Tag: mathematics and statistics

Quizzes (38) Questions (955)

Questions Related to mathematics and statistics

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

The area of a triangle, whose vertices are $(3, 2), (5, 2)$ and the point of intersection of the lines $x = a$ and $y = 5$, is $3$ square units. What is the value of $a$?

  1. $2$
  2. $3$
  3. $4$
  4. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$A(3,2) B(5,2)$ intersection of $x=a,y=5$   $(a,5)$
$\cfrac { 1 }{ 2 } \left| \triangle  \right| =3$
$\triangle =6$
$=\begin{vmatrix} 1 & 1 & 1 \\ 3 & 5 & a \\ 2 & 2 & 5 \end{vmatrix}$
$=25-2a-15+2a+6-10$
$=6$
For any value of a area $=3sq.units$
Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

If $P=(x _{1}, y _{1}), Q=(x _{2}, y _{2})$ and $R=(x _{3}, y _{3})$ are three points of a triangle in $\mathbb{R}^{2}$. Then, area of a $\triangle PQR$ in terms of determinant of matrix $M=\begin{bmatrix} 1& 1 & 1 \ x _{1} & x _{2} & x _{3} \ y _{1} & y _{2} & y _{3}\end{bmatrix}$ is

  1. $-|det(M)|$
  2. $|det(M)|$
  3. $\dfrac{1}{2}|det(M)|$
  4. $2|det(M)|$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$P=\left( { x } _{ 1 },{ y } _{ 1 } \right) ,Q=\left( { x } _{ 2 },{ y } _{ 2 } \right) ,R\left( { x } _{ 3 },{ y } _{ 3 } \right) $
$\left| M \right| =\triangle =\begin{vmatrix} 1 & 1 & 1 \\ { x } _{ 1 } & { x } _{ 2 } & { x } _{ 3 } \\ { y } _{ 1 } & { y } _{ 2 } & { y } _{ 3 } \end{vmatrix}$
Area of triangle$=\cfrac { 1 }{ 2 } det(M)$
Answer: Option C
Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

If $\triangle _1,\triangle _2$ be the areas of two triangles with vertices $(b,c), (c,a), (a,b)$, and $ (ac-b^2, ab-c^2),(ba-c^2, bc-a^2), (cb-a^2, ca-b^2)$, then $\ \dfrac{\triangle _1}{\triangle _2}=(a+b+c)^2$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$T _1 = (b, c)\ (c, a)\ (a, b) = (x _1, y _1)\ (x _2, y _2)\ (x _3, y _3)$

$T _2 = [(ac - b^2), (ab  - c^2)], [(ba  - c^2), (bc - a^2)], [(cb - a^2), (ca - b^2)]$
$T _1 = \dfrac{1}{2} [c(c - a) + a(a - b) + b(b - c)]$
$= \dfrac{1}{2} [c^2 - ac + a^2 - ab + b^2 - bc]$
$T _2 = \dfrac{1}{2} [(ab - c^2) [ba - c^2 - cb + a^2] + (bc - a^2) [cb - a^2 - ac + b^2] + (ca - b^2) [ac - b^2 - ba + c^2)]$
$= \dfrac{1}{2} [[(ab - c^2)(a - c) (a + b + c)] + (bc - a^2) [(b - a)(a + b + c)] + (ca - b^2) [(c - b) (a + b + c)]]$
$= \dfrac{(a + b + c)}{2} [a^2b - abc - ac^2 + c^3 + b^2 c - abc - a^2b + c^2a - abc - b^2 c + b^3]$
$= \dfrac{(a + b + c)}{2} (a^3 + b^3 + c^3 - 3abc)$
$= \left(\dfrac{a + b + c}{2}\right) [a + b + c] [ a^2 + b^2 + c^2 - ab - bc - ca]$
$\therefore \ \dfrac{\Delta _1}{\Delta _2} = \dfrac{(a + b + c)^2 [a^2 + b^2 + c^2 - bc - ba - ca]}{(a^2 + b^2 + c^2 - bc - ba - ca)}$
$= (a + b + c)^2$

