Tag: congruence and inequalities of triangles

Questions Related to congruence and inequalities of triangles

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. $9, 20, 8$

Show answer
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. $27$

Show answer
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. $CD > CA$

Show answer
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

Show answer
Correct Option: B
Explanation:

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