Mix Examples - Triangles Questions in English

(C) Given that $\Delta ABC \sim \Delta DEF$ for the correspondence $ABC \leftrightarrow DEF$.
Since the triangles are similar,the ratios of their corresponding sides are equal.
Therefore,$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$.
Given $3 AB = 5 DE$,we can write $\frac{AB}{DE} = \frac{5}{3}$.
Using the similarity ratio,we have $\frac{AC}{DF} = \frac{AB}{DE}$.
Substituting the given values,$\frac{AC}{9} = \frac{5}{3}$.
$AC = \frac{5}{3} \times 9 = 5 \times 3 = 15$.
Thus,$AC = 15$.

(B) Given the correspondence $PQR \leftrightarrow YZX$ is a similarity,the corresponding angles are equal.
Therefore,$m \angle P = m \angle Y$,$m \angle Q = m \angle Z$,and $m \angle R = m \angle X$.
Given $m \angle X = 120^\circ$,it follows that $m \angle R = 120^\circ$.
In $\Delta PQR$,the sum of angles is $m \angle P + m \angle Q + m \angle R = 180^\circ$.
Substituting the given values: $m \angle P + m \angle Q + 120^\circ = 180^\circ$,which implies $m \angle P + m \angle Q = 60^\circ$.
Since $m \angle P = 2 m \angle Q$,we substitute $m \angle P$ in the equation: $2 m \angle Q + m \angle Q = 60^\circ$.
$3 m \angle Q = 60^\circ \implies m \angle Q = 20^\circ$.
Then,$m \angle P = 2 \times 20^\circ = 40^\circ$.
Since $m \angle Y = m \angle P$,we have $m \angle Y = 40^\circ$.

(C) According to the Angle Bisector Theorem,in $\Delta PQR$,if $PS$ is the bisector of $\angle P$,then $\frac{PQ}{PR} = \frac{QS}{SR}$.
Given that $\frac{PQ}{PR} = \frac{5}{4}$ and $SR = 5.6 \text{ cm}$.
Substituting the values,we get $\frac{5}{4} = \frac{QS}{5.6}$.
Therefore,$QS = \frac{5 \times 5.6}{4} = 5 \times 1.4 = 7 \text{ cm}$.
Now,$QR = QS + SR = 7 + 5.6 = 12.6 \text{ cm}$.

(C) In $\Delta ABC$, $P$ and $Q$ are the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively.
Therefore, by the Midpoint Theorem, $PQ \parallel BC$ and $PQ = \frac{1}{2} BC$.
This implies that $\Delta APQ \sim \Delta ABC$ by the $AA$ similarity criterion.
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Thus, $\frac{\text{Area}(\Delta APQ)}{\text{Area}(\Delta ABC)} = \left( \frac{AP}{AB} \right)^2$.
Since $P$ is the midpoint of $AB$, $\frac{AP}{AB} = \frac{1}{2}$.
Therefore, $\frac{12 \sqrt{3}}{\text{Area}(\Delta ABC)} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Hence, $\text{Area}(\Delta ABC) = 12 \sqrt{3} \times 4 = 48 \sqrt{3}$.

(C) The diagonals of a rhombus bisect each other at right angles.
This division creates four congruent triangles within the rhombus.
Therefore,$\Delta OAB \cong \Delta OCB \cong \Delta OCD \cong \Delta OAD$.
Since these four triangles are congruent,their areas are equal:
$\text{Area}(\Delta OAB) = \text{Area}(\Delta OCB) = \text{Area}(\Delta OCD) = \text{Area}(\Delta OAD)$.
The total area of the rhombus $ABCD$ is the sum of the areas of these four triangles:
$\text{Area}(\square ABCD) = \text{Area}(\Delta OAB) + \text{Area}(\Delta OCB) + \text{Area}(\Delta OCD) + \text{Area}(\Delta OAD)$.
Substituting the equal areas,we get:
$\text{Area}(\square ABCD) = 4 \times \text{Area}(\Delta OAB)$.
Thus,$\text{Area}(\Delta OAB) = \frac{1}{4} \times \text{Area}(\square ABCD)$.

