Tag: triangle inequality

Questions Related to triangle inequality

All equilateral triangles are ____.

maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

  1. congruent

  2. will be congruent if their length of sides are equal

  3. never congruent

  4. none of these

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Correct Option: B
Explanation:

Equilateral triangles have all the three angles equal to $\displaystyle { 60 }^{ o }$. 

But the length of sides can be different for different equilateral triangles. 
Thus if the length is also fixed, then all the equilateral triangles having sides lengths same are congruent.

Find the length of the third side of the triangle inequality, if sides of a triangle have $a = 4$ and $b = 8$.

maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

  1. $c = 10$

  2. $c = 2$

  3. $c = 3$

  4. $c = 4$

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Correct Option: A
Explanation:

The Triangle Inequality theorem states that the sum of any $2$ sides of a triangle must be greater than the measure of the third side.
Take option A: $a = 4, b = 8, c = 10$
$4 + 8 > 10 (a + b > c)$
$8 + 10 > 4 (b + c > a)$
$4 + 10 < 8 (a + c > b)  $
Therefore, the third side, $c = 10$ will satisfy the triangle inequality
Similarly check for option 2: $c = 2$
$4 + 8 > 2 (a + b > c)$
$8 + 2 > 10 (b + c = a)$
$4 + 2 < 8 (a + c < b)  $
Hence, the second option will not satisfying the triangle inequality.
Similarly check for option $3$ and $4$.
Here, the third side of the triangle inequality, $c = 10$.

In triangle $PQR$, an exterior angle at $A$ measures $160^o$, and $\angle$ $Q = 70 ^o$. Which is the longest side of the triangle?

maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

  1. $\overline{PR}$

  2. $\overline{PA}$

  3. $\overline{PQ}$

  4. $\overline{QR}$

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Correct Option: C
Explanation:

Exterior angle $A = 180 -$ $\angle$ $P = 180-160$
$\angle$ $P = 20$
$\angle$ $Q = 70$
Interior angle $= 20 + 70 +$ $\angle$ $R = 180$
$\angle$ $R = 90$
In a triangle inequality theorem,the largest side is across from the longest angle.
So, $90^o$ is the longest angle in the triangle, $\overline{PQ}$, across from it, is the largest side.
Therefore, $\overline{PQ}$ is the largest side.

Mark the triplet that can be the lengths of the sides of a triangle.

maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

  1. $2, 3, 5$

  2. $1, 4, 2$

  3. $7, 4, 4$

  4. $5, 6, 12$

  5. $9, 20, 8$

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Correct Option: C
Explanation:

for triplet to be the length of triangle, it must satisfy triangle property.

sum of any two sides must be greater than third side
So, $7,4,4$ is correct answer.
If we consider remaining options, observe that sum of two sides is less than third side, so they cannot form triangle.

Two sides of a triangle have lengths $5$ and $8$, and the length of the third side is an integer. What is the greatest possible value of the perimeter of the triangle?

maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

  1. $22$

  2. $24$

  3. $25$

  4. $26$

  5. $27$

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Correct Option: C
Explanation:

The third side rule says that the length of the third side of the triangle in this case must be less than $5+8=13$
Since the length of the third side is an integer, the length must be $12$. 

The biggest possible perimeter is then $5+8+12=25$
As we know, perimeter of a triangle is $=1^{st}$ side $+2^{nd}$ side $+ 3^{rd}$ side.

In a triangle $ABC$, $(a+b+c)(b+c-a)=k$$bc$ if:

maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

  1. $k< 0$

  2. $k> 6$

  3. $0< k< 4$

  4. $k> 4$

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Correct Option: C

$D$ is a point on the side $BC$ of a $\Delta$ $ABC$, such that $AD$ bisects $\angle $ $BAC$. Then:

maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

  1. $BA = CD$

  2. $BA > BD$

  3. $BD > BA$

  4. $CD > CA$

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Correct Option: B
Explanation:

The definition of the angle bisector of a triangle is a line segment that bisects one of the vertex angles of a triangle. In general, an angle bisector is equidistant from the sides of the angle when measured along a segment perpendicular to the sides of the angle 

Hence, by definition,$BD+DC=BC$
Therefore, $BA>BD$

Let a,b,c be the sides of a triangle. No two of them are equal and $\lambda  \in R.$ If the roots of the equation 

maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

  1. $\lambda < \frac{4}{3}$

  2. $\lambda < \frac{5}{3}$

  3. $\lambda \in \left( {\frac{1}{3},\frac{5}{3}} \right)$

  4. $\lambda \in \left( {\frac{4}{3},\frac{5}{3}} \right)$

Show answer
Correct Option: A
Explanation:
The question must be Let $a,b,c$ be the sides of a triangle.No two of them are equal and $\lambda\in R$. If the roots of the eqn ${x}^{2}+2\left(a+b+c\right)x + 3\lambda\left(ab+bc+ca\right) = 0$ are real, then $\lambda\in\,R$

