Inequalities of a triangle - class-VIII

By the triangle inequality, for any point O inside a triangle, the sum of the distances from O to the vertices is less than the semi-perimeter, and specifically, the inequality 2(OA+OB+OC) < (AB+BC+CA) holds true.

Since PQ is parallel to BC, triangle APQ is similar to triangle ABC. Therefore, PQ/BC = AP/AB. Substituting the values: 5/BC = 4/12. This simplifies to 5/BC = 1/3, so BC = 15.

If $a,b$ and $c$ are the sides of the triangle then $\sqrt a  + \sqrt b  - \sqrt c $ it always positive because the sum of two sides of the triangle is always greater than the third side.

For any $\triangle ABC$,
the difference between two sides of the triangle should be less than the third side.
Thus, $AB - AC < BC$

In $\triangle PQR$, $\angle R > \angle Q$
Hence, $PQ > PR$ (Sides opposite larger angles is large)

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. $ 

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.

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 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 $\Delta ABC$ 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 $AD + BE + CF < AB + BC + CA$

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

Let the sides be a=2001, b=1002, and c be the third side. By the triangle inequality, |a-b| < c < a+b. So 999 < c < 3003. The number of possible integer values for c is 3003 - 999 - 1 = 2003.

Let a, b, c be sides of a triangle. By the triangle inequality, (a+b-c) > 0, (b+c-a) > 0, and (c+a-b) > 0. The expression (a+b-c)(b+c-a)(c+a-b) - abc is known to be non-negative for all triangles.

$If  \ two &nbsp; sides  \ of &nbsp; a &nbsp; triangle &nbsp; are &nbsp; 8 cm &nbsp; and &nbsp; 13 cm.$
$Then  \ the  \ length &nbsp; of  &nbsp; third &nbsp; side  &nbsp; is$
$more &nbsp; than &nbsp; 13-8 \ =  5cm$
$and  \ less &nbsp; than &nbsp; 13+8  =  21cm$
$a = 5cm , b = 21cm$

This can be taken as a fact that the median corresponding to the longer side has the larger median.

Equilateral triangles have all the three angles equal to $\displaystyle { 60 }^{ o }$. 

The Triangle Inequality theorem states that the sum of any $2$ sides of a triangle must be greater than the measure of the third side.
Take option A: $a = 4, b = 8, c = 10$
$4 + 8 > 10 (a + b > c)$
$8 + 10 > 4 (b + c > a)$
$4 + 10 < 8 (a + c > b)  $
Therefore, the third side, $c = 10$ will satisfy the triangle inequality
Similarly check for option 2: $c = 2$
$4 + 8 > 2 (a + b > c)$
$8 + 2 > 10 (b + c = a)$
$4 + 2 < 8 (a + c < b)  $
Hence, the second option will not satisfying the triangle inequality.
Similarly check for option $3$ and $4$.
Here, the third side of the triangle inequality, $c = 10$.

Exterior angle $A = 180 -$ $\angle$ $P = 180-160$
$\angle$ $P = 20$
$\angle$ $Q = 70$
Interior angle $= 20 + 70 +$ $\angle$ $R = 180$
$\angle$ $R = 90$
In a triangle inequality theorem,the largest side is across from the longest angle.
So, $90^o$ is the longest angle in the triangle, $\overline{PQ}$, across from it, is the largest side.
Therefore, $\overline{PQ}$ is the largest side.

for triplet to be the length of triangle, it must satisfy triangle property.

                            	<span class="ng-binding">The third side rule says that the length of the third side of the triangle in this case must be less than $5+8=13$<span class="MathJax_Preview"></span><span class="MathJax"><span class="math"><span><span><span class="mrow"><span class="mn"></span><span class="mo"></span><span class="mn"></span><span class="mo"></span><span class="mn"></span></span></span></span></span></span></span><span class="ng-binding"><br> Since the length of the third side is an integer, the length must be $12$<span class="MathJax"><span class="math"><span><span><span class="mrow"><span class="mn"></span></span></span></span></span></span></span><span class="ng-binding">.&nbsp;</span><div><span class="ng-binding">The biggest possible perimeter is then $5+8+12=25$<br>As we know, perimeter of a triangle is $=1^{st}$ side $+2^{nd}$ side $+ 3^{rd}$ side<span class="MathJax_Preview"></span><span class="MathJax"><span class="math"><span><span><span></span></span></span><span></span></span></span>.</span></div>

The expression (a+b+c)(b+c-a) = (b+c)^2 - a^2 = b^2 + c^2 + 2bc - a^2. By the Law of Cosines, a^2 = b^2 + c^2 - 2bc cos(A). Substituting this, we get 2bc + 2bc cos(A) = 2bc(1+cos(A)). Since 0 < 1+cos(A) < 2, the value k must be between 0 and 4.

The definition of the angle bisector of a triangle is a line segment that bisects one of the vertex angles of a triangle. In general, an angle bisector is equidistant from the sides of the angle when measured along a segment perpendicular to the sides of the angle 

In ABC,
Sum of the two sides is always greater than third side.
So, AB + BC > AC