Tag: congruency of triangles

Quizzes (5) Questions (48)

Questions Related to congruency of triangles

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


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

  1. $\geq||z _{1}|-|z _{2}||$
  2. $\leq|z _{1}|-|z _{2}|$
  3. $=|\mathrm{z} _{1}|-|\mathrm{z} _{2}|$
  4. $\geq|\mathrm{z} _{1}|-|\mathrm{z} _{2}|$
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $+1\geq cos(\theta _{1}-\theta _{2})\geq -1$


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


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


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


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


 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

  1. $8$
  2. $18$
  3. $10$
  4. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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.

  1. 15, 6, 11

  2. 9, 6, 4

  3. 13, 5, 8

  4. 25, 7, 24
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The area of triangles formed by a point inside a parallelogram with vertices A, B, C, D follows specific geometric constraints. The sum of areas of opposite triangles (AMD and BMC) equals half the area of the parallelogram, and the sum of the other two (AMB and DMC) also equals half the area.

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


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

  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
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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 

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

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

  1. 20

  2. 25

  3. 30

  4. 35
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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


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

  1. $\sqrt{7}-1$
  2. $\sqrt{7}+1$
  3. $\sqrt{7}$
  4. $\displaystyle \frac{\sqrt{7}}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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

  1. $0$
  2. $4$
  3. $10$
  4. $6$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The condition |z + 4| <= 3 describes a disk centered at -4 with radius 3. We want to maximize |z - (-1)|, which is the distance from z to -1. The maximum distance from a point in the disk to -1 occurs at the point furthest from -1, which is -4 - 3 = -7. The distance from -7 to -1 is |-7 - (-1)| = 6.

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

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

  1. $2, 1$
  2. $6, 5$
  3. $4, 3$
  4. $7, 3$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

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

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

  3. 1

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $=\frac{\sqrt{3}}{4}\cdot 3 =\frac{3\sqrt{3}}{4}$

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

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: $|z _1\pm z _2| \leq |z _1|+|z _2|$.

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