Tag: mathematics and statistics

Quizzes (38) Questions (955)

Questions Related to mathematics and statistics

Multiple choice mathematics and statistics angle and its measurement directed angles

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

  1. $R$
  2. $P$
  3. $O$

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $O$.

Multiple choice mathematics and statistics angle and its measurement directed angles

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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} $.
Multiple choice mathematics and statistics angle and its measurement directed angles

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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°

Multiple choice mathematics and statistics angle and its measurement directed angles

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

  1. One obtuse and one acute angle

  2. One right angle and one acute angle

  3. Two acute angles

  4. Two right angles
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice mathematics and statistics angle and its measurement directed angles

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:

  1. 90- $\frac{1}{2} \angle A$
  2. 90+ $\frac{1}{2} \angle A$
  3. $120^{\circ}$ + $\frac{1}{2} \angle A$
  4. 120 - $\frac{1}{2} \angle A$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)
$\angle BOC = 180^{circ}-  (\angle 1 + \angle 2)$
=$180^{circ}- [180^{circ}-\dfrac{1}{2}$ ($\angle B + \angle C$)]
=$\dfrac{1}{2}[\angle B + \angle C] = \dfrac{1}{2} (180^{circ} - \angle A) = 90^{circ}-\dfrac{1}{2} \angle A$

Multiple choice mathematics and statistics angle and its measurement directed angles

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

  1. $-90^0, +90^0$
  2. $+180^0, -180^0$
  3. $+360^0, -360^0$
  4. $-270^0, +270^0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A half turn is a rotation of 180 degrees. In either clockwise or counter-clockwise direction, this is represented as +180 or -180 degrees.

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 $\begin{vmatrix} x _1 & y _1 & 1 \ x _2 & y _2 & 1 \ x _3 & y _3 & 1\end{vmatrix}=\begin{vmatrix} a _1 & b _1 & 1\ a _2 & b _2 & 1 \ a _3 & b _3 & 1\end{vmatrix}$, 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.

  1. True

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

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

  1. $-5$ or $35$
  2. $5$ or $-35$
  3. $15$ or $-5$
  4. $-5$ or $-25$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $-35$
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 $\displaystyle \left | \begin{matrix}x _{1} &y _{1}  &1 \ x _{2} &y _{2}  &1 \ x _{3} &y _{3}  &1 \end{matrix} \right |=\left | \begin{matrix}1 &1  &1 \ b _{1} &b _{2}  &b _{3} \ a _{1} &a _{2}  &a _{3}\end{matrix} \right |$ 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

  1. congruent

  2. similar

  3. equal in area

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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| $

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