Mix Examples - Triangles Questions in English

(N/A) $1$. Extend $AD$ to a point $E$ such that $AD = DE$. Join $EC$.
$2$. In $\Delta ADB$ and $\Delta EDC$:
- $AD = DE$ (by construction)
- $\angle ADB = \angle EDC$ (vertically opposite angles)
- $BD = DC$ ($AD$ is the median to $BC$)
$3$. By $SAS$ congruence criterion,$\Delta ADB \cong \Delta EDC$.
$4$. Therefore,$AB = EC$ (by $CPCT$).
$5$. In $\Delta AEC$,by the triangle inequality theorem,the sum of any two sides is greater than the third side:
- $AC + EC > AE$
$6$. Since $AE = AD + DE = AD + AD = 2AD$ and $EC = AB$,we substitute these into the inequality:
- $AC + AB > 2AD$.
$7$. Thus,$AB + AC > 2AD$ is proved.

(N/A) $1$. Extend $BP$ to intersect $AC$ at point $D$.
$2$. In $\Delta ABD$,by the triangle inequality theorem,the sum of two sides is greater than the third side: $AB + AD > BD$.
$3$. Since $BD = BP + PD$,we can write: $AB + AD > BP + PD$ --- (Equation $1$).
$4$. In $\Delta PDC$,by the triangle inequality theorem: $PD + DC > PC$ --- (Equation $2$).
$5$. Adding Equation $1$ and Equation $2$: $(AB + AD) + (PD + DC) > (BP + PD) + PC$.
$6$. Simplifying the inequality: $AB + (AD + DC) + PD > BP + PD + PC$.
$7$. Since $AD + DC = AC$,we get: $AB + AC + PD > BP + PC + PD$.
$8$. Subtracting $PD$ from both sides,we obtain: $AB + AC > PB + PC$,which is equivalent to $PB + PC < AB + AC$.

(N/A) In a triangle,the sum of any two sides is always greater than the third side.
Consider the triangles formed by the diagonals of the quadrilateral $ABCD$:
$1$. In $\triangle ABC$,$AB + BC > AC$ (Equation $1$).
$2$. In $\triangle ADC$,$AD + CD > AC$ (Equation $2$).
$3$. In $\triangle ABD$,$AB + AD > BD$ (Equation $3$).
$4$. In $\triangle BCD$,$BC + CD > BD$ (Equation $4$).
Adding Equations $1$ and $2$ gives: $(AB + BC + AD + CD) > 2AC$.
Adding Equations $3$ and $4$ gives: $(AB + AD + BC + CD) > 2BD$.
Adding these two resulting inequalities: $2(AB + BC + CD + DA) > 2(AC + BD)$.
Dividing by $2$,we get: $AB + BC + CD + DA > AC + BD$.

(N/A) Let $O$ be the intersection point of the diagonals $AC$ and $BD$.
In $\triangle ABC$,by the triangle inequality,$AB + BC > AC$.
In $\triangle ADC$,by the triangle inequality,$CD + DA > AC$.
Adding these two inequalities,we get: $(AB + BC + CD + DA) > 2AC$.
Similarly,in $\triangle ABD$,$AB + AD > BD$.
In $\triangle BCD$,$BC + CD > BD$.
Adding these two inequalities,we get: $(AB + AD + BC + CD) > 2BD$.
Now,consider the triangles formed by the intersection point $O$:
In $\triangle OAB$,$AB < OA + OB$.
In $\triangle OBC$,$BC < OB + OC$.
In $\triangle OCD$,$CD < OC + OD$.
In $\triangle ODA$,$DA < OD + OA$.
Summing these four inequalities: $AB + BC + CD + DA < 2(OA + OB + OC + OD)$.
Since $OA + OC = AC$ and $OB + OD = BD$,we have $AB + BC + CD + DA < 2(AC + BD)$.