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

If ${ \left( { x } _{ 1 }-{ { x } _{ 2 } } \right)  }^{ 2 }+{ \left( { y } _{ 1 }-{ y } _{ 2 } \right)  }^{ 2 }={ a }^{ 2 }$, ${ \left( x _{ 2 }-{ x } _{ 3 } \right)  }^{ 2 }+{ \left( { y } _{ 2 }-{ y } _{ 3 } \right)  }^{ 2 }={ b }^{ 2 }$, ${ \left( { x } _{ 3 }-{ x } _{ 1 } \right)  }^{ 2 }+{ \left( { y } _{ 3 }-{ y } _{ 1 } \right)  }^{ 2 }={ c }^{ 2 }$ and $k\begin{vmatrix} { x } _{ 1 } & { y } _{ 1 } & 1 \ { x } _{ 2 } & { y } _{ 2 } & 1 \ { x } _{ 3 } & { y } _{ 3 } & 1 \end{vmatrix}=(a+b+c)(b+c-a)(c+a-b)\times (a+b-c)$, then the value of $k$ is

  1. $1$
  2. $2$
  3. $4$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given : $k\begin{vmatrix} { x } _{ 1 } & { y } _{ 1 } & 1 \\ { x } _{ 2 } & { y } _{ 2 } & 1 \\ { x } _{ 3 } & { y } _{ 3 } & 1 \end{vmatrix}^{2}=(a+b+c)(b+c-a)(c+a-b)\times (a+b-c)$ ...... $(i)$

Area of any triangle $PQR$ with vertices $(x _1, y _1) , (x _2, y _2) $ and $(x _3, y _3)$ is given by
$\Delta = \dfrac{1}{2} \begin{vmatrix}x _1 & y _1 & 1\\ x _2 & y _2 & 1\\ x _3 & y _3 & 1\end{vmatrix}$ ......... $(ii)$

${ \left( { x } _{ 1 }-{ { x } _{ 2 } } \right)  }^{ 2 }+{ \left( { y } _{ 1 }-{ y } _{ 2 } \right)  }^{ 2 }={ a }^{ 2 }$, ${ \left( x _{ 2 }-{ x } _{ 3 } \right)  }^{ 2 }+{ \left( { y } _{ 2 }-{ y } _{ 3 } \right)  }^{ 2 }={ b }^{ 2 }$, ${ \left( { x } _{ 3 }-{ x } _{ 1 } \right)  }^{ 2 }+{ \left( { y } _{ 3 }-{ y } _{ 1 } \right)  }^{ 2 }={ c }^{ 2 }$

$\implies a,b,c$ are the length of sides of a $\triangle PQR$

Also, area of $\triangle PQR$ with sides $a, b, c$ is

$\Delta = \sqrt{s (s - a) (s - b) (s - c)}$ ......... $\left[\text{Where } s=\dfrac{a+b+c}{2}\right]$

$= \displaystyle \sqrt{\frac{1}{16}(2s)(2s - 2a)(2s - 2b)(2s - 2c)}$

$\Delta = \displaystyle \sqrt{\left [ \frac{(a + b + c)(b + c - a)(c + a - b)(a + b - c)}{16}\right ]}$ ........... $(iii)$

From $(ii)$ and $(iii)$, we get

$\displaystyle \frac{1}{2} \begin{vmatrix} x _1&  y _1& 1\\ x _2 & y _2 & 1\\ x _3 & y _3 & 1\end{vmatrix} = \sqrt{\left [\frac{(a+b+c)(b+c- a)(c + a - b)(a + b - c)}{16} \right ]}$

Squaring both sides, we have

$\Rightarrow 4 \begin{vmatrix}x _1 & y _1 & 1\\ x _2 & y _2 & 1\\ x _3 & y _3 & 1\end{vmatrix}^2 = (a+b+c)(b+c-a)(c+a- b)(a+b-c)$

Comparing above equation with $(i)$, we get $k=4$

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

$(x _1 - x _2)^2 + (y _1 - y _2)^2 = a^2$;
$(x _2 - x _3)^2 + (y _2 - y _3)^2 = b^2$;
$(x _3 - x _1)^2 + (y _3 - y _1)^2 = c^2$;
then find $4 \begin{vmatrix}x _1 & y _1 & 1\ x _2 & y _2 & 1\ x _3 & y _3 & 1\end{vmatrix}^2 = $

