Tag: congruency of triangles

Questions Related to congruency of triangles

$\begin{array} { l } { \text { If } z _ { 1 } \text { and } z _ { 2 } \text { are complex numbers, then } \left| z _ { 1 } + z _ { 2 } \right| ^ { 2 } = \left| z _ { 1 } \right| ^ { 2 } + \left| z _ { 2 } \right| ^ { 2 } \text { if and only if } z _ { 1 } \overline { z } _ { 2 } \text { is } } \ { \text { purely imaginary. } } \end{array}$

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. True

  2. False

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Correct Option: A

If $|z| < \sqrt 2-1$, then $|z^2+2z cos\alpha|$ is

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. Less than 1

  2. $\sqrt 2+1$

  3. $\sqrt 2-1$

  4. None of these

Show answer
Correct Option: A
Explanation:

$|z^2+2z cs \alpha| \leq |z^2|+|2z cos \alpha|$
$=|z^2|+|2z^2| |cos\alpha|$
$\leq |z|^2 + 2 |z| $
$ < (\sqrt 2-1)^2 + 2(\sqrt 2-1) =1$

If $P$ and $Q$ are represented by complex numbers $z _{1}$ and $z _{2}$ such that $\left| \dfrac { 1 }{ { z } _{ 1 } } +\dfrac { 1 }{ { z } _{ 2 } }  \right| =\left| \dfrac { 1 }{ { z } _{ 1 } } -\dfrac { 1 }{ { z } _{ 2 } }  \right| $ then the circumference of $\triangleOPQ(O is origin)$ is

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $\dfrac{{ z } _{ 1 } -{ z } _{ 2 } }{2}$

  2. $\dfrac{{ z } _{ 1 } +{ z } _{ 2 } }{2}$

  3. $\dfrac{{ z } _{ 1 } +{ z } _{ 2 } }{3}$

  4. ${ z } _{ 1 } +{ z } _{ 2 } $

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Correct Option: A

The sum of all sides of a quadrilateral is lessthan the sum of its diagonals.

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. True

  2. False

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Correct Option: B

If $\left| {z - 1} \right| + \left| {z + 3} \right| \le 8$ then the range of values of $\left| {z - 4} \right|$

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $[1,\,7]$

  2. $[1,\,8]$

  3. $[1,\,9]$

  4. $[2,\,5]$

Show answer
Correct Option: A
Explanation:

$|z-1|+|z+3|\le 8$

Using the triangle inequality 
$|z _1\pm z _2|\le |z _1|+|z _2|$
We have $|z-1|+|z+3|\le 8$
$\implies |z-1+z+3|\le 8$
$\implies |z+1|\le 4$
Using triangle inequality again 
$|z|+1\le 4\implies |z|\le 3$
So, the maximum value of $|z _1+ _2|$ is $|z _1|+|z _2|$
And  the minimum value of $|z _1+ _2|$ is $|z _1|-|z _2|$
Hence the maximum value of $|z-4|$ is $|z|+|4|=3+4=7$
 the minimum value of $|z-4|$ is $|z|-|4|=3-4=-1$
Hence the range of $|z-4|$
$1\le|z-4|\le 7$

If $\left|z\right| <\sqrt{2} -1$, then $\left|z^2 + 2 z  cos  \alpha \right|$ is

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. less than 1

  2. $\sqrt{2} + 1$

  3. $\sqrt{2} -1$

  4. none of these

Show answer
Correct Option: A
Explanation:

$\left| z \right| <\sqrt { 2 } -1\ \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| \le \left| { z }^{ 2 } \right|+ \left| 2z\cos { \alpha  }  \right| \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ...\left{ \quad \because \quad \left| { z } _{ 1 }+{ z } _{ 2 } \right| \le \left| { z } _{ 1 } \right| +\left| { z } _{ 2 } \right|  \right} \ \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| \le |z|(|z|+2) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ...\left{ \quad \because \quad \left| \cos { \alpha  }  \right| \le 1\quad  \right} \ \Rightarrow \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| <(\sqrt{2}-1){ \left( \sqrt { 2 } +1 \right)  }<1\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ...\left{ \quad \because \quad \left| z \right| <\sqrt { 2 } -1\quad  \right} \ \therefore \quad \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| <1\ $
Hence, option 'A' is correct.

