Tag: mathematics and statistics

Questions Related to mathematics and statistics

In $\displaystyle \angle ROP,$ the vertex is at:

mathematics and statistics angle and its measurement directed angles

  1. $R$

  2. $P$

  3. $O$

  4. None of the above

Show answer
Correct Option: C
Explanation:

Here, the angle is written as $\angle{ROP}$ and the angle is made by the intersection of two lines. 

So, here $RO$ and $OP$ are two lines which makes the angle at point $O$.
Hence, the vertex is at $O$.

The common end point where the two rays meet is called:

mathematics and statistics angle and its measurement directed angles

  1. arm

  2. vertex

  3. ray

  4. line

Show answer
Correct Option: B
Explanation:

The common end point where the two rays meet is called as a vertex.

Hence, the answer is vertex.

In $\displaystyle \angle PRQ $, the two arms are:

mathematics and statistics angle and its measurement directed angles

  1. $\displaystyle \overrightarrow{PR} $ and $\displaystyle \overrightarrow{RQ} $

  2. $\displaystyle \overrightarrow{RP} $ and  $\displaystyle \overrightarrow{PQ} $

  3. $\displaystyle \overrightarrow{QR} $ and $\displaystyle \overrightarrow{QP} $

  4. None of the above

Show answer
Correct Option: A
Explanation:
An angle is made by the intersection of two lines and that lines are also called as arms.
In $\angle{PRO}$, the angle is formed by the intersection of $\overrightarrow{PR}$ and $\overrightarrow{RO}$ at $R$.
Hence, the two arms are $\displaystyle \overrightarrow{PR} $ and $\displaystyle \overrightarrow{RQ} $.

To draw an angle of $150^o$ using a pair of compass and ruler ______.

mathematics and statistics angle and its measurement directed angles

  1. Bisect angle between $120^o$ and $180^o$

  2. Bisect angle between $60^o$ and $120^o$

  3. Bisect angle between $0^o$ and $160^o$

  4. None of these

Show answer
Correct Option: A
Explanation:

To draw an angle of 150° using a pair of compass and 

rule we bisect an angle between 120° and 180° 
$\rightarrow$ Since 120°<150°<180° we bisect angle
     between 120° and 180°

If the sum of two angles is equal to an obtuse angle, then which of the following is NOT possible?

mathematics and statistics angle and its measurement directed angles

  1. One obtuse and one acute angle

  2. One right angle and one acute angle

  3. Two acute angles

  4. Two right angles

Show answer
Correct Option: D
Explanation:

Obtuse angles are those angles whose measure is more than$90°$ but less than $180°$.


Since, sum of two right angles is $90°+90°=180°$.

Hence sum of two right angles can not be an obtuse angle.

Choose the correct answers from the alternatives given.
In $\Delta $ABC, the sides AB and AC are produced to P and Q respectively. The bisectors of $\angle PBC  \, and \,  \angle QCB $ intersect at a point 0. then $\angle BOC$ is equal to:

mathematics and statistics angle and its measurement directed angles

  1. 90- $\frac{1}{2} \angle A$

  2. 90+ $\frac{1}{2} \angle A$

  3. $120^{\circ}$ + $\frac{1}{2} \angle A$

  4. 120 - $\frac{1}{2} \angle A$

Show answer
Correct Option: A
Explanation:

2 $\angle 1 + \angle B = 180^{circ}$         (linear pair)
$\angle 1 = 90^{\circ}-\dfrac{1}{2}\angle B$          (1)
Similarly,
$\angle 2 = 90^{\circ} - \dfrac{1}{2} \angle C$   (2)
$\angle BOC = 180^{circ}-  (\angle 1 + \angle 2)$
=$180^{circ}- [180^{circ}-\dfrac{1}{2}$ ($\angle B + \angle C$)]
=$\dfrac{1}{2}[\angle B + \angle C] = \dfrac{1}{2} (180^{circ} - \angle A) = 90^{circ}-\dfrac{1}{2} \angle A$

A half turn about O is a rotation through angel of ____ or ____

mathematics and statistics angle and its measurement directed angles

  1. $-90^0, +90^0$

  2. $+180^0, -180^0$

  3. $+360^0, -360^0$

  4. $-270^0, +270^0$

Show answer
Correct Option: B

If $\begin{vmatrix} x _1 & y _1 & 1 \ x _2 & y _2 & 1 \ x _3 & y _3 & 1\end{vmatrix}=\begin{vmatrix} a _1 & b _1 & 1\ a _2 & b _2 & 1 \ a _3 & b _3 & 1\end{vmatrix}$, then the two triangles with vertices $(x _1, y _1), (x _2, y _2), (x _3, y _3)$ and $(a _1,b _1)$, $(a _2, b _2)$, $(a _3, b _3)$ must be congruent.