(B) In $\Delta ABC$,$D$,$E$,and $F$ are the midpoints of sides $BC$,$CA$,and $AB$ respectively. By the midpoint theorem,the triangle is divided into four congruent triangles: $\Delta AFE$,$\Delta FBD$,$\Delta EDC$,and $\Delta DEF$.
Therefore,$\text{Area}(\Delta AFE) = \text{Area}(\Delta FBD) = \text{Area}(\Delta EDC) = \text{Area}(\Delta DEF) = \frac{1}{4} \text{Area}(\Delta ABC)$.
The quadrilateral $BDEF$ is composed of $\Delta FBD$ and $\Delta DEF$.
Thus,$\text{Area}(BDEF) = \text{Area}(\Delta FBD) + \text{Area}(\Delta DEF) = \frac{1}{4} \text{Area}(\Delta ABC) + \frac{1}{4} \text{Area}(\Delta ABC) = \frac{1}{2} \text{Area}(\Delta ABC)$.

(D) In $\Delta ABC$,since $P, Q,$ and $R$ are the midpoints of sides $BC, CA,$ and $AB$ respectively,the triangle is divided into four congruent triangles: $\Delta ARQ, \Delta RBP, \Delta QPC,$ and $\Delta PQR$.
Therefore,$Area(\Delta ARQ) = Area(\Delta RBP) = Area(\Delta QPC) = Area(\Delta PQR) = \frac{1}{4} Area(\Delta ABC)$.
The quadrilateral $BCQR$ is composed of three of these triangles: $\Delta RBP, \Delta QPC,$ and $\Delta PQR$.
Thus,$Area(BCQR) = Area(\Delta RBP) + Area(\Delta QPC) + Area(\Delta PQR) = \frac{1}{4} Area(\Delta ABC) + \frac{1}{4} Area(\Delta ABC) + \frac{1}{4} Area(\Delta ABC) = \frac{3}{4} Area(\Delta ABC)$.

(A) The area of a triangle can be calculated using the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
For $\Delta ABC$,we can express the area in two ways:
$1. \text{Area} = \frac{1}{2} \times BC \times AM$
$2. \text{Area} = \frac{1}{2} \times AB \times CN$
Equating both expressions for the area:
$\frac{1}{2} \times BC \times AM = \frac{1}{2} \times AB \times CN$
$BC \times AM = AB \times CN$
Substitute the given values: $15 \times 9.6 = 12 \times CN$
$144 = 12 \times CN$
$CN = \frac{144}{12} = 12$
Thus,$CN = 12$.

(B) In trapezium $ABCD$,$\overline{ AD } \| \overline{ BC }$ and $\overline{ AC } \cap \overline{ BD } = \{ P \}$.
Since $\overline{ AD } \| \overline{ BC }$,the alternate interior angles are equal,i.e.,$\angle PAD = \angle PCB$ and $\angle PDA = \angle PBC$.
Therefore,by $AA$ similarity criterion,$\triangle PAD \sim \triangle PCB$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{ PA }{ PC } = \frac{ PD }{ PB }$
Substituting the given values:
$\frac{5}{ PC } = \frac{9}{7.2}$
$PC = \frac{5 \times 7.2}{9} = \frac{36}{9} = 4$.
Now,$AC = PA + PC = 5 + 4 = 9$.

(C) In trapezium $ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{P\}$.
Since $\overline{AB} \parallel \overline{CD}$,the alternate interior angles are equal,i.e.,$\angle PAB = \angle PCD$ and $\angle PBA = \angle PDC$.
By the $AA$ similarity criterion,$\triangle PAB \sim \triangle PCD$.
Therefore,the ratios of their corresponding sides are equal:
$\frac{PA}{PC} = \frac{PB}{PD}$
Substituting the given values:
$\frac{10}{15} = \frac{PB}{12}$
$\frac{2}{3} = \frac{PB}{12}$
$PB = \frac{2 \times 12}{3} = 8$
Now,the length of diagonal $BD$ is:
$BD = PB + PD = 8 + 12 = 20$.