(A) In $\Delta ABC$,let $AD$,$BE$,and $CF$ be the medians to sides $BC$,$AC$,and $AB$ respectively.
By the property of triangles,the sum of two sides of a triangle is greater than the third side.
In $\Delta ABD$,$AB + BD > AD$ $(1)$.
In $\Delta ACD$,$AC + CD > AD$ $(2)$.
Adding $(1)$ and $(2)$: $AB + AC + (BD + CD) > 2AD$.
Since $BD + CD = BC$,we have $AB + AC + BC > 2AD$ $(3)$.
Similarly,for medians $BE$ and $CF$:
$AB + BC + AC > 2BE$ $(4)$.
$AB + BC + AC > 2CF$ $(5)$.
Adding $(3)$,$(4)$,and $(5)$:
$3(AB + BC + AC) > 2(AD + BE + CF)$.
This implies $AB + BC + AC > \frac{2}{3}(AD + BE + CF)$.
However,the standard inequality for medians is $AB + BC + AC > AD + BE + CF$. This is derived from the fact that in any triangle,the sum of the medians is less than the perimeter $(AD + BE + CF < AB + BC + AC)$.

(N/A) In $\Delta PQS$,by the triangle inequality theorem,the sum of any two sides is greater than the third side:
$PQ + QS > PS$ ---$(1)$
In $\Delta PRS$,by the triangle inequality theorem:
$PR + RS > PS$ ---$(2)$
Adding equations $(1)$ and $(2)$:
$(PQ + QS) + (PR + RS) > PS + PS$
$PQ + PR + (QS + RS) > 2PS$
Since $S$ is a point on $QR$,$QS + RS = QR$.
Therefore,$PQ + PR + QR > 2PS$.

(N/A) $1$. Extend $QS$ to meet $PR$ at point $T$.
$2$. In $\Delta PQT$,by the triangle inequality theorem,the sum of two sides is greater than the third side: $PQ + PT > QT$. This can be written as $PQ + PT > QS + ST$ (Equation $1$).
$3$. In $\Delta SRT$,by the triangle inequality theorem,$ST + TR > SR$ (Equation $2$).
$4$. Adding Equation $1$ and Equation $2$: $PQ + PT + ST + TR > QS + ST + SR$.
$5$. Simplifying by canceling $ST$ from both sides: $PQ + (PT + TR) > QS + SR$.
$6$. Since $PT + TR = PR$,we get $PQ + PR > QS + SR$,which is equivalent to $SQ + SR < PQ + PR$.

(N/A) Given: In $\Delta ABC$,$\angle A = 90^{\circ}$ and $AB = AC$. $AD$ is the bisector of $\angle A$.
Since $AB = AC$,$\Delta ABC$ is an isosceles triangle. Therefore,$\angle B = \angle C$.
In $\Delta ABC$,$\angle A + \angle B + \angle C = 180^{\circ}$.
$90^{\circ} + \angle B + \angle B = 180^{\circ} \implies 2 \angle B = 90^{\circ} \implies \angle B = 45^{\circ}$. Thus,$\angle C = 45^{\circ}$.
Since $AD$ is the angle bisector of $\angle A$,$\angle BAD = \angle CAD = 45^{\circ}$.
In $\Delta ABD$,$\angle B = 45^{\circ}$ and $\angle BAD = 45^{\circ}$. Since two angles are equal,$AD = BD$.
In $\Delta ACD$,$\angle C = 45^{\circ}$ and $\angle CAD = 45^{\circ}$. Since two angles are equal,$AD = CD$.
Since $D$ lies on $BC$,$BC = BD + CD$.
Substituting $BD = AD$ and $CD = AD$,we get $BC = AD + AD = 2 AD$.
Hence,$BC = 2 AD$ is proved.

(N/A) Let $ABC$ be a triangle with sides $AB$,$BC$,and $AC$. Let $AD$ be the median to the side $BC$,such that $BD = DC = \frac{1}{2} BC$.
Extend $AD$ to a point $E$ such that $AD = DE$. Join $CE$.
In $\triangle ABD$ and $\triangle ECD$:
$1$. $AD = DE$ (by construction)
$2$. $\angle ADB = \angle EDC$ (vertically opposite angles)
$3$. $BD = DC$ (since $AD$ is the median)
Thus,$\triangle ABD \cong \triangle ECD$ by $SAS$ congruence criterion.
Therefore,$AB = EC$ (by $CPCT$).
In $\triangle ACE$,by the triangle inequality theorem,the sum of any two sides is greater than the third side:
$AC + CE > AE$
Since $AE = AD + DE = 2AD$ and $CE = AB$,we substitute these into the inequality:
$AC + AB > 2AD$
Hence,the sum of two sides is greater than twice the median to the third side.

