Tag: maths

Questions Related to maths

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

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

  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 $

Show answer
Correct Option: A,B
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

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

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

$\displaystyle \Rightarrow m=\frac{-2\pm \sqrt 7}{\sqrt 3}$

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

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

  1. $0$

  2. $-1$

  3. $1$

  4. $-6$

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Correct Option: D
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 $c = -6ah = -6$

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

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

  1. $3$

  2. $11$

  3. $\displaystyle \frac{3}{11}$

  4. $\displaystyle \frac{11}{3}$

Show answer
Correct Option: C
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
$9a+3a-3-a=0$
$11a=3$
$\therefore a=\dfrac{3}{11}$

Consider the railway platform which is in square shape having side length $2\ km$. Then area of the platform is $4$ ____.

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

  1. $m$

  2. $km$

  3. $km^{2}$

  4. $m^{2}$

Show answer
Correct Option: C
Explanation:

Area of square platform$=2\times 2=4 km^2$

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

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

  1. $5\dfrac {1}{3}m$

  2. $5\dfrac {1}{2}m$

  3. $5\dfrac {2}{5}m$

  4. $5\dfrac {1}{4}m$

Show answer
Correct Option: B
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)

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

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

  1. $8\dfrac {25}{27}m$

  2. $8\dfrac {24}{27}m$

  3. $8\dfrac {26}{27}m$

  4. $8\dfrac {22}{27}m$

Show answer
Correct Option: C
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)

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 $m^2$)

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

  1. $19.03$

  2. $18.02$

  3. $18.03$

  4. $17.03$

Show answer
Correct Option: C
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)

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

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

  1. $16.04$ metres

  2. $16.02$ metres

  3. $16.06$ metres

  4. $16.08$ metres

Show answer
Correct Option: B
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)

By converting the $5.6 m^2$ into the $cm^2$, the answer will be

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

  1. $0.0056cm^2$

  2. $5600cm^2$

  3. $56000cm^2$

  4. $560cm^2$

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Correct Option: B

Which would be weight closest to 800 kg?

maths metric system converting between units of areas the metric system units of area and their conversions

  1. Feather

  2. Ball

  3. Cow

  4. None of these

Show answer
Correct Option: C
Explanation:

Weight of feather is very low about $0.082$ grams.


Weight of ball is also normally about $150$ grams.

And it's well known that cow is heavy and it weighs about $800$ kg. In fact weight of a bull(male) is about $1100$ kg.