(C) An equilateral triangle has $3$ sides of equal length and $3$ angles of $60^{\circ}$ each.
Let the vertices of the equilateral triangle be $A, B,$ and $C$.
$A$ correspondence of a triangle with itself is a permutation of its vertices.
The total number of permutations of $3$ vertices is given by $3! = 3 \times 2 \times 1 = 6$.
These $6$ correspondences are:
$1. (A \to A, B \to B, C \to C)$
$2. (A \to A, B \to C, C \to B)$
$3. (A \to B, B \to A, C \to C)$
$4. (A \to B, B \to C, C \to A)$
$5. (A \to C, B \to A, C \to B)$
$6. (A \to C, B \to B, C \to A)$
Since all sides are equal and all angles are $60^{\circ}$,any permutation of the vertices results in a correspondence that preserves the ratios of corresponding sides (all ratios are $1$) and the equality of corresponding angles. Thus,all $6$ correspondences are similarities.

(C) Given that $\Delta ABC \sim \Delta XYZ$ for the correspondence $ABC \leftrightarrow XYZ$.
By the property of similar triangles,the ratios of their corresponding sides are equal.
Therefore,$\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$.
From the given equation $\frac{AB}{4} = \frac{XY}{5}$,we can rearrange the terms to find the ratio $\frac{AB}{XY}$.
Dividing both sides by $XY$ and multiplying by $4$,we get $\frac{AB}{XY} = \frac{4}{5}$.
Since $\frac{BC}{YZ} = \frac{AB}{XY}$,it follows that $\frac{BC}{YZ} = \frac{4}{5}$.

(B) The perimeter of $\Delta ABC = AB + BC + AC = 4 + 8 + 10 = 22$.
Since $\Delta ABC \sim \Delta PQR$,the ratio of their perimeters is equal to the ratio of their corresponding sides.
Therefore,$\frac{\text{Perimeter of } \Delta ABC}{\text{Perimeter of } \Delta PQR} = \frac{AC}{PR}$.
Substituting the given values: $\frac{22}{\text{Perimeter of } \Delta PQR} = \frac{10}{15}$.
Simplifying the fraction: $\frac{22}{\text{Perimeter of } \Delta PQR} = \frac{2}{3}$.
Thus,the perimeter of $\Delta PQR = \frac{22 \times 3}{2} = 11 \times 3 = 33$.

(B) Given that $\Delta ABC \sim \Delta XYZ$ for the correspondence $ABC \leftrightarrow XYZ$.
By the property of similar triangles,the ratios of their corresponding sides are equal:
$\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ} = k$
Using the property of equal ratios (componendo),we have:
$\frac{AB + BC}{XY + YZ} = \frac{AC}{XZ}$
Substituting the given values:
$\frac{7}{10.5} = \frac{5}{XZ}$
$XZ = \frac{5 \times 10.5}{7}$
$XZ = \frac{52.5}{7} = 7.5$
Therefore,$XZ = 7.5$.

(D) Given that in $\Delta ABC$,$\overline{MN} \parallel \overline{BC}$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Therefore,$\frac{AM}{AB} = \frac{AN}{AC}$.
Given $\frac{AM}{AB} = \frac{2}{3}$ and $AC = 15$.
Substituting the values,we get $\frac{2}{3} = \frac{AN}{15}$.
$AN = \frac{2 \times 15}{3} = 10$.
Since $A-N-C$,we have $AC = AN + NC$.
$15 = 10 + NC$.
$NC = 15 - 10 = 5$.

(D) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
$m \angle B = 180^{\circ} - m \angle A - m \angle C = 180^{\circ} - 30^{\circ} - 60^{\circ} = 90^{\circ}$.
Now,comparing $\Delta ABC$ and $\Delta PQR$:
$m \angle A = 30^{\circ}$ and $m \angle Q = 30^{\circ}$,so $m \angle A = m \angle Q$.
$m \angle B = 90^{\circ}$ and $m \angle P = 90^{\circ}$,so $m \angle B = m \angle P$.
By the $AA$ similarity criterion,the vertices must correspond in the order of their equal angles.
Therefore,the correspondence $ABC \leftrightarrow QPR$ is a similarity.

