Tag: triangle inequality

Questions Related to triangle inequality

Two sides of a triangle have lengths $7$ and $9$. Which of the following could not be the length of the third side?

  1. $4$

  2. $5$

  3. $7$

  4. $11$

  5. $16$


Correct Option: E
Explanation:

An important rule to remember about triangles is called the third side rule: the length of the third side of a triangle is less than the sum of the lengths of the other two sides and greater than the (positive) difference of the lengths of the other two sides. 

For this triangle, the length of the third side must be greater than $9-7=2$97=2 and less than $9+7=16$9+7=16. All the answers are possible except for answer E, which is equal to $16$16 but not less than $16$16.

In a triangle, the difference of any two sides is ____ than the third side.

  1. smaller

  2. equal

  3. greater

  4. cannot be determined


Correct Option: A
Explanation:

In a triangle, the difference of any two sides is smaller than the third side.

Which of the following sets of measurements can be used to construct a triangle?

  1. $4\ cm, 5\ cm, 6\ cm$

  2. $4\ cm, 3\ cm, 8\ cm$

  3. $5\ cm, 6\ cm, 12\ cm$

  4. $6\ cm, 3\ cm, 10\ cm$


Correct Option: A
Explanation:

Because the sum of the length of any $2$ sides of the triangle should be greater than the third side, which is only satisfied by option 1.

$(4+5)>6$
$(4+6)>5$
$(5+6)>4$

Two sides of an acute-angled triangle are $6\ cm$ and $2\ cm$ respectively. Which one of the following represents the correct range of the third side in cm?

  1. $(4, 8)$

  2. $(4, 2\sqrt {10})$

  3. $(4\sqrt {2}, 8)$

  4. $(4\sqrt {2}, 2\sqrt {10})$


Correct Option: A
Explanation:

If two sides of the triangle are 'a' and 'b' units, then the range of the third side is $(a - b, a + b)$.
So the range of the third side is $(4, 8)$.

In $\Delta ABC, AD \bot BC; BE \bot AC; CF \bot AB.$ then which of the following option is correct:

  1. $AD - BE + CF < AB + BC - CA$

  2. $AD + BE - CF < AB - BC - CA$

  3. $AD + BE - CF < AB - BC + CA$

  4. $AD + BE + CF < AB + BC + CA$


Correct Option: D
Explanation:

In triangle ADB,
$AD + BD > AB$
In triangle ADC,
$AD + DC > AC$
In tringle BEC,
$BE + EC > BC$
In tringle BEA,
$BE + AE > AB$
In traingle, CFA,
$CF + FA > AC$
In traingle, CFB,
$CF + FB > BC$
Add all the inequalities,
$2(AD + BE + CF) + AB + BC + CA > 2(AB + BC+ CA)$
$2 (AD + BE + CF) > AB + BC + CA$
or $AD + BE + CF < AB + BC + CA$

In a $\Delta ABC$, which of the following relation is correct where A, B, C are the vertices and D, E, F are the corresponding mid-points?

  1. $AD + BE + CF < AB + BC + CA$

  2. $AB + BE + CF < AD + BC + CA$

  3. $AD + BC + CF < AB + BE + CA$

  4. $AD + BE + CF > AB + BC + CA$


Correct Option: A
Explanation:

In a $\Delta ABC$ let perimeter equal to $AB + BC + CA$ and sum of altitudes is equal to $AD + BE + CF$. Now since $AD\bot BC$
$\therefore AD < AB, AD < AC$
$\therefore AD + AD < AB + AC$
$ 2AD < AB + AC$
$\therefore BE \bot CA$
$\Rightarrow BE < BC, BE < BA$
$\Rightarrow BE + BE < BC + BA$
$2BE < BC +BA.........(2)$
$CF \bot AB$
$\therefore CF < CA, CF < CB$
$\therefore CF + CF < CA + CB$
$2CF < CA + CB.......(3)$
On adding (1), (2) and (3), we get
$2(AD+BE+CF)< AB+ AC + BC+ BA + CA + CB < 2AB + 2BC + 2 CA < 2(AB+BC+CA)$
Hence $AD + BE + CF < AB + BC + CA$

The longest side of a triangle is three times the shortest side and the third side is $2$ cm shorter than the longest side. If the perimeter of the triangle is at least $61$ cm, find the minimum length of the shortest-side.

  1. $9$ cm

  2. $11$ cm

  3. $16$ cm

  4. $61$ cm


Correct Option: A
Explanation:

Let the shortest side $= s$
Longest side $= 3s$
Third side $= 3s -2$
Now perimeter $= s + 3s +3s -2 = 7s -2 \ge 61$
$\therefore 7s \ge 63$
$\therefore s \ge 9$
Thus minimum length of the shortest side$= 9$ cm

If the inequality $\left( m-2 \right) { x }^{ 2 }+8x+m+4>0$ is satisfied for all $x\epsilon R$, then least integral m is

  1. $4$

  2. $5$

  3. $6$

  4. None of these


Correct Option: A
Explanation:
$(m-2){ x }^{ 2 }+8x+m+4>0$
${ b }^{ 2 }-4ac>0$
${ (8) }^{ 2 }-4(m-2)(m+4)>0$
$64-4({ m }^{ 2 }-2m-8)>0$
$64-4{ m }^{ 2 }+8m+32>0$
$-4{ m }^{ 2 }+8m+96>0$
$-{ m }^{ 2 }+2m+24>0$
$-{ m }^{ 2 }-6m+4m+24>0$
$-m(m+6)+4(m+6)>0$
$(4-m)(m+6)>0$
The least integral of m is $4$.

Triangle ABC has integral sides AB, BC measuring $2001$ unit and $1002$ units respectively. Then the number of such triangles, is?

  1. $3002$

  2. $2003$

  3. $1003$

  4. None of these


Correct Option: B

$O$ is any point in the interior of $\Delta ABC.$ then 
$OA + OB + OC < \frac{1}{2}\left( {AB + BC + CA} \right)$ statement is ?

  1. True

  2. False


Correct Option: B