Tag: triangle inequality

Questions Related to triangle inequality

O if any point in the interior of a triangle ABC. 

then  $2(OA + OB + OC) < (AB + BC + AC)$. that relation is right then true  and otherwise false?

  1. True

  2. False


Correct Option: B

Which of the following will form the sides of a triangle?

  1. $23\,cm,\,17\,cm,\,8\,cm$

  2. $12\,cm,\,10\,cm,\,25\,cm$

  3. $6\,cm,\,7\,cm,\,16\,cm$

  4. $8\,cm,\,7\,cm,\,16\,cm$


Correct Option: A
Explanation:
Triangle Inequality Theorem, states that the sum of two side lengths of a triangle is always greater than the third side. 
$(a)17+8>23$ is  true
Hence the sides with the measure $23\ cm,17\ cm,8\ cm$ will  form the sides of a triangle.
$(b)10+12>25$ is not true
Hence the sides with the measure $10\ cm,12\ cm,25\ cm$ will not form the sides of a triangle.
$(c)6+7>16$ is not true
Hence the sides with the measure $6\ cm,7\ cm,16\ cm$ will not form the sides of a triangle.
$(d)8+7>16$ is not true
Hence the sides with the measure $8\ cm,7\ cm,16\ cm$ will not form the sides of a triangle.

The length  of altitude through $A$ of the triangle $ABC$, where $A = (-3,\ 0);\ B = (4,\ -1);\ C = (5,\ 2).$

  1. $\dfrac{2}{\sqrt{10}}$

  2. $\dfrac{4}{\sqrt{10}}$

  3. $\dfrac{11}{\sqrt{10}}$

  4. $\dfrac{22}{\sqrt{10}}$


Correct Option: D
Explanation:
Equation of line BC 
$\frac { y-2 }{ x-5 } =\frac { 2-(-1) }{ 5-4 } \\ \Rightarrow { (y-2) }\times { (5-4) }={ (x-5) }\times { (2+1) }\\ \Rightarrow { 3x }-{ y }-{ 13 }={ 0 }$
Length of altitude from point A to the line BC
$\\ =\frac { \left| 3\times (-3)-1\times 0-13 \right|  }{ \sqrt { { 3 }^{ 2 }+{ (-1) }^{ 2 } }  } \\ =\frac { \left| -9-13 \right|  }{ \sqrt { 10 }  } \\ =\frac { 22 }{ \sqrt { 10 }  } $
Corrct answer is D

In triangle ABC, (a +b -c )(b+c -a)(c+a-b)-abc is always ;

  1. non negative

  2. non positive

  3. negative

  4. positive


Correct Option: A

In a triangle $ABC, \angle C=90^{o}, a=3, b=4$ and $D$ is a point on $AB$, so that $\angle BCD=30^{o}$. Then the length of $CD$ is 

  1. $\dfrac{18-24\sqrt{3}}{25}$

  2. $\dfrac{18+24\sqrt{3}}{25}$

  3. $\dfrac{8-24\sqrt{3}}{25}$

  4. $None\ of\ these$


Correct Option: D
Explanation:
Area of $\triangle BCD$ + Area of $\triangle ACD$ = Area of $\triangle ABC$

$\dfrac{1}{2}\cdot 3CD\cdot \sin 30+\dfrac{1}{2}\cdot 4CD\cdot \sin 60=\dfrac{1}{2}3\times 4$

$3CD\times \dfrac{1}{2}+4CD \times \dfrac{\sqrt{3}}{2}=12$

$\dfrac{CD}{2}(3+4\sqrt{3})=12$

$\therefore CD=\dfrac{24}{3+4\sqrt{3}}$

In a triangle $ABC, \angle ABC=50^{o}, \angle BAC=30^{o}$, then the shortest sides is 

  1. $AB$

  2. $BC$

  3. $CA$

  4. $None$


Correct Option: B
Explanation:
The side opposite to the largest angle is the longest side of the triangle and the side opposite to the smallest angle is the shortest side of the triangle.
By angle sum property,$ \angle{A}+\angle{B}+\angle{C}={180}^{\circ}$
$\Rightarrow {30}^{\circ}+{50}^{\circ}+\angle{C}={180}^{\circ}$
$\Rightarrow \angle{C}={180}^{\circ}-{80}^{\circ}={100}^{\circ}$
The side opposite to the smallest angle $\left({30}^{\circ}\right)$ is the side $BC$

If two sides of a triangle are 8 cm and 13 cm, then the length of the third side is between a cm and b cm. Find the values of a and b such  that a is less than b.

  1. a = 2 cm and b =9 cm

  2. a = 7 cm and b = 12 cm

  3. a = 5 cm and b = 21 cm

  4. a = 8 cm and b = 24 cm


Correct Option: C
Explanation:

$If  \ two \  sides  \ of \  a \  triangle \  are \  8 cm \  and \  13 cm.$
$Then  \ the  \ length \  of  \  third \  side  \  is$
$more \  than \  13-8 \ =  5cm$
$and  \ less \  than \  13+8  =  21cm$
$a = 5cm , b = 21cm$

Two sides of a triangle are $7$ and $10$ units, which of the following length can be the length of the third side?

  1. $19$ units

  2. $17$ units

  3. $13$ units

  4. $3$ units


Correct Option: C
Explanation:
The sum of any two sides is greater than the third side.
and difference between two sides should be lesser than the third side.
For option $A,$ $7+10\ngtr 19$ 
For option $B,$ $7+10\ngtr 17$ 
For option $C,$ $7+10 > 13$ 
Also, $ |10-7| < 13$
For option $D,$ $7+10 > 3$ 
but $|7-10| \nless 3$
Only, $C$ is correct.

If AB > AC and length of median from C is 4 units, then the length of median from B can be

  1. 2

  2. 3

  3. 4

  4. 5


Correct Option: D
Explanation:

This can be taken as a fact that the median corresponding to the longer side has the larger median.

Or you can even take a rough triangle to deduce the result.

If $k$ is an integer and $2 < k < 7$, for how many different values of $k$ is there a triangle with sides of lengths $2, 7$, and $k$?

  1. One

  2. Two

  3. Three

  4. Four

  5. Five


Correct Option: A
Explanation:
In a triangle, the sum of the smaller two sides must be larger than the largest side.
For $k$ values $3, 4, 5$, and $6$, the only triangle possible is $2, 7$, and $k = 6$ because only $2 + 6 > 7$. For $k$ values $3, 4$, and $5$, the sum of the smaller two sides is not larger than the third side; thus, $6$ is the only possible value of $k$ that satisfies the conditions.
The correct answer is A.