Tag: triangle inequality

Questions Related to triangle inequality

O is a point that lies in the interior of $\Delta ABC$. Then $2(OA - OB -OC) > \text{Perimeter}\ of\ \Delta ABC$.

  1. True

  2. False


Correct Option: B
Explanation:
From the $\triangle ABC,$ by triangle inequality,
$ OA+OB>AB$ ....... $(i)$
$ OB+OC>BC$ ........ $(ii)$
$ OA+OC>AC$ ........ $(iii)$
By adding $(i),(ii)$ and $(iii)$
$ 2(OA+OB+OC)>AB+BC+AC$
$ \therefore 2(OA+OB+OC)>\text{Perimeter of triangle } ABC$
Hence, the statement is false.

Sum of the length of any two sides of a triangle is always greater than the length of third side.

  1. True

  2. False


Correct Option: A
$\dfrac{\sin 2x}{2\cos x}=\tan x \ \ ?$
  1. True

  2. False


Correct Option: A
Explanation:
LHS

$\dfrac{\sin 2x}{2\cos x}$

$\dfrac{2\sin x \cos x}{2\cos x}$

$\implies \dfrac{\sin x}{\cos x}$

$\implies \tan x $

Hence proved.

If $O$ is any point in the interior of $\Delta ABC$. then "$2(OA+OB+OC)=(AB+BC+CA)$" the statement  is?

  1. True

  2. False


Correct Option: A

In $\triangle {ABC},ABC,APQ$ and $\overline { PQ } \parallel \overline { BC } $. If $PQ=5,AP=4,AB=12$, then $BC=$_____

  1. $9.6$

  2. $20$

  3. $15$

  4. $5$


Correct Option: C

If a,b,c are the sides of a triangle ABC, then $\sqrt{a} + \sqrt{b} - \sqrt{c} $  is always:

  1. negative

  2. Positive

  3. non - negative

  4. non - positive


Correct Option: B
Explanation:

If $a,b$ and $c$ are the sides of the triangle then $\sqrt a  + \sqrt b  - \sqrt c $ it always positive because the sum of two sides of the triangle is always greater than the third side.

 

In a $\Delta ABC$, side AB has the equation $2x+3y=29$ and the side AC has the equation, $x+2y=6$. If the mid-point of BC is (5, 6), then the equation of BC is

  1. $x-y=-1$

  2. $5x-2y=13$

  3. $21x+31y=291$

  4. $3x-4y=-9$


Correct Option: A

In a triangle ABC , if AB , BC and AC are the three sides of the triangle , then which of the following statements is necessarily true ?

  1. $\displaystyle AB + BC < AC$

  2. $\displaystyle AB + BC > AC$

  3. $\displaystyle AB + BC = AC$

  4. $\displaystyle AB^2 + BC^2 = AC^2$


Correct Option: B
Explanation:
The sum of any two sides of a triangle is greater than the third side .

In $\triangle ABC, AB, BC$ and $AC$ are the three sides ,

Now ,

$AB + BC > AC$

For a triangle ABC, which of the following is true?

  1. ${ BC }^{ 2 }-{ AB }^{ 2 }={ AC }^{ 2 }$

  2. $AB-\,AC=BC$

  3. $(AB-\,AC)>BC$

  4. $ AB - AC< BC$


Correct Option: D
Explanation:

For any $\triangle ABC$,
the difference between two sides of the triangle should be less than the third side.
Thus, $AB - AC < BC$

In $\triangle PQR$, if $\angle R\displaystyle>\angle Q$, then

  1. $QR>PR$

  2. $PQ>PR$

  3. $PQ

  4. $QR


Correct Option: B
Explanation:

In $\triangle PQR$, $\angle R > \angle Q$
Hence, $PQ > PR$ (Sides opposite larger angles is large)

option B is correct