  1. $(a+b+c) (b+c - a) (c + a - b) (a + b - c)$
  2. $-(a+b+c) (b+c - a) (c + a - b) (a + b - c)$
  3. $-(a+b+c) (b+c - a) (c + a - b) (a + b - c)/2$
  4. $(a+b+c) (b+c - a) (c + a - b) (a + b - c)/2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Area of the triangle PQR with vertices $(x _1, y _1) , (x _2, y _2) $ and $(x _3, y _3)$ is

$\therefore

\Delta = \dfrac{1}{2} \begin{vmatrix}x _1 & y _1 & 1\ x _2 &

y _2 & 1\ x _3 & y _3 & 1\end{vmatrix}$          ....(1)

Now the area of $\Delta$ $PQR$ with sides $a, b, c$ is

$\Delta = \sqrt{s (s - a) (s - b) (s - c)}$

$= \displaystyle \sqrt{\frac{1}{16}(2s)(2s - 2a)(2s - 2b)(2s - 2c)}$

$\Delta = \displaystyle \sqrt{\left [ \frac{(a + b + c)(b + c - a)(c + a - b)(a + b - c)}{16}\right ]}$          ....(2)

From (1) and (2), we get

$\displaystyle

\frac{1}{2} \begin{vmatrix} x _1&  y _1& 1\ x _2 & y _2 &

1\ x _3 & y _3 & 1\end{vmatrix} = \sqrt{\left [

\frac{(a+b+c)(b+c- a)(c + a - b)(a + b - c)}{16} \right ]}$

Squaring both sides, we have

$\Rightarrow

4 \begin{vmatrix}x _1 & y _1 & 1\ x _2 & y _2 & 1\ x _3

& y _3 & 1\end{vmatrix}^2 = (a+b+c)(b+c-a)(c+a- b)(a+b-c)$

Multiple choice mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

The value of $ \displaystyle 36^{\circ} $ in radians is 

  1. $ \displaystyle \frac{\pi }{2} $
  2. $ \displaystyle \frac{2\pi }{5} $
  3. $ \displaystyle \frac{\pi }{5} $
  4. $ 3\pi $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle 180^{\circ}      $    =  $\displaystyle \pi       $ radians 


$\displaystyle  \therefore  1^{\circ}=\frac{\pi }{180}    radians     $
$\displaystyle  \Rightarrow 36^{\circ}=\frac{\pi }{180}\times 36 \ radians = \frac{\pi }{5}radians      $

Multiple choice mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

A unit radian is approximately equal to

  1. $57^{\circ} 17' 43"$
  2. $57^{\circ} 17' 45"$
  3. $57^{\circ} 17' 47"$
  4. $57^{\circ} 17' 49"$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
1 radian x 180 degrees per pi radians = 57.295779513082 degrees.

0.295779513082 degrees x 60 minutes per 1 degree = 17.74677078492 minutes.
0.74677078492 minutes x 60 seconds per 1 minute = 44.8062470952 seconds.

Answer : 57 degrees, 17.75 minutes or

57 degrees, 17 minutes, 45 seconds
Multiple choice mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

The value of $\cot 15^{\circ} \cot 20^{\circ} \cot 70^{\circ} \cot 75^{\circ}$ is equal to

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

The value of $\cot 15^{\circ}.\cot 20^{\circ}.\cot 70^{\circ}.\cot 75^{\circ}$ is
$= (\cot 15^{\circ} . \cot 75^{\circ}) (\cot 20^{\circ} \cot 70^{\circ})$

Now $\cot 75^{\circ} = \tan 15^{\circ}$ and $\cot 70^{\circ} = \tan 20^{\circ}$   ....................... $\cot(90-\theta)=\tan\theta$

Therefore, the given expression can be written as 
$ (\cot 15^{\circ} \tan 15^{\circ})(\cot 20^{\circ} \tan 20^{\circ})$
$= 1\times 1 = 1$

Multiple choice mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

Consider the following statements :
1. $1^o$ in radian measure is less than 0.02 radians.
2. 1 radian in degree measure is greater than $45^o$ 
Which of the above statements is/are correct ?

  1. 1 only

  2. 2 only

  3. Both 1 and 2

  4. Neither 1 nor 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

1 degree in radians $=1\times \dfrac { \pi  }{ 180 } =0.0174$  which is less than $0.02$ radians.


1 radians in degree $=1\times \dfrac { 180 }{ \pi  } =57.32$ which is greater than $45^{o}$.
Hence, both are correct.