Tag: inequalities in triangles

Questions Related to inequalities in triangles

The maximum value of |z| where z satisfies the condition $\displaystyle \left | z + \frac{2}{z} \right | = 2$ is

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

  1. $\sqrt{3} -1$

  2. $\sqrt{3} +1$

  3. $\sqrt{3} $

  4. $\sqrt{2} +\sqrt{3} $

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Correct Option: B
Explanation:

Given, $\displaystyle \left | z + \frac{2}{z} \right | = 2   $

$   \Rightarrow |z| - \dfrac{2}{|z|} \leq 2      $
$ \Rightarrow |z|^2 - 2 |z| - 2 \leq 0$
$\Rightarrow |z| \leq \displaystyle \frac{2 \pm \sqrt{4 + 8}}{2} \leq 1 \pm \sqrt 3$
Hence, max. value of |z| is $1 + \sqrt 3$

If $|z| \leq 1$ then the minimum and maximum value of |z - 3| are

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

  1. 4, 2

  2. 3, 4

  3. 4, 6

  4. 2, 4

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Correct Option: D


$|\mathrm{z} _{1}-\mathrm{z} _{2}|=$

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

  1. $\geq||z _{1}|-|z _{2}||$

  2. $\leq|z _{1}|-|z _{2}|$

  3. $=|\mathrm{z} _{1}|-|\mathrm{z} _{2}|$

  4. $\geq|\mathrm{z} _{1}|-|\mathrm{z} _{2}|$

Show answer
Correct Option: A
Explanation:

Let $ argz _1=\theta _1  \quad argz _2=\theta _2$
we know that
$|z _{1}-z _{2}|^{2}=|z _{1}|^{2}+|z _{2}|^{2}-2|z _{1}||z _{2}|cos(\theta _{1}-\theta _{2})$


now, $+1\geq cos(\theta _{1}-\theta _{2})\geq -1$


$-2|z _{1}||z _{2}|cos(\theta _{1}-\theta _{2})\geq -2|z _{1}||z _{2}|$


$\therefore |z _{1}|^{2}+|z _{2}|^{2}-2|z _{1}||z _{2}|cos(\theta _{1}-\theta _{2})\geq |z _{1}|^{2}+|z _{2}|^{2}-2|z _{1}||z _{2}|$


$\therefore |z _{1}-z _{2}|^{2}\geq (|z _{1}|-|z _{2}|)^{2}\Rightarrow |z _{1}-z _{2}|\geq ||z _{1}|-|z _{2}||$



 lf $|\mathrm{z} _{1}|=2,\ |\mathrm{z} _{2}|=3$, then $|\mathrm{z} _{1}+\mathrm{z} _{2}+5+12\mathrm{i}|$ is less than or equal to

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

  1. $8$

  2. $18$

  3. $10$

  4. $5$

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Correct Option: B
Explanation:

We know that $|z _{1}+z _{2}+ z _3 |\leq |z _{1}|+|z _{2}| + | z _3| $
$|z _{1}|+|z _{2}|=5$
$z _3 = 5+12 i $

$|z _3 | = 13 $
$\therefore 18\geq |z _{1}+z _{2}+5+12i|$
Hence, option B is correct

A point M is taken inside a parallelogram ABCD, then area of $\displaystyle \Delta AMD,$ $\displaystyle \Delta AMB,$ $\displaystyle \Delta AMC$ can take which of of the following values, respectively.

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

  1. 15, 6, 11

  2. 9, 6, 4

  3. 13, 5, 8

  4. 25, 7, 24

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Correct Option: C


 Let $z _{1}=24+7i$ and $z _{2}$ be complex number whose magnitude is unity, then

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

  1. Maximum value of $|z _{1}+z _{2}|$ is 26

  2. Maximum value of $|z _{1}+z _{2}|$ is 31

  3. Minimum value of $|z _{1}+z _{2}|$ is 24

  4. Minimum value of $|z _{1}+z _{2}|$ is 19

Show answer
Correct Option: A,C
Explanation:

$ |z _1 | =  25 $

$| z _2 | = 1$

We have, 
$\left| |z _1| - |z _2| \right| \leq |z _1+z _2 | \leq \left | |z _1| + |z _2| \right |$
$\Rightarrow 24 \leq |z _1+z _2| \leq 26$
Hence, options A and C are correct 

The complex number $z$ satisfies the condition $\left|\displaystyle {z}-\frac{25}{z}\right|=24$. Then the maximum distance from the origin to the point '$z$' in the argand plane is

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

  1. 20

  2. 25

  3. 30

  4. 35

Show answer
Correct Option: B
Explanation:

$|z| = |z - \frac{25}{z} + \frac{25}{z}| \leq 24 + \frac{25}{|z|}$
$\therefore$ $|z|^2 - 24|z| - 25 \leq 0.$
$\therefore$ $(|z| - 25)(|z| + 1) \leq 0., \Rightarrow |z| \leq 25.$
Hence maximum distance of z from origin is 25.


If $\left |z-\displaystyle \frac{6}{z}\right|=2$, then the greatest value of $|z|$ is

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

  1. $\sqrt{7}-1$

  2. $\sqrt{7}+1$

  3. $\sqrt{7}$

  4. $\displaystyle \frac{\sqrt{7}}{2}$

Show answer
Correct Option: B
Explanation:

we have,
$|z|=\left |z-\dfrac{6}{z}+\dfrac{6}{z} \right |\leq \left |z-\dfrac{6}{z} \right |+\left |\dfrac{6}{z} \right|$
$|z|\leqslant 2+\left |\dfrac{6}{z} \right|$
$\Rightarrow |z|^{2}-2 |z|-6\leqslant 0$
Hence, 

$ {1 - \sqrt{7}} \leq |z| \leq {1+\sqrt 7} $
$0 < |z| \leq {1+\sqrt 7} $
Hence, option B is correct

If $|z+4|\leq 3$, then the maximum value of $|{z}+1|$ is

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

  1. $0$

  2. $4$

  3. $10$

  4. $6$

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Correct Option: D

A point $'z'$ moves on the curve $|z - 4 - 3i| = 2$ in an argand plane. The maximum and minimum values of $|z|$ are

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

  1. $2, 1$

  2. $6, 5$

  3. $4, 3$

  4. $7, 3$

Show answer
Correct Option: D
Explanation:
Let $w = 4 + 3i$. 
We can write, $|z| = |(z-w) + w|$. Hence by triangle inequality ($|z _1+z _2| \leq |z _1| + |z _2|$), we can write $|z| \leq |z-w| + |w|$. It is given in the question that, $|z-w| = 2$ and $|w| = \sqrt{4^2 + 3^2} = 5$.
Putting the values, we get $|z| \leq 7$. 
Using another result of triangle inequality ($\big||z _1| - |z _2| \big| \leq |z _1 + z _2|$), we can write $\big||z-w| - |w|\big| \leq |z - w + w|$.
Hence, we get $|z| \geq 3$. The minimum value is 3 and maximum value is 7.
Hence, (D) is the correct option