(A) Let the sides of the triangle be $a$,$b$,and $c$,and the angles opposite to them be $A$,$B$,and $C$ respectively.
Assume $a$ is the longest side,so $a > b$ and $a > c$.
By the property of triangles,the angle opposite to the longest side is the largest angle,so $A > B$ and $A > C$.
We know that the sum of angles in a triangle is $A + B + C = 180^{\circ}$.
Since $A > B$ and $A > C$,we have $A + A + A > A + B + C$,which implies $3A > 180^{\circ}$.
Therefore,$A > 60^{\circ}$.
$A$ right angle is $90^{\circ}$,and $\frac{2}{3}$ of a right angle is $\frac{2}{3} \times 90^{\circ} = 60^{\circ}$.
Thus,$A > 60^{\circ}$ proves that the angle opposite to the longest side is greater than $\frac{2}{3}$ of a right angle.

(B) $(1)$ False: For two triangles to be congruent by the $SAS$ (Side-Angle-Side) congruence rule,the angle must be included between the two sides. Here,in $\Delta ABC$,the angle between $AB$ and $BC$ is $\angle B$. In $\Delta PQR$,the angle between $PQ$ and $PR$ is $\angle P$. However,the given condition is $BC = PR$,but the sides forming $\angle P$ are $PQ$ and $PR$. Since $BC$ is not the side $QR$,the $SAS$ criterion is not satisfied.
$(2)$ True: The sum of an interior angle and its corresponding exterior angle is $180^{\circ}$. If an exterior angle is acute $(< 90^{\circ})$,the corresponding interior angle must be obtuse $(> 90^{\circ})$. $A$ triangle can have at most one obtuse angle. Therefore,it is impossible for two exterior angles to be acute,as that would imply the existence of two obtuse interior angles,which violates the angle sum property of a triangle.

(A) $(1)$ False. According to the theorem,in any triangle,the angle opposite to the longer side is larger. Since $XY > XZ$,the angle opposite to $XY$ (which is $\angle Z$) must be greater than the angle opposite to $XZ$ (which is $\angle Y$). Therefore,$\angle Z > \angle Y$.
$(2)$ True. Given $\frac{AB}{PR} = \frac{BC}{QP} = \frac{CA}{RQ} = 1$,it implies $AB = PR$,$BC = QP$,and $CA = RQ$. By the $SSS$ (Side-Side-Side) congruence criterion,$\Delta ABC \cong \Delta RPQ$.

(B) The exterior angle of a triangle is equal to the sum of the two interior opposite angles.
For $\Delta ABC$,the exterior angle $\angle ABD = \angle BAC + \angle ACB = 110^{\circ}$.
The exterior angle $\angle ACE = \angle BAC + \angle ABC = 130^{\circ}$.
Since $\angle ACE > \angle ABD$,we have $(\angle BAC + \angle ABC) > (\angle BAC + \angle ACB)$,which implies $\angle ABC > \angle ACB$.
In a triangle,the side opposite to the larger angle is longer. Therefore,the side opposite to $\angle ABC$ (which is $AC$) must be greater than the side opposite to $\angle ACB$ (which is $AB$).
Thus,$AC > AB$,which means $AB < AC$.
Therefore,the statement $AB > AC$ is False.

(B) Given that in $\Delta ABC$ and $\Delta PQR$:
$1$. $AB = PR$
$2$. $BC = RQ$
$3$. $\angle B = \angle R$
By the $SAS$ (Side-Angle-Side) congruence criterion,the two triangles are congruent.
We match the vertices based on the given equalities:
- Vertex $A$ corresponds to vertex $P$
- Vertex $B$ corresponds to vertex $R$
- Vertex $C$ corresponds to vertex $Q$
Therefore,$\Delta ABC \cong \Delta PRQ$.

