Tag: inequalities in triangle

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. $ 

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. 

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

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.