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

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

The two determinants denote twice the area of $\Delta^s$ whose vertices are $(x _1,y _1), (x _2, y _2)$, $(x _3, y _3)$ and $(a _1, b _1)$, $(a _2, b _2), (a _3, b _3)$. This the equality of two determinants implies that their areas are equal. But equality of the areas of two triangles does not imply that they are congruent.

If the area of the triangle with vertices $(2, 5), (7, k)$ and $(3, 1)$ is $10$, then find the value of $k$.

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

  1. $-5$ or $35$

  2. $5$ or $-35$

  3. $15$ or $-5$

  4. $-5$ or $-25$

Show answer
Correct Option: B
Explanation:
If $(x _1,y _1), (x _2, y _2)$ ans $(x _3, y _3)$ are the vertices of a triangle, then its area is given by $\pm \dfrac {1}{2}[x _1(y _2-y _3)+x _2(y _3-y _1)+x _3(y _1-y _2)]$ 
Given vertices are $(2,5), (7,k), (3,1)$ and area is $10$.
Therefore, $\pm 10 = \dfrac {1}{2}[2(k-1)+7(1-5)+3(5-k)]$
$\Rightarrow \pm 20=2k-2-28+15-3k$
$\Rightarrow \pm 20=-k-15$
$\Rightarrow k = 5$ or $-35$

If $\displaystyle \left | \begin{matrix}x _{1} &y _{1}  &1 \ x _{2} &y _{2}  &1 \ x _{3} &y _{3}  &1 \end{matrix} \right |=\left | \begin{matrix}1 &1  &1 \ b _{1} &b _{2}  &b _{3} \ a _{1} &a _{2}  &a _{3}\end{matrix} \right |$ then the two triangles whose vertices are $\displaystyle \left ( x _{1},y _{1} \right ), \left ( x _{2},y _{2} \right ), ( \left ( x _{3},y _{3} \right ) $ and $\displaystyle\left ( a _{1},b _{1} \right ), \left ( a _{2},b _{2} \right ), \left ( a _{13},b _{3} \right ),$ are

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

  1. congruent

  2. similar

  3. equal in area

  4. none of these

Show answer
Correct Option: C
Explanation:

If $\left( x _{ 1 },y _{ 1 } \right) ,\left( x _{ 2 },y _{ 2 } \right) ,(\left( x _{ 3 },y _{ 3 } \right) $ are the vertices of triangle , then its area is 

$A _{1}=\dfrac { 1 }{ 2 } \left| \begin{matrix} x _{ 1 } & y _{ 1 } & 1 \ x _{ 2 } & y _{ 2 } & 1 \ x _{ 3 } & y _{ 3 } & 1 \end{matrix} \right| $

If $\left( a _{ 1 },b _{ 1 } \right) ,\left( a _{ 2 },b _{ 2 } \right) ,\left( a _{ 3 },b _{ 3 } \right) $ are the vertices of triangle , then its area is

$A _{2}=\dfrac { 1 }{ 2 } \left| \begin{matrix} a _{ 1 } & b _{ 1 } & 1 \ a _{ 2 } & b _{ 2 } & 1 \ a _{ 3 } & b _{ 3 } & 1 \end{matrix} \right| $

$A _{2}=\dfrac { 1 }{ 2 } \left| \begin{matrix} a _{ 1 } & a _{ 2 } & a _{ 3 } \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ 1 & 1 & 1 \end{matrix} \right|    (\because |A|=|A^{T}|)$


$A _{2}=-\dfrac{1}{2}\left| \begin{matrix} 1 & 1 & 1 \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ a _{ 1 } & a _{ 2 } & a _{ 3 } \end{matrix} \right| $

Since, area is positive,

$A _{2}=\dfrac{1}{2}\left| \begin{matrix} 1 & 1 & 1 \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ a _{ 1 } & a _{ 2 } & a _{ 3 } \end{matrix} \right| $

Given, $\left| \begin{matrix} x _{ 1 } & y _{ 1 } & 1 \ x _{ 2 } & y _{ 2 } & 1 \ x _{ 3 } & y _{ 3 } & 1 \end{matrix} \right| =\left| \begin{matrix} 1 & 1 & 1 \ b _{ 1 } & b _{ 2 } & b _{ 3 } \ a _{ 1 } & a _{ 2 } & a _{ 3 } \end{matrix} \right| $

$\Rightarrow A _{1}=A _{2}$