(C) Given that in $\Delta PQR$ and $\Delta XYZ$:
$1$. $\angle P = \angle Z$
$2$. $\angle Q = \angle Y$
$3$. $PQ = YZ$
By the $ASA$ (Angle-Side-Angle) congruence criterion,the correspondence must match the vertices in the same order.
Since $\angle P$ corresponds to $\angle Z$,$\angle Q$ corresponds to $\angle Y$,and the side $PQ$ corresponds to $ZY$ (or $YZ$),the remaining vertex $R$ must correspond to $X$.
Therefore,the correspondence is $P \leftrightarrow Z$,$Q \leftrightarrow Y$,and $R \leftrightarrow X$.
Thus,$\Delta PQR \cong \Delta ZYX$.

(D) Given that in $\Delta ABC$ and $\Delta XYZ$:
$1$. $\angle A = \angle X$
$2$. $\angle C = \angle Z$
$3$. $AB = XY$
By the Angle-Angle-Side $(AAS)$ congruence criterion,since two angles and the included side between them are not equal,we look at the correspondence.
Here,$\angle A$ corresponds to $\angle X$,$\angle C$ corresponds to $\angle Z$,and the side $AB$ corresponds to $XY$.
Since $\angle A = \angle X$ and $\angle C = \angle Z$,the third angle $\angle B$ must be equal to $\angle Y$ (by Angle Sum Property).
Thus,the correspondence is $A \leftrightarrow X$,$B \leftrightarrow Y$,and $C \leftrightarrow Z$.
Therefore,$\Delta ABC \cong \Delta XYZ$.

(A) Given that $\frac{AB}{PQ} = \frac{BC}{RQ} = \frac{AC}{PR} = 1$.
This implies $AB = PQ$,$BC = RQ$,and $AC = PR$.
By the $SSS$ (Side-Side-Side) congruence criterion,if three sides of one triangle are equal to the three corresponding sides of another triangle,then the triangles are congruent.
Matching the vertices based on the equality of sides:
$AB = PQ$ (Vertex $A$ corresponds to $P$,$B$ to $Q$)
$BC = RQ$ (Vertex $B$ corresponds to $R$,$C$ to $Q$)
$AC = PR$ (Vertex $A$ corresponds to $P$,$C$ to $R$)
Thus,$\Delta ABC \cong \Delta PQR$ is incorrect based on the ratio provided. Let's re-evaluate the correspondence: $A \leftrightarrow P$,$B \leftrightarrow Q$,$C \leftrightarrow R$ is not possible since $BC=RQ$ and $AC=PR$.
Actually,$AB=PQ$,$BC=RQ$,$AC=PR$ implies $\Delta ABC \cong \Delta PQR$ is not the correct order. Let's check: $A$ corresponds to $P$,$B$ corresponds to $Q$,$C$ corresponds to $R$. Wait,$BC=RQ$ means $B$ corresponds to $R$ and $C$ corresponds to $Q$. $AC=PR$ means $A$ corresponds to $P$ and $C$ corresponds to $R$. $AB=PQ$ means $A$ corresponds to $P$ and $B$ corresponds to $Q$.
Matching: $A \leftrightarrow P$,$B \leftrightarrow Q$,$C \leftrightarrow R$. Therefore,$\Delta ABC \cong \Delta PQR$.

(B) Given: In $\Delta ABC$ and $\Delta PQR$,$\angle B = \angle P = 90^{\circ}$.
$AC = RQ$ (Hypotenuse of $\Delta ABC$ = Hypotenuse of $\Delta PQR$).
$AB = RP$ (One side of $\Delta ABC$ = One side of $\Delta PQR$).
By the $RHS$ (Right angle-Hypotenuse-Side) congruence criterion,the triangles are congruent.
Matching the vertices: $A$ corresponds to $R$,$B$ corresponds to $P$,and $C$ corresponds to $Q$.
Therefore,$\Delta ABC \cong \Delta RPQ$.

(C) In $\Delta ABC$,it is given that $AB = AC$.
According to the Isosceles Triangle Theorem,the angles opposite to equal sides of a triangle are equal.
Therefore,the angle opposite to side $AB$ is $\angle C$ and the angle opposite to side $AC$ is $\angle B$.
Since $AB = AC$,it follows that $\angle C = \angle B$.
Given that $\angle B = 75^{\circ}$,we have $\angle C = 75^{\circ}$.

