Tag: congruence and inequalities of triangles

Questions Related to congruence and inequalities of triangles

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

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