Tag: triangle inequality

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

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

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

$b < c + a$
$\because$ Sum of any two sides is greater than the third side.

$\because$ All given statements are true

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

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

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

$\displaystyle AB+BC>AC$ (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
$\displaystyle 2(AB+BC+CD+DA)>2(AC+BD)$
$\displaystyle AB+BC+CD+DA>AC+BD$

(6 + 5) 11 > 3 Inequality property
(5 + 3) 8 > 6
(3 + 6) 9 > 5
They can form a $\displaystyle \Delta. $