(D) For two similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Given,the ratio of corresponding sides $= 4:9 = \frac{4}{9}$.
Therefore,the ratio of their areas $= \left(\frac{4}{9}\right)^2 = \frac{16}{81} = 16:81$.

(A) In the given correspondence $ABC \leftrightarrow RPQ$,the vertices are mapped as follows:
$A$ corresponds to $R$
$B$ corresponds to $P$
$C$ corresponds to $Q$
Therefore,$\angle B$ corresponds to $\angle P$.

(C) In $\Delta DEF$,the sum of angles is $180^{\circ}$.
Given $m \angle E + m \angle F = 130^{\circ}$.
Therefore,$m \angle D = 180^{\circ} - (m \angle E + m \angle F) = 180^{\circ} - 130^{\circ} = 50^{\circ}$.
Since the correspondence $ABC \leftrightarrow FDE$ represents a similarity,the corresponding angles are equal.
Specifically,$\angle B$ corresponds to $\angle D$.
Therefore,$m \angle B = m \angle D = 50^{\circ}$.

(B) Given the ratio of the angles in $\Delta ABC$ is $m \angle A : m \angle B : m \angle C = 2 : 3 : 5$.
Let the angles be $2x, 3x,$ and $5x$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $2x + 3x + 5x = 180^{\circ}$.
$10x = 180^{\circ} \implies x = 18^{\circ}$.
Therefore,$m \angle A = 2(18^{\circ}) = 36^{\circ}$,$m \angle B = 3(18^{\circ}) = 54^{\circ}$,and $m \angle C = 5(18^{\circ}) = 90^{\circ}$.
Given the similarity correspondence $ABC \leftrightarrow QRP$,the corresponding angles are equal.
Thus,$m \angle A = m \angle Q$,$m \angle B = m \angle R$,and $m \angle C = m \angle P$.
Since $m \angle B = 54^{\circ}$,it follows that $m \angle R = 54^{\circ}$.

(B) Given that $\Delta XYZ \sim \Delta ABC$ for the correspondence $XYZ \leftrightarrow ABC$.
By the property of similar triangles,the ratios of their corresponding sides are equal:
$\frac{XY}{AB} = \frac{YZ}{BC} = \frac{XZ}{AC}$.
We are given that $\frac{XY}{AB} = \frac{3}{5}$.
Therefore,$\frac{YZ}{BC} = \frac{3}{5}$.
To find $\frac{BC}{YZ}$,we take the reciprocal of both sides:
$\frac{BC}{YZ} = \frac{5}{3}$.

(B) Given that $\Delta PQR \sim \Delta XYZ$ for the correspondence $PQR \leftrightarrow XYZ$.
Since the triangles are similar,the ratios of their corresponding sides are equal.
Given $PQ : QR : PR = 3 : 5 : 7$,it follows that $XY : YZ : XZ = 3 : 5 : 7$.
Let $XY = 3t$,$YZ = 5t$,and $XZ = 7t$ for some constant $t > 0$.
The perimeter of $\Delta XYZ$ is given as $22.5$.
Therefore,$XY + YZ + XZ = 22.5$.
Substituting the values in terms of $t$: $3t + 5t + 7t = 22.5$.
$15t = 22.5$.
$t = \frac{22.5}{15} = 1.5$.
Now,calculating $YZ$: $YZ = 5t = 5 \times 1.5 = 7.5$.

(B) Given that $\Delta ABC \sim \Delta MNP.$
For similar triangles,the ratio of their perimeters is equal to the ratio of their corresponding sides.
Therefore,$\frac{\text{Perimeter of } \Delta ABC}{\text{Perimeter of } \Delta MNP} = \frac{AB}{MN}.$
Given: Perimeter of $\Delta ABC = 18,$ Perimeter of $\Delta MNP = 27,$ and $AB = 8.$
Substituting the values: $\frac{18}{27} = \frac{8}{MN}.$
Simplifying the fraction $\frac{18}{27}$ gives $\frac{2}{3}.$
So,$\frac{2}{3} = \frac{8}{MN}.$
$2 \times MN = 8 \times 3.$
$2 \times MN = 24.$
$MN = \frac{24}{2} = 12.$
Thus,$MN = 12.$