(D) Given that in $\Delta PQR$,$PQ = PR$.
Since the sides opposite to equal angles are equal,the angles opposite to equal sides are also equal.
Therefore,$\angle Q = \angle R$.
Given $\angle R = 40^{\circ}$,so $\angle Q = 40^{\circ}$.
The sum of all angles in a triangle is $180^{\circ}$.
So,$\angle P + \angle Q + \angle R = 180^{\circ}$.
Substituting the values: $\angle P + 40^{\circ} + 40^{\circ} = 180^{\circ}$.
$\angle P + 80^{\circ} = 180^{\circ}$.
$\angle P = 180^{\circ} - 80^{\circ} = 100^{\circ}$.

(A) Given that in $\Delta XYZ$,$XY = XZ$.
Since two sides are equal,$\Delta XYZ$ is an isosceles triangle.
Therefore,the angles opposite to the equal sides are equal,which means $\angle Y = \angle Z$.
We know that the sum of all interior angles in a triangle is $180^{\circ}$.
So,$\angle X + \angle Y + \angle Z = 180^{\circ}$.
Substituting the given values: $80^{\circ} + \angle Y + \angle Y = 180^{\circ}$.
$80^{\circ} + 2\angle Y = 180^{\circ}$.
$2\angle Y = 180^{\circ} - 80^{\circ} = 100^{\circ}$.
$\angle Y = 100^{\circ} / 2 = 50^{\circ}$.

(B) In any triangle,the side opposite to the larger angle is longer than the side opposite to the smaller angle.
Given that $\angle A > \angle B$ in $\Delta ABC$.
The side opposite to $\angle A$ is $BC$.
The side opposite to $\angle B$ is $AC$.
Since $\angle A > \angle B$,it follows that the side opposite to $\angle A$ must be greater than the side opposite to $\angle B$.
Therefore,$BC > AC$,which is equivalent to $AC < BC$.

(C) According to the Exterior Angle Theorem of a triangle,the measure of an exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.
In $\Delta ABC$,$\angle ACD$ is the exterior angle at vertex $C$.
The interior opposite angles are $\angle A$ and $\angle B$.
Therefore,$\angle ACD = \angle A + \angle B$.
Given $\angle A = 50^{\circ}$ and $\angle B = 65^{\circ}$.
Substituting these values,we get $\angle ACD = 50^{\circ} + 65^{\circ} = 115^{\circ}$.

(D) In $\Delta ABC$,we are given that $AB = AC$.
According to the property of an isosceles triangle,angles opposite to equal sides are equal.
Therefore,$\angle C = \angle B = 70^{\circ}$.
The sum of angles in a triangle is $180^{\circ}$.
So,$\angle A + \angle B + \angle C = 180^{\circ}$.
$\angle A + 70^{\circ} + 70^{\circ} = 180^{\circ}$.
$\angle A + 140^{\circ} = 180^{\circ}$,which gives $\angle A = 40^{\circ}$.
The exterior angle property states that the exterior angle of a triangle is equal to the sum of the two interior opposite angles.
Thus,$\angle ACD = \angle A + \angle B$.
$\angle ACD = 40^{\circ} + 70^{\circ} = 110^{\circ}$.

(A) An equilateral triangle is a triangle in which all three sides are equal and all three interior angles are equal.
Each interior angle of an equilateral triangle is $60^{\circ}$.
The sum of an interior angle and its corresponding exterior angle is $180^{\circ}$ (linear pair).
Therefore,the exterior angle = $180^{\circ} - 60^{\circ} = 120^{\circ}$.
Alternatively,the sum of all exterior angles of any polygon is $360^{\circ}$. Since an equilateral triangle has $3$ equal exterior angles,each exterior angle = $360^{\circ} / 3 = 120^{\circ}$.

(B) Given that in $\Delta ABC$, $\angle B = \angle C$.
According to the property of triangles, sides opposite to equal angles are equal.
Therefore, the side opposite to $\angle B$ is $AC$ and the side opposite to $\angle C$ is $AB$.
Thus, $AC = AB$.
Given $AB = 5 \text{ cm}$, so $AC = 5 \text{ cm}$.
We are also given $BC = 8 \text{ cm}$.
The perimeter of $\Delta ABC = AB + BC + AC$.
Perimeter $= 5 \text{ cm} + 8 \text{ cm} + 5 \text{ cm} = 18 \text{ cm}$.

