Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

If $\displaystyle y=mx$ bisects the angle between the lines $\displaystyle x^{2}\left ( \tan ^{2}\theta +\cos ^{2}\theta  \right )+2xy\tan \theta -y^{2}\sin ^{2}\theta =0$  when $\displaystyle \theta =\dfrac\pi3$ the value of $m$ is

  1. $\displaystyle \frac{-2- \sqrt 7}{ \sqrt 3}$
  2. $\displaystyle \frac{ \sqrt 7-2}{ \sqrt 3}$
  3. $\displaystyle 2 \sqrt 7 $
  4. $\displaystyle 2 \sqrt 3 $
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation
Equation of the bisectors of the angles between the given lines is

$\displaystyle \dfrac{x _2-y _2}{a-b}=\dfrac{xy}{h }$

Equation of the bisectors of the angles between the given lines is
$\displaystyle \frac{x^{2}-y^{2}}{\tan ^{2}\theta +\cos ^{2}\theta +\sin ^{2}\theta }=\frac{xy}{\tan \theta }$

$\displaystyle \Rightarrow \frac{x^{2}-y^{2}}{1+\tan ^{2}\theta  }=\frac{xy}{\tan \theta }$

$\displaystyle \Rightarrow \frac{x^{2}-y^{2}}{1+3  }=\frac{xy}{\sqrt 3 } \ when \ \ \theta=\pi/3$

Which satisfied by $y=mx $ if

$\displaystyle \frac{1-m^{2}}{4}=\frac{m}{\sqrt 3}$

$\displaystyle \Rightarrow \sqrt 3 m^{2}+4m-\sqrt 3=0$

$\displaystyle \Rightarrow m=\frac{-2\pm \sqrt 7}{\sqrt 3}$
Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

If two of the lines represented by $ x^{4} + x^{3} y + cx^{2}y^{2} -xy^{3} + y^{4} =0$ bisect the angle between the other two, then the value of $c$ is

  1. $0$
  2. $-1$
  3. $1$
  4. $-6$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since the product of the slopes of the four lines represented by the given equation is $1$ and a pair of lines represent the bisectors of the angles between the other two, the product of the slopes of each pair is $-1$. So let the equation of one pair be $ax^{2} + 2hxy -ay^{2} = 0$

The equation of its bisectors is $ \displaystyle \frac{x^{2}-y^{2}}{2a}=\frac{xy}{h} $

By hypothesis $ x^{4} +x^{3}y + cx^{2} y^{2}-xy^{3} + y^{4} $ $= (ax^{2} + 2hxy -ay^{2}) (hx^{2} -2axy -hy^{2})$ 

$ = ah(x^{4} + y^{4}) + 2(h^{2} -a^{ 2}) (x^{3}y- xy^{3}) -6ahx^{2}y^{2} $ 

Comparing the respective coefficients we get

$ah = 1 $ and $c = -6ah = -6$

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

The line $y=3x$ bisects the angle between the lines $ax^{2}+2axy+y^{2}=0$ if ${a}=$ 

  1. $3$
  2. $11$
  3. $\displaystyle \frac{3}{11}$
  4. $\displaystyle \frac{11}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Pair of angle bisectors represented by
$ax^2+2hxy+by^2=0$   is given by
$\dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h}$
$\therefore ax^2+2axy+y^2=0$
pair of angle bisector is,
$\dfrac{x^2-y^2}{a-1}=\dfrac{xy}{a}$
$ax^2-ay^2=(a-1)xy$
Given  $ y=3x$  is one of angle bisector of given lines
$m=\dfrac{y}{x},$        $am^2+(a-1)x-a=0$
$m=3$ is satisfied to this equation
$9a+3a-3-a=0$
$11a=3$
$\therefore a=\dfrac{3}{11}$

Multiple choice maths perimeter and area converting between units of areas the metric system metric system

The area of a square field is $30\dfrac {1}{4}m^2$. Calculate the length of the side of the square.

  1. $5\dfrac {1}{3}m$
  2. $5\dfrac {1}{2}m$
  3. $5\dfrac {2}{5}m$
  4. $5\dfrac {1}{4}m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let side of square field be $x$.


Then area of square field $= x^2$


According to question

$x^2 = 30 \dfrac{1}{4} m^2$

$x^2 = \dfrac{121}{4} m^2$

$x = \dfrac{\sqrt{121}}{\sqrt{4}} m$

$x = \dfrac{11}{2}$

Hence side of square field is $5 \dfrac{1}{2} m$

Option (B)

Multiple choice maths perimeter and area converting between units of areas the metric system metric system

The area of a square field is $80\dfrac {244}{729}$ square metres. Find the length of each sides field.

  1. $8\dfrac {25}{27}m$
  2. $8\dfrac {24}{27}m$
  3. $8\dfrac {26}{27}m$
  4. $8\dfrac {22}{27}m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given area of square field $= 80 \dfrac{244}{729} m^2$


Let us assume that side of square field is x 


Then area of field $= x^2$

According to question

$x^2 = 80 \dfrac{244}{729} m^2$

$x^2 = \dfrac{58,564}{729} m^2$

$x = \dfrac{\sqrt{58,564}}{\sqrt{729}} m$

$x = \dfrac{242}{27} m$

Hence side of field is $8 \dfrac{26}{27} m$.

Option (C)

Multiple choice maths perimeter and area converting between units of areas the metric system metric system

The area of a square field is $325\ m^2$. Find the approximate length of one side of the field. (upto 2 places of decimals) (in $m^2$)

  1. $19.03$
  2. $18.02$
  3. $18.03$
  4. $17.03$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know area of any square is its side x side 

Let us assume that side of square field is x 
Then area of square field $= x^2$
According to question

$x^2 = 325 m^2$

$\Rightarrow x^2 = 325$

$x = \sqrt{325}$

So $x = 18.03$

Hence side of square field is $18.03 m$

option (C)

Multiple choice maths perimeter and area converting between units of areas the metric system metric system

The area of a square playground is $256.6404$ square metres. Find the length of one side of the playground.

  1. $16.04$ metres
  2. $16.02$ metres
  3. $16.06$ metres
  4. $16.08$ metres
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let side of square play ground be x

Then area of square play ground will be $x^2$
According to question

$x^2 = 256.6404 m^2$

$x = \sqrt{256.6404} m$

$x = \dfrac{\sqrt{2566404}}{\sqrt{10000}} m$

$= \dfrac{1602}{100}$

Hence side of square is $16.02 m$

Option (B)