Tag: inequalities in triangle

Questions Related to inequalities in triangle


 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

If $|{z _1}| = |{z _2}| = |{z _3}| = 1$ and ${z _1} + {z _2} + {z _3} = 0$ then the area of the triangle whose vertices are $z _1, z _2, z _3$ is

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

  1. $\frac{3\sqrt{3}}{4}$

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

  3. 1

  4. None of these

Show answer
Correct Option: A
Explanation:

given  $|Z _1|=|Z _2|=|Z _3|=1$     and     $Z _1+Z _2+Z _3=0$
$\Rightarrow |Z _1 -Z _2|=2 (Cos 30)=\sqrt{3}$
$\Rightarrow  area =\frac{\sqrt{3}}{4}a^2$    &     $ a=\sqrt{3}$
So, area $=\frac{\sqrt{3}}{4}\cdot 3 =\frac{3\sqrt{3}}{4}$

Statement 1: $|z _1-a| < a, |z _2-b| < b, |z _3-c| < c$, where a, b, c are positive real numbers, then $|z _1+z _2+z _3|$ is greater than $2|a+b+c|$.
Statement 2: $|z _1\pm z _2| \leq |z _1|+|z _2|$.

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

  1. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.

  2. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

  3. Statement 1 is true and Statement 2 is false.

  4. Statement 1 is false and Statement 2 is true.

Show answer
Correct Option: D
Explanation:

$|z _1+z _2+z _3| = |z _1-a+z _2-b+z _3-c+(a+b+c)|$
$\leq |z _1-a|+|z _2-b|+|z _3-c|+|a+b+c|$
$\leq 2|a+b+c| $
Hence, $|z _1+z _2+z _3|$ is less than $2|a+b+c|$.
Statement 1 is false and Statement 2 is true

$z _0$ is a root of the equation $z^n cos \theta _o+z^{n-1} cos\theta _1+....+z cos\theta _{n-1}+cos\theta _n=2$, where $\theta, \epsilon R$, then

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

  1. $|z _0| > 1$

  2. $|z _0| > \dfrac {1}{2}$

  3. $|z _0| > \dfrac {1}{4}$

  4. $|z _0| > \dfrac {3}{2}$

Show answer
Correct Option: A,B
Explanation:

$z^n cos\theta _0+z^{n-1} cos\theta _1+.....+z cos\theta _{n-1}+cos\theta _n=2$

or $2=|z _0^n cos\theta _0+z _0^{n-1} cos\theta _1+....+z _0 cos\theta _{n-1}+cos\theta _n|$

or $2\leq |z _0|^n |cos\theta _0|+|z|^{n-1}|cos\theta _1|+....+|z _0||cos\theta _{n-1}|+|cos\theta _n|$

or $2\leq |z _0|^n+|z _0|^{n-1}+|z _0|^{n-2}+.....+|z _0|+1$

which is clearly satisfied for $|z _0| \geq 1$. If $|z _0| < 1$, then

$2 < 1+|z _0|+|z _0|^2+.....+|z|^n+....\infty$

$\Rightarrow 2 < \dfrac {1}{1-|z _0|}$

$\Rightarrow |z _0| > \dfrac {1}{2}$