Tag: triangle inequality

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Questions Related to triangle inequality

Multiple choice maths construction of parallel lines and triangles triangle inequality inequalities in triangle inequalities in triangles

The points $\left( 0,\dfrac { 8 }{ 3 }  \right),(1,3)$ and $(82,30)$ are the vertices of:

  1. an equilateral triangle

  2. an isosceles triangle

  3. a right angled triangle

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to the problem :

$AB^2=(0-1)^2+(\dfrac{8}{3}-3)^2$
$=1+\dfrac{1}{9}=\dfrac{10}{9}=1.11$

Similarly,
$BC^2=(82-1)^2+(30-3)^2=7290$
and
$AC^2=(82-0)^2+(30-\dfrac{8}{3})^2=7471.11$

Therefore,
$AB^2+BC^2<AC^2$

Hence the answer is acute-angled triangle.

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 } $