(B) Given that in $\Delta ABC$ and $\Delta PQR$,we have $m \angle A = m \angle R$ and $m \angle B = m \angle Q$.
By the Angle-Angle $(AA)$ similarity criterion,if two angles of one triangle are equal to two angles of another triangle,then the two triangles are similar.
The correspondence must match the equal angles in order.
Since $\angle A = \angle R$,$\angle B = \angle Q$,and by the angle sum property,$\angle C = \angle P$.
Therefore,the correspondence $ABC \leftrightarrow RQP$ is a similarity.

(C) Given that $\Delta XYZ \sim \Delta DEF$ under the correspondence $XYZ \leftrightarrow DEF$.
This implies that the corresponding angles are equal: $\angle X = \angle D$,$\angle Y = \angle E$,and $\angle Z = \angle F$.
In $\Delta XYZ$,the sum of the angles is $180^{\circ}$.
Therefore,$m \angle X + m \angle Y + m \angle Z = 180^{\circ}$.
Substituting the given values: $50^{\circ} + 75^{\circ} + m \angle Z = 180^{\circ}$.
$125^{\circ} + m \angle Z = 180^{\circ}$.
$m \angle Z = 180^{\circ} - 125^{\circ} = 55^{\circ}$.
Since $\angle Z = \angle F$,we have $m \angle F = 55^{\circ}$.

(B) Given that $\Delta ABC \sim \Delta XYZ$ for the correspondence $ABC \leftrightarrow XYZ$.
Since the triangles are similar,the ratios of their corresponding sides are equal.
Therefore,$\frac{AB}{XY} = \frac{AC}{XZ}$.
Substituting the given values: $\frac{4}{5} = \frac{6}{XZ}$.
By cross-multiplying,we get $4 \times XZ = 5 \times 6$.
$4 \times XZ = 30$.
$XZ = \frac{30}{4} = 7.5$.

(D) Given the similarity correspondence $\Delta DEF \sim \Delta QPR$,the ratios of corresponding sides are equal.
Therefore,$\frac{DE}{QP} = \frac{DF}{QR}$.
From the given equation $2 DE = 3 PQ$,we get $\frac{DE}{PQ} = \frac{3}{2}$.
Substituting the values into the ratio: $\frac{3}{2} = \frac{DF}{8}$.
Solving for $DF$: $DF = \frac{3 \times 8}{2} = 12$.

(C) The correspondence $XYZ \leftrightarrow EFD$ indicates that the triangles are similar.
According to the theorem of areas of similar triangles,the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\text{Area}(\Delta XYZ)}{\text{Area}(\Delta DEF)} = \left(\frac{XY}{EF}\right)^2$
Given that $\text{Area}(\Delta XYZ) = 18$ and $\frac{XY}{EF} = \frac{2}{3}$.
Substituting these values into the formula:
$\frac{18}{\text{Area}(\Delta DEF)} = \left(\frac{2}{3}\right)^2$
$\frac{18}{\text{Area}(\Delta DEF)} = \frac{4}{9}$
$\text{Area}(\Delta DEF) = \frac{18 \times 9}{4}$
$\text{Area}(\Delta DEF) = \frac{162}{4} = 40.5$

(B) Given that $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow QPR.$
This implies that $\angle A = \angle Q,$ $\angle B = \angle P,$ and $\angle C = \angle R.$
We are given the ratio $m \angle P : m \angle Q : m \angle R = 2 : 3 : 4.$
Substituting the corresponding angles:
$m \angle B : m \angle A : m \angle C = 2 : 3 : 4.$
Rearranging the order to $m \angle A : m \angle B : m \angle C,$ we get $3 : 2 : 4.$
Therefore,the correct ratio is $3 : 2 : 4.$

(D) For two triangles $\Delta DEF$ and $\Delta PQR$ to be similar,the ratio of their corresponding sides must be equal,and the included angles between those sides must be equal.
Given that $m \angle D = m \angle R$,the sides forming $\angle D$ are $DE$ and $DF$,and the sides forming $\angle R$ are $RQ$ and $RP$.
According to the $SAS$ (Side-Angle-Side) similarity criterion,the triangles are similar if $\frac{DE}{RQ} = \frac{DF}{RP}$.
Rearranging this,we get $\frac{DE}{RP} = \frac{DF}{RQ}$,which matches option $D$.

