sin}^{2}{\dfrac{\pi}{8}}$

  3. ${\cos}^{2}{\dfrac{\pi}{8}}$

  4. ${\tan}^{2}{\dfrac{\pi}{8}}$

Show answer
Correct Option: C
Explanation:

Inscribed circle of a regular polygon of $n$ sides
${A} _{1}=n{r}^{2}\tan{\dfrac{\pi}{n}}$
Here $n=8$
$\therefore {A} _{1}=8{r}^{2}\tan{\dfrac{\pi}{8}}$
Circumscribed circle of a regular polygon of $n$ sides is
${A} _{2}=\dfrac{n{R}^{2}}{2}\sin{\dfrac{2\pi}{n}}$
For $n=8$ we have
${A} _{2}=\dfrac{8{R}^{2}}{2}\sin{\dfrac{2\pi}{8}}$
  $=\dfrac{8{R}^{2}}{2}\sin{\dfrac{\pi}{4}}$
  $=\dfrac{8{r}^{2}}{2}2\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}$ (for $R=r$)
  $=8{r}^{2}\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}$ 
$\therefore \dfrac{{A} _{2}}{{A} _{1}}=\dfrac{8{r}^{2}\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}}{8{r}^{2}\tan{\dfrac{\pi}{8}}}$
$=\dfrac{\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}}{\dfrac{\sin{\dfrac{\pi}{8}}}{\cos{\dfrac{\pi}{8}}}}$
$={\cos}^{2}{\dfrac{\pi}{8}}$