Tag: triangle inequality

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Questions Related to triangle inequality

Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

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

  1. Greater

  2. Lesser

  3. Equal

  4. May be greater or lesser
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

Two sides of a $\Delta$ le are $7$ and $10$ units. Which of the following length can be the length of the third side?

  1. $19$ cm
  2. $17$ cm
  3. $13$ cm
  4. $3$ cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that the two sides of triangle  are $7$ and $10$.

Sum of two sides $= 17$
Difference between two sides $= 3$
Therefore, the third side should be between $3$ and $17$ and only one option satisfies it i.e Option C.

Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

If A is the area of a triangle in em", whose sides are 9 em, 10 cm and 11 em, then which one of the following is correct?

  1. $A < 40\:cm^2$
  2. $40\:cm^2 < A < 45\:cm^2$
  3. $45\:cm^2 < A < 50\:cm^2$
  4. $A>50\:cm^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle s=\frac{1}{2}(9+10+11):cm=15:cm$

$\therefore\Delta=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{15\times6\times5\times4}:cm^2$

$=30\sqrt{2}=30\times1.4=42:cm^2$
which lies between $40:cm^2$ and $45:cm^2$.

Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

ABCD is a quadrilateral. Then which of the following is true? 

  1. $\displaystyle AC+BC<(AB+BC+CD+DA)$
  2. $\displaystyle AC+BD<\frac { 1 }{ 2 } \left( AB+BC+CD+DA \right) $
  3. $\displaystyle AC+BD>\frac { 1 }{ 4 } \left( AB+BC+CD+DA \right) $
  4. $\displaystyle AC+BD<\frac { 1 }{ 4 } \left( AB+BC+CD+DA \right) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle AB+BC>AC$ (considering $\displaystyle \Delta ABC$)
$\displaystyle BC+CD>BD$ (considering $\displaystyle \Delta BCD$)
$\displaystyle CD+DA>AC$ (considering $\displaystyle \Delta ADC$)
$\displaystyle DA+AB>BD$ (considering $\displaystyle \Delta ABD$)
Adding all four inequalities, we get
$\displaystyle 2(AB+BC+CD+DA)>2(AC+BD)$
$\displaystyle AB+BC+CD+DA>AC+BD$