(A) Given the ratio of the sides of the two triangles: $\frac{XY}{MN} = \frac{XZ}{NO} = \frac{YZ}{MO}$.
According to the $SSS$ (Side-Side-Side) similarity criterion,if the sides of two triangles are proportional,then the triangles are similar.
By observing the ratios:
$XY$ corresponds to $MN$
$XZ$ corresponds to $NO$
$YZ$ corresponds to $MO$
Therefore,the vertices must correspond in the same order: $X \leftrightarrow M$,$Y \leftrightarrow N$,and $Z \leftrightarrow O$.
Thus,the correspondence is $\Delta XYZ \sim \Delta MNO$.

(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{DE} \parallel \overline{BC}$,we have $\frac{AD}{DB} = \frac{AE}{EC}$.
Given $AB:AC = 3:4$,let $AB = 3k$ and $AC = 4k$.
By the property of similar triangles,$\Delta ADE \sim \Delta ABC$.
Therefore,$\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$.
From the given ratio,$\frac{AD}{AB} = \frac{AE}{AC} = \frac{3}{4}$.
This implies $AD = 3x$ and $AB = 4x$ (where $AB$ is the total length),but here $AB$ is given as $3k$. Thus,$AD = \frac{3}{4} AB$ and $AE = \frac{3}{4} AC$.
Then $BD = AB - AD = AB - \frac{3}{4} AB = \frac{1}{4} AB$.
Similarly,$EC = AC - AE = AC - \frac{3}{4} AC = \frac{1}{4} AC$.
Now,calculating the ratio $\frac{EC}{BD} = \frac{\frac{1}{4} AC}{\frac{1}{4} AB} = \frac{AC}{AB} = \frac{4}{3}$.
Thus,$EC:BD = 4:3$.

(C) According to the Angle Bisector Theorem,in $\Delta ABC$,if the bisector of $\angle A$ intersects $\overline{BC}$ at $D$,then $\frac{AB}{AC} = \frac{BD}{DC}$.
Given $\frac{AB}{AC} = \frac{3}{4}$ and $BD = 4.5$.
Substituting the values: $\frac{3}{4} = \frac{4.5}{DC}$.
Therefore,$DC = \frac{4.5 \times 4}{3} = 1.5 \times 4 = 6$.
Since $BC = BD + DC$,we have $BC = 4.5 + 6 = 10.5$.

(B) In $\Delta ABC$,$\overline{BE}$ is a median to side $\overline{AC}$.
Therefore,$E$ is the midpoint of $\overline{AC}$,which implies $AE = EC$,or $AC = 2EC$.
In $\Delta CEB$,$D$ is the midpoint of $\overline{BC}$ (since $\overline{AD}$ is a median) and line $m$ (passing through $D$) is parallel to $\overline{BE}$.
By the Converse of the Midpoint Theorem in $\Delta CEB$,since $D$ is the midpoint of $\overline{BC}$ and $\overline{DK} \parallel \overline{BE}$,$K$ must be the midpoint of $\overline{EC}$.
Therefore,$EK = KC = 2.5$.
Thus,$EC = EK + KC = 2.5 + 2.5 = 5$.
Finally,$AC = 2EC = 2 \times 5 = 10$.

(D) Since $G$ is the centroid of $\Delta ABC$,it divides the median $\overline{AD}$ in the ratio $2:1$. Therefore,$\frac{GD}{AD} = \frac{1}{3}$.
In $\Delta ADE$,since $\overline{GK} \parallel \overline{DE}$,by the Basic Proportionality Theorem,we have $\frac{GD}{AD} = \frac{EK}{AE} = \frac{1}{3}$.
This implies $AE = 3 \times EK = 3 \times 1.8 = 5.4$.
Since $\overline{BE}$ is a median to $\overline{AC}$,$E$ is the midpoint of $\overline{AC}$.
Therefore,$AC = 2 \times AE = 2 \times 5.4 = 10.8$.

