Tag: inequalities in triangle

Questions Related to inequalities in triangle

The triangle inequality theorem states that 

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

  1. The sum of the lengths of the $2$ sides of a triangle is equal than the third side of the triangle

  2. The sum of the lengths of the $2$ sides of a triangle is less than the third side of the triangle

  3. The sum of the lengths of the $2$ sides of a triangle is more than the third side of the triangle

  4. None of these

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Correct Option: C
Explanation:

The triangle inequality theorem states that the sum of the lengths of the $2$ sides of a triangle is greater than the third side of the triangle.

State the following statement is True or False
It is possible to have a triangle of sides $3,4,8$

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

  1. True

  2. False

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Correct Option: B
Explanation:

It is not possible to have a triangle of sides $3,4,8$

Since, sum of the $2$ sides ($3$ and $4$) is not greater than the third side that is $8$.
$3+4<8$
According to triangle inequality theorem, it is not possible to construct such triangle.

State the following statement is True or False
The triangle inequality theorem states that the sum of the lengths of the $2$ sides of a triangle is equal than the third side of the triangle

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

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

The triangle inequality theorem states that the sum of the lengths of the $2$ sides of a triangle is greater than the third side of the triangle.

State whether the following statement is True or False.
It is possible to have a triangle of sides $8,10,14$.

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

It is possible to have a triangle of sides $8,10,14$

Since, sum of any $2$ sides of the triangle is greater than the $3^{rd}$ side.
$8+14>10\ 10+14>8\ 10+8>14$

A triangle cannot be drawn with the following three sides:

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

  1. $2m, 3m, 4m$

  2. $3m, 4m, 8m$

  3. $4m, 6m, 9m$

  4. $5m, 7m, 10m$

Show answer
Correct Option: B
Explanation:

A triangle with three sides a,b and c will be possible when:

$a+b>c$
$ b+c>a$
$ a+c>b$
$ Here,a=2,b=3,c=4$
$ 2+3>4$
$ 3+4>2$
$ 2+4>3$
$ \therefore A)is\quad possible.$
$ Here,a=3,b=4,c=8$
$ 3+4=7$
$7<8$
$ \therefore a+b>c\quad is\quad not\quad satisfied.$
$ C)&amp; D)\quad will\quad also\quad be\quad possible.$
$ \therefore B)Correct\quad answer.$

The complex number z having least positive argument which satisfies the condition $|z - 25i| \le 15$   is:

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

  1. $25i$

  2. $12+5i$

  3. $16+12i$

  4. $12+16i$

Show answer
Correct Option: D
Explanation:
Solution:

$|z-25 i| \leq 15$

Let $z= r(cos \theta + i \, sin \theta)$

$\theta$ must be minimum

$| r \, cos \theta +i ( r\, sin \theta-25)|\leq 15$

$|\sqrt{r^2cos^2\theta+r^2 sin^2 \theta+ 625- 50 r \, sin \theta} \,|\leq 15$

square both side

$r^2 (cos^2 \theta+ sin^2 \theta)+625 - 50 r \, sin \theta \leq 225$

$r^250 r \, sin \theta \leq - 400$

$f(r)=\dfrac {400+r^2}{50 \, r}\leq sin \theta $

Find maximum value of $f(r)=\dfrac {400+r^2}{50\, r}$

$f'(r)=\dfrac {100 r^2- 50(400+r^2)}{2500 r^2}=0$

$50 r^2- 50 \times 400=0$

$r= 20$

$f(r=20)=\dfrac {800}{1000}\leq sin \theta $

$\dfrac {4}{5}\leq  sin \theta $

Least value of $sin \theta $ is $4/5$

$ tan \, \theta = 4/3 \,\,\,\,\,\,\,\,\,\, cos \theta  = 3/5$

$z= 20(3/5+4/5 \,i)$

$z= 12+16\, i$

D is correct.

If $|z^2-3|=3|z|$, then the maximum value of |z| is

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

  1. $1$

  2. $\displaystyle \frac {3+\sqrt {21}}{2}$

  3. $\displaystyle \frac {\sqrt {21}-3}{2}$

  4. $None\ of\ these$

Show answer
Correct Option: B
Explanation:

By the law of inequality, 
$|{ z }^{ 2 }-3|\ge { |z| }^{ 2 }-3$
$ \Longrightarrow 3|z|\ge { |z| }^{ 2 }-3\ \Longrightarrow { |z| }^{ 2 }-3|z|-3\le 0\ \Longrightarrow 0\le |z|\le \displaystyle\frac { 3+\sqrt { 21 }  }{ 2 } $
Hence the maximum value of $|z|=\displaystyle\frac { 3+\sqrt { 21 }  }{ 2 } $

If $z$ is a complex number satisfying the equation $\left| z+i \right| +\left| z-i \right| =8$, on the complex plane then maximum value of $\left| z \right| $ is

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

  1. $2$

  2. $4$

  3. $6$

  4. $8$

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Correct Option: B

Suppose z and $\omega$ are two complex numbers such that $|z| \leq 1, |\omega| \leq 1$, and $|z+i\omega|=|z-i\omega|=2$.

Which of the following is true about $|z|$ and $|\omega|$?

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

  1. $|z|=|\omega|=\frac {1}{2}$

  2. $|z|=\frac {1}{2}, |\omega|=\frac {3}{4}$

  3. $|z|=|\omega|=\frac {3}{4}$

  4. $|z|=|\omega|=1$

Show answer
Correct Option: D
Explanation:

We have,


$|z+i\omega|\le |z|+|i\omega|\le |z|+|i||\omega|\le 2$

$|z-i\omega|\le |z|+|-i\omega|\le |z|+|i||-\omega|=|z|+|i||\omega|\le 2$

Also,

$|z+i\omega|=|z-i\omega|$

and

$ |z+i\omega|=2$ 

Hence, 

$|z|  = |\omega| = 1 $

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

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

  1. True

  2. False

Show answer
Correct Option: A