Tag: inequalities in triangle

Questions Related to inequalities in triangle

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

The perimeter of a triangle is $.........$ than the sum of its medians.

  1. Greater

  2. Lesser

  3. Equal

  4. May be greater or lesser


Correct Option: A
Explanation:

Given: $\triangle ABC$, AD, BE and CF are medians from A, B and C respectively on the corresponding sides.

We know that sum of any two sides of the triangle is greater than twice the median bisecting the third side 
Hence, $AB + AC > 2 AD$ (1) 
$AB + BC > 2 BE$ (2)
$BC + AC > 2 CF$ (3)
Adding the three equations:
$2 (AB + BC + AC) > 2 (AD + BE + CF)$
$AB + BC + AC > AD + BE + CF$
Hence, the perimeter of the triangle is greater than the sum of the medians.

In a right triangle, the hypotenuse is the $.......$ side

  1. Largest

  2. Shortest

  3. Can be largest or shortest

  4. None of these


Correct Option: A
Explanation:

In a right triangle, using Pythagoras theorem,
$Hypotenuse^2 = perpendicular^2 + base^2$
Thus, Hypotenuse is the largest side.

The sum of the three altitudes of a triangle is $......$ than its perimeter

  1. Less

  2. More

  3. Equal

  4. Less than or more than


Correct Option: A
Explanation:

Given: $\triangle ABC$, AD, BE and CF are perpendiculars from A, B and C respenctively on the corresponding sides.

Now, In $\triangle ADB$,
$AD < AB$ (Hypotenuse is the longest side)
Similarly in $\triangle ADC$,
$AD < AC$ (Hypotenuse is the longest side)
Hence, $2AD < AB  + AC$ (1)
Similarly we can say, $2BE < BC + AB$ (2)
and $2 CF < AC + BC$ (3)
or adding (1), (2), (3)
$2 (AC + AB + BC) > 2(AD + BE + CF)$
$AC + AB + BC > AD + BE + CF$
Thus, perimeter of the triangle is greater than the sum of altitudes