(C) In $\Delta XYZ$,we are given that $\angle Y = 90^{\circ}$.
Since the sum of all angles in a triangle is $180^{\circ}$,we have $\angle X + \angle Y + \angle Z = 180^{\circ}$.
Substituting the value of $\angle Y$,we get $\angle X + 90^{\circ} + \angle Z = 180^{\circ}$,which implies $\angle X + \angle Z = 90^{\circ}$.
We are also given that $XY = YZ$. In a triangle,angles opposite to equal sides are equal.
Therefore,$\angle Z = \angle X$.
Substituting $\angle Z = \angle X$ into the equation $\angle X + \angle Z = 90^{\circ}$,we get $\angle X + \angle X = 90^{\circ}$.
This simplifies to $2\angle X = 90^{\circ}$,which gives $\angle X = 45^{\circ}$.

(D) In $\Delta ABC$,the interior angles are $\angle ABC$,$\angle ACB$,and $\angle BAC$ (or $\angle A$).
Since $\angle ABD$ is an exterior angle at vertex $B$,we have $\angle ABC + \angle ABD = 180^{\circ}$ (linear pair).
$\angle ABC = 180^{\circ} - 110^{\circ} = 70^{\circ}$.
Since $\angle ACE$ is an exterior angle at vertex $C$,we have $\angle ACB + \angle ACE = 180^{\circ}$ (linear pair).
$\angle ACB = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
In $\Delta ABC$,the sum of interior angles is $180^{\circ}$,so $\angle A + \angle ABC + \angle ACB = 180^{\circ}$.
$\angle A + 70^{\circ} + 30^{\circ} = 180^{\circ}$.
$\angle A + 100^{\circ} = 180^{\circ}$.
$\angle A = 180^{\circ} - 100^{\circ} = 80^{\circ}$.

(A) Let the vertices of the first triangle be $A, B, C$ and the vertices of the second triangle be $P, Q, R$.
$A$ one-one correspondence is a mapping between the sets of vertices of the two triangles.
The number of ways to map $3$ vertices of the first triangle to $3$ vertices of the second triangle is given by the number of permutations of $3$ elements,which is $3!$.
$3! = 3 \times 2 \times 1 = 6$.
Therefore,there are $6$ possible one-one correspondences between two triangles.

(B) In the given correspondence $ABC \leftrightarrow PRQ$,the vertices correspond as follows:
$A$ corresponds to $P$
$B$ corresponds to $R$
$C$ corresponds to $Q$
Therefore,the side $AB$ corresponds to the side formed by the corresponding vertices of $A$ and $B$,which is $PR$.

(C) Given that $\Delta ABC \cong \Delta YZX$.
By the property of congruent triangles,the corresponding parts of congruent triangles are equal $(CPCT)$.
Comparing the vertices: $A$ corresponds to $Y$,$B$ corresponds to $Z$,and $C$ corresponds to $X$.
Therefore,side $AB$ corresponds to side $YZ$.
Thus,side $AB = YZ$.

(D) Given that $\Delta PRQ \cong \Delta ZXY$.
By the property of congruent triangles,the corresponding parts of congruent triangles are equal $(CPCT)$.
Comparing the vertices of the two triangles:
$P$ corresponds to $Z$
$R$ corresponds to $X$
$Q$ corresponds to $Y$
Therefore,$\angle P = \angle Z$,$\angle R = \angle X$,and $\angle Q = \angle Y$.
Since the question asks for the angle in $\Delta XYZ$ that is equal to $\angle R$,we look at the correspondence: $\angle R$ corresponds to $\angle X$.
Thus,$\angle R = \angle X$.

