Tag: congruence and inequalities of triangles

Quizzes (1) Questions (47)

Questions Related to congruence and inequalities of triangles

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

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

  1. $2, 3, 5$
  2. $1, 4, 2$
  3. $7, 4, 4$
  4. $5, 6, 12$
  5. $9, 20, 8$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in 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?

  1. $22$
  2. $24$
  3. $25$
  4. $26$
  5. $27$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. $k< 0$
  2. $k> 6$
  3. $0< k< 4$
  4. $k> 4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression (a+b+c)(b+c-a) = (b+c)^2 - a^2 = b^2 + c^2 + 2bc - a^2. By the Law of Cosines, a^2 = b^2 + c^2 - 2bc cos(A). Substituting this, we get 2bc + 2bc cos(A) = 2bc(1+cos(A)). Since 0 < 1+cos(A) < 2, the value k must be between 0 and 4.

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

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

  1. $BA = CD$
  2. $BA > BD$
  3. $BD > BA$
  4. $CD > CA$
Reveal answer Fill a bubble to check yourself
B Correct answer
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, $BA>BD$

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

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 

  1. $\lambda < \frac{4}{3}$
  2. $\lambda < \frac{5}{3}$
  3. $\lambda \in \left( {\frac{1}{3},\frac{5}{3}} \right)$
  4. $\lambda \in \left( {\frac{4}{3},\frac{5}{3}} \right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
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,

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

$\therefore\,\lambda<\dfrac{4}{3}$
Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

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

  1. $\dfrac {1}{12}$
  2. $\dfrac { 1 } { 2 \sqrt { 5 } }$
  3. $\dfrac { 2 } { \sqrt { 3 } }$
  4. $\dfrac { \sqrt { 3 } } { 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)  } $
    $=\dfrac { 1 }{ 12 } { unit }^{ 2 }$