Tag: inequalities in triangle

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

The triangle inequality theorem states that 

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. True

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

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

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

  1. True

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

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

A triangle cannot be drawn with the following three sides:

  1. $2m, 3m, 4m$
  2. $3m, 4m, 8m$
  3. $4m, 6m, 9m$
  4. $5m, 7m, 10m$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.\$

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

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

  1. $25i$
  2. $12+5i$
  3. $16+12i$
  4. $12+16i$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

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

  1. $1$
  2. $\displaystyle \frac {3+\sqrt {21}}{2}$
  3. $\displaystyle \frac {\sqrt {21}-3}{2}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $

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

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

  1. $2$
  2. $4$
  3. $6$
  4. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation |z+i| + |z-i| = 8 represents an ellipse with foci at (0, -1) and (0, 1). The sum of distances to foci is 2a = 8, so a = 4. The center is at (0,0). The maximum distance from the origin is the semi-major axis length, which is 4.

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

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 $|\omega|$?

  1. $|z|=|\omega|=\frac {1}{2}$
  2. $|z|=\frac {1}{2}, |\omega|=\frac {3}{4}$
  3. $|z|=|\omega|=\frac {3}{4}$
  4. $|z|=|\omega|=1$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $

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

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

  1. True

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

|z1 + z2|^2 = (z1 + z2)(conjugate(z1) + conjugate(z2)) = |z1|^2 + |z2|^2 + z1*conjugate(z2) + conjugate(z1)*z2. For this to equal |z1|^2 + |z2|^2, the cross terms must sum to zero: z1*conjugate(z2) + conjugate(z1*conjugate(z2)) = 0. This means 2*Re(z1*conjugate(z2)) = 0, so z1*conjugate(z2) must be purely imaginary.