(A) Given that $\Delta ABC \cong \Delta RPQ$.
By the property of congruent triangles,corresponding parts of congruent triangles are equal $(CPCT)$.
This implies that the sides of $\Delta ABC$ are equal to the corresponding sides of $\Delta RPQ$.
Specifically,$AB = RP$,$BC = PQ$,and $AC = RQ$.
The perimeter of a triangle is the sum of its three sides.
Perimeter of $\Delta ABC = AB + BC + AC = 18 \, cm$.
Since $\Delta ABC \cong \Delta RPQ$,the perimeter of $\Delta RPQ$ is equal to the perimeter of $\Delta ABC$.
Perimeter of $\Delta RPQ = RP + PQ + RQ = AB + BC + AC = 18 \, cm$.
Note that the perimeter of $\Delta PQR$ is the same as the perimeter of $\Delta RPQ$ because the set of side lengths is identical.
Therefore,the perimeter of $\Delta PQR$ is $18 \, cm$.

(B) According to the triangle inequality theorem,the sum of the lengths of any two sides of a triangle is always greater than the length of the third side. Therefore,the correct word is 'greater'.

(B) According to the triangle inequality theorem,the difference between the lengths of any two sides of a triangle is always less than the length of the third side. If the sides of a triangle are $a$,$b$,and $c$,then $|a - b| < c$,$|a - c| < b$,and $|b - c| < a$.

(D) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
$\angle A + 50^{\circ} + 85^{\circ} = 180^{\circ}$.
$\angle A + 135^{\circ} = 180^{\circ}$.
$\angle A = 180^{\circ} - 135^{\circ} = 45^{\circ}$.
We know that in a triangle,the side opposite to the larger angle is longer.
Here,$\angle C = 85^{\circ}$ and $\angle B = 50^{\circ}$.
Since $\angle C > \angle B$,the side opposite to $\angle C$ $(AB)$ is greater than the side opposite to $\angle B$ $(AC)$.
Thus,$AB > AC$.

(A) According to the triangle inequality theorem,the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
In $\Delta ABC$,the sum of sides $AB$ and $BC$ must be greater than $AC$.
Therefore,$AB + BC > AC$.
Substituting the given values: $6 \text{ cm} + 9 \text{ cm} > AC$.
$15 \text{ cm} > AC$,which means $AC < 15 \text{ cm}$.
Thus,the length of side $AC$ must be less than $15 \text{ cm}$.

(B) According to the triangle inequality theorem,the sum of any two sides of a triangle is always greater than the third side.
Also,the difference between any two sides of a triangle is always less than the third side.
Let the sides be $a = 8 \, cm$,$b = 5 \, cm$,and $c = AC$.
According to the property: $|a - b| < c < a + b$.
Substituting the values: $|8 - 5| < AC < 8 + 5$.
$3 < AC < 13$.
Therefore,$AC$ must be greater than $3 \, cm$.

(C) In $\Delta PQR$,we are given that $\angle Q = \angle R$.
According to the theorem,sides opposite to equal angles of a triangle are equal.
Therefore,the side opposite to $\angle Q$ (which is $PR$) must be equal to the side opposite to $\angle R$ (which is $PQ$).
Thus,$PR = PQ$.
Given that $PQ = 6.5 \, cm$,it follows that $PR = 6.5 \, cm$.

(D) Given that in $\Delta ABC$,$AB = AC$.
According to the property of an isosceles triangle,angles opposite to equal sides are equal.
Therefore,the angle opposite to side $AB$ is $\angle C$ and the angle opposite to side $AC$ is $\angle B$.
Since $AB = AC$,it follows that $\angle B = \angle C$.
Given that $\angle C = 75^{\circ}$.
Thus,$\angle B = 75^{\circ}$.

(A) According to the triangle inequality theorem,the sum of any two sides of a triangle must be greater than the third side.
Let the third side be $AC = x \text{ cm}$.
Then,$AB + BC > AC \implies 8 + 10 > x \implies x < 18$.
Also,$AB + AC > BC \implies 8 + x > 10 \implies x > 2$.
And $BC + AC > AB \implies 10 + x > 8$,which is true for all $x > 0$.
So,$2 < x < 18$.
The perimeter $P = AB + BC + AC = 8 + 10 + x = 18 + x$.
Since $x < 18$,the perimeter $P < 18 + 18 = 36 \text{ cm}$.
Therefore,the perimeter is less than $36 \text{ cm}$.