If z be a complex number for which $|2z  cos  \theta + z^2| = 1$, then the minimum value of |z|
 is ......................

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $\sqrt{3} -1$

  2. $\sqrt{3} +1$

  3. $\sqrt{2} -1$

  4. $\sqrt{2} +1$

Show answer
Correct Option: C
Explanation:

$|z^{2}+2zcos\theta|$
$=|z(z+2cos\theta)|$
$=|z|.|z+2cos\theta|$
$=1$
Now 
$|z|=1$ and 
$|z+2cos\theta|=1$
Now 
$|z+2cos\theta|\leq |z|+|2cos\theta|$
Considering 
$|z+2\cos\theta|=|z|+|2cos\theta|=1$
Hence
$|z|=|2cos\theta|\pm1$
Considering $z=|2cos\theta|-1$ we get the minimum value at multiples of $\theta=45^{0}$
Hence
$z=\sqrt{2}-1$.

$sin^{-1}\left { \frac{1}{i} (z-1)\right }$ ,Where Z is non - real, can be the angle  of a triangle, if 

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $Re(z)=1, Im(z)=2$

  2. $Re(z)=1,-1\leq Im(z)\leq 1$

  3. $Re(z)=1,Im(z)=0$

  4. $Re(z)=1,Im(z)=-2$

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Correct Option: A

Let $z$ be any point in $\displaystyle A\cap B\cap C$ and let $w$ be any point satisfying $\displaystyle \left | w-2-i \right |< 3.$ Then, $\displaystyle \left | z \right |-\left | w \right |+3$ lies between

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $-6$ and $3$

  2. $-3$ and $6$

  3. $-6$ and $6$

  4. $-3$ and $9$

Show answer
Correct Option: D
Explanation:

$\left| w-\left( 2+i \right)  \right| <3\Rightarrow \left| w \right| -\left| 2+i \right| <3\ \Rightarrow -3+\sqrt { 5 } <\left| w \right| <3+\sqrt { 5 } $
$\Rightarrow -3-\sqrt { 5 } <-\left| w \right| <3-\sqrt { 5 } $   ...(1)
Also, $\left| z-\left( 2+i \right)  \right| =3$
$\Rightarrow -3+\sqrt { 5 } <-\left| z \right| \le 3+\sqrt { 5 } $   ...(2)
$\therefore -3<\left| z \right| -\left| w \right| +3<9$

If $z=a+ib$ where $a>0,b>0$, then

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $\displaystyle \left| z \right| \ge \frac { 1 }{ \sqrt { 2 }  } \left( a-b \right) $

  2. $\displaystyle \left| z \right| \ge \frac { 1 }{ \sqrt { 2 }  } \left( a+b \right) $

  3. $\displaystyle \left| z \right| < \frac { 1 }{ \sqrt { 2 }  } \left( a+b \right) $

  4. None of these

Show answer
Correct Option: B
Explanation:

As ${ \left( a-b \right)  }^{ 2 }\ge 0,{ a }^{ 2 }+{ b }^{ 2 }\ge 2ab$   ...(1)

But $\left| z \right| =\sqrt { { a }^{ 2 }+{ b }^{ 2 } } ;$ si from (1), ${ \left| z \right|  }^{ 2 }\ge 2ab$
$\therefore { \left| z \right|  }^{ 2 }+{ a }^{ 2 }+{ b }^{ 2 }\ge { a }^{ 2 }+{ b }^{ 2 }+2ab\ \Rightarrow { \left| z \right|  }^{ 2 }+{ \left| z \right|  }^{ 2 }\ge { \left( a+b \right)  }^{ 2 }\Rightarrow 2{ \left| z \right|  }^{ 2 }\ge { \left( a+b \right)  }^{ 2 }$
$\Rightarrow \sqrt { 2 } \left| z \right| \ge a+b$ as $\left| z \right| $ is positive
$\displaystyle \left| z \right| \ge \frac { 1 }{ \sqrt { 2 }  } \left( a+b \right) $