Given, roots of equation ${x}^{2}+2\left(a+b+c\right)x + 3\lambda\left(ab+bc+ca\right) = 0$ are real,
So,$D\ge\,0$

$\Rightarrow\,{\left[2\left(a+b+c\right)\right]}^{2}-4\times 1\times 3\lambda\left(ab+bc+ca\right)\ge 0$

$\Rightarrow\,4{\left(a+b+c\right)}^{2}-12\lambda\left(ab+bc+ca\right)\ge 0$

$\Rightarrow\,4{\left(a+b+c\right)}^{2}\ge\,12\lambda\left(ab+bc+ca\right)$

$\Rightarrow\,\lambda\le\dfrac{4{\left(a+b+c\right)}^{2}}{12\lambda\left(ab+bc+ca\right)}$

$\Rightarrow\,\lambda\le\,\dfrac{4\left({a}^{2}+{b}^{2}+{c}^{2}\right)}{12\lambda\left(ab+bc+ca\right)}+\dfrac{8\lambda\left(ab+bc+ca\right)}{12\lambda\left(ab+bc+ca\right)}$

$\Rightarrow\,\lambda\le\,\dfrac{4\left({a}^{2}+{b}^{2}+{c}^{2}\right)}{12\lambda\left(ab+bc+ca\right)}+\dfrac{8}{12}$

$\Rightarrow\,\lambda\le\,\dfrac{4\left({a}^{2}+{b}^{2}+{c}^{2}\right)}{12\lambda\left(ab+bc+ca\right)}+\dfrac{2}{3}$      .......$(1)$

Now, we know that,

$\left|a-b\right|<c\Rightarrow\,{a}^{2}+{b}^{2}-2ab<{c}^{2}$

$\left|b-c\right|<a\Rightarrow\,{b}^{2}+{c}^{2}-2bc<{a}^{2}$

$\left|c-a\right|<c\Rightarrow\,{c}^{2}+{a}^{2}-2ac<{b}^{2}$

On adding,

${a}^{2}+{b}^{2}-2ab+{b}^{2}+{c}^{2}-2bc+{c}^{2}+{a}^{2}-2ac<{a}^{2}+{b}^{2}+{c}^{2}$

${a}^{2}+{b}^{2}+{c}^{2}<2ab+2bc+2ca$

$\Rightarrow\,dfrac{{a}^{2}+{b}^{2}+{c}^{2}}{ab+bc+ca}<2$

So, eqn$(1)$ becomes,

$\lambda<\dfrac{2}{3}+\dfrac{2}{3}$

$\therefore\,\lambda<\dfrac{4}{3}$

The area of the triangle formed by the lines  $x ^ { 2 } - 3 x y + y ^ { 2 } = 0$  and  $x + y + 1 = 0$  is square units. is

maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

  1. $\dfrac {1}{12}$

  2. $\dfrac { 1 } { 2 \sqrt { 5 } }$

  3. $\dfrac { 2 } { \sqrt { 3 } }$

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

Show answer
Correct Option: A
Explanation:
${ x }^{ 2 }-3xy+2{ y }^{ 2 }=0$
$\Rightarrow \left( x-y \right) \left( x-2y \right) =0$
Hence three sides are
$x-y=0\quad \longrightarrow \left( i \right) $
$x-2y=0\quad \longrightarrow \left( ii \right) $
$x+y+1=0\quad \longrightarrow \left( iii \right) $
Solving $(i)$, $(ii)$ & $(iii)$
three vertices are $A\left( 0,0 \right) ,\quad B\left( \dfrac { -2 }{ 3 } ,\dfrac { -1 }{ 3 }  \right) ,\quad C\left( \dfrac { -1 }{ 2 } ,\dfrac { -1 }{ 2 }  \right) $
$AB=\sqrt { \dfrac { 1 }{ 9 } +\dfrac { 4 }{ 9 }  } =\sqrt { \dfrac { 5 }{ 9 }  } =\dfrac { \sqrt { 5 }  }{ 3 } $
$BC=\sqrt { \dfrac { 1 }{ 36 } +\dfrac { 1 }{ 36 }  } =\dfrac { \sqrt { 2 }  }{ 6 } $
$AC=\sqrt { \dfrac { 1 }{ 4 } +\dfrac { 1 }{ 4 }  } =\dfrac { 1 }{ \sqrt { 2 }  } =\dfrac { \sqrt { 2 }  }{ 2 } $
$\therefore$   area using heron's formula
$\Delta =\sqrt { S\left( S-AB \right) \left( S-BC \right) \left( S-CA \right)  } $
    $=\dfrac { 1 }{ 12 } { unit }^{ 2 }$

In a triangle ABC. The relation which is true for its sides is-

maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

  1. AB + BC < AC

  2. AB + BC > AC

  3. AB +BC = CA

  4. AB = BC + AC

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Correct Option: B
Explanation:

In ABC,
Sum of the two sides is always greater than third side.
So, AB + BC > AC