(B) In any triangle,the sum of the lengths of any two sides must be greater than the length of the third side (Triangle Inequality Theorem).
Let the sides of $\Delta XYZ$ be $XY = 7 \, cm$,$YZ = 10 \, cm$,and $XZ = z \, cm$.
According to the triangle inequality theorem:
$XY + YZ > XZ \implies 7 + 10 > z \implies z < 17$
$XY + XZ > YZ \implies 7 + z > 10 \implies z > 3$
$YZ + XZ > XY \implies 10 + z > 7 \implies z > -3$ (which is always true for a length).
Thus,the third side $z$ must satisfy $3 < z < 17$.
The perimeter $P$ of the triangle is given by $P = XY + YZ + XZ = 7 + 10 + z = 17 + z$.
Since $z > 3$,we have $P = 17 + z > 17 + 3 = 20$.
Therefore,the perimeter of $\Delta XYZ$ is greater than $20 \, cm$.

(C) Two triangles are said to be congruent if they are identical in shape and size.
If $\Delta ABC \cong \Delta DEF$,it implies that all corresponding sides and angles are equal.
As a consequence of congruence,the areas of two congruent triangles must be equal.
Given that the area of $\Delta ABC = 58 \ cm^2$.
Therefore,the area of $\Delta DEF = 58 \ cm^2$.

(D) In any triangle,the angle opposite to the longest side is the largest angle.
Given the side lengths satisfy the inequality $AC > AB > BC$.
The side $AC$ is the longest side of $\Delta ABC$.
The angle opposite to side $AC$ is $\angle B$.
Therefore,$\angle B$ is the greatest angle of the triangle.

(A) According to the triangle inequality theorem,the sum of any two sides of a triangle is always greater than the third side,and the difference of any two sides is always less than the third side.
Given sides are $PQ = 7.5 \, cm$ and $QR = 6.2 \, cm$.
Let the third side be $PR$.
$1$. The difference of the two sides must be less than the third side: $PR > |PQ - QR| = |7.5 - 6.2| = 1.3 \, cm$.
$2$. The sum of the two sides must be greater than the third side: $PR < PQ + QR = 7.5 + 6.2 = 13.7 \, cm$.
Therefore,$1.3 < PR < 13.7$.
Comparing this with $a < PR < b$,we get $a = 1.3$ and $b = 13.7$.

(B) In any triangle,the side opposite to the largest angle is the longest,and the side opposite to the smallest angle is the shortest.
Given the angles are $\angle Y > \angle X > \angle Z$.
The side opposite to $\angle Y$ is $XZ$.
The side opposite to $\angle X$ is $YZ$.
The side opposite to $\angle Z$ is $XY$.
Since $\angle Z$ is the smallest angle,the side opposite to it,which is $XY$,must be the smallest side of the triangle.

(C) Given that $\Delta ABC \cong \Delta RPQ$.
By the property of congruent triangles,the corresponding parts of congruent triangles are equal $(CPCT)$.
The vertices correspond as follows: $A \leftrightarrow R$,$B \leftrightarrow P$,and $C \leftrightarrow Q$.
Therefore,the side $AB$ of $\Delta ABC$ corresponds to the side $RP$ of $\Delta RPQ$.
Since $\Delta RPQ$ is the same as $\Delta PQR$ (as the set of vertices is the same),the side $RP$ is the same as side $PR$ in $\Delta PQR$.
Thus,side $AB$ is equal to side $RP$.

(D) Let the third side of the triangle be $AC = b$.
According to the triangle inequality theorem,the sum of any two sides of a triangle is greater than the third side,and the difference of any two sides is less than the third side.
$1$. $AB + BC > AC \implies 9 + 12 > b \implies b < 21$.
$2$. $BC - AB < AC \implies 12 - 9 < b \implies b > 3$.
So,$3 < b < 21$.
The perimeter $P = AB + BC + AC = 9 + 12 + b = 21 + b$.
Since $3 < b < 21$,we add $21$ to all parts of the inequality:
$3 + 21 < 21 + b < 21 + 21$
$24 < P < 42$.
Comparing this with $x < P < y$,we get $x = 24$ and $y = 42$.