(D) Given that $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$.
For similar triangles,the ratio of their perimeters is equal to the ratio of their corresponding sides.
$\therefore \frac{\text{Perimeter of } \Delta ABC}{\text{Perimeter of } \Delta PQR} = \frac{AC}{PR}$
Given,Perimeter of $\Delta ABC = 35$,Perimeter of $\Delta PQR = 28$,and $PR = 4\sqrt{10}$.
Substituting the values:
$\frac{35}{28} = \frac{AC}{4\sqrt{10}}$
Simplifying the fraction $\frac{35}{28}$ by dividing by $7$,we get $\frac{5}{4}$.
$\frac{5}{4} = \frac{AC}{4\sqrt{10}}$
$AC = \frac{5}{4} \times 4\sqrt{10}$
$AC = 5\sqrt{10}$

(D) In $\Delta ABC$ and $\Delta XYZ$,we are given $m\angle A = m\angle X$ and $m\angle B = m\angle Y$.
By the $AA$ (Angle-Angle) similarity criterion,$\Delta ABC \sim \Delta XYZ$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AB}{XY} = \frac{AC}{XZ} = \frac{BC}{YZ}$.
Given $\frac{AB}{XY} = \frac{2}{3}$ and $AC = 7.2$,we substitute these values into the ratio:
$\frac{2}{3} = \frac{7.2}{XZ}$.
Cross-multiplying to solve for $XZ$:
$2 \times XZ = 7.2 \times 3$.
$2 \times XZ = 21.6$.
$XZ = \frac{21.6}{2} = 10.8$.

(D) Given that the correspondence $ABC \leftrightarrow QPR$ is a similarity.
Also,the correspondence $PQR \leftrightarrow YZX$ is a similarity.
By the property of symmetry of similarity,if $PQR \leftrightarrow YZX$ is a similarity,then $QPR \leftrightarrow ZYX$ is also a similarity.
Now,we have $ABC \leftrightarrow QPR$ and $QPR \leftrightarrow ZYX$.
By the transitivity property of similarity,the correspondence $ABC \leftrightarrow ZYX$ is a similarity.

(A) According to the Angle Bisector Theorem,in $\Delta ABC$,if the bisector of $\angle B$ intersects $\overline{AC}$ at $D$,then the ratio of the segments of the opposite side is equal to the ratio of the other two sides of the triangle.
Therefore,$\frac{AB}{BC} = \frac{AD}{DC}$.
Given that $AB = 7.5$ and $\frac{AD}{DC} = \frac{3}{4}$,we substitute these values into the equation:
$\frac{7.5}{BC} = \frac{3}{4}$
By cross-multiplying,we get:
$3 \times BC = 7.5 \times 4$
$3 \times BC = 30$
$BC = \frac{30}{3}$
$BC = 10$

(C) In $\Delta ABC$,$E$ is the midpoint of $\overline{AC}$.
Therefore,$AC = 2EC$ $(1)$.
In $\Delta BEC$,line $DK$ passes through the midpoint $D$ of $\overline{BC}$ and is parallel to $\overline{BE}$.
By the Converse of the Midpoint Theorem,$K$ is the midpoint of $\overline{EC}$.
Therefore,$EC = 2EK$ $(2)$.
Substituting $(2)$ into $(1)$,we get $AC = 2(2EK) = 4EK$.

(D) By the Midpoint Theorem,the segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
Thus,$PQ = \frac{1}{2} AC$,$QR = \frac{1}{2} AB$,and $RP = \frac{1}{2} BC$.
$1$. The ratio of the sides of $\Delta PQR$ to $\Delta ABC$ is $1:2$. Thus,the perimeter of $\Delta PQR = \frac{1}{2} \times$ Perimeter of $\Delta ABC$. (Statement $C$ is true).
$2$. The area of a triangle formed by joining the midpoints is $\frac{1}{4}$ of the area of the original triangle. Thus,Area of $\Delta PQR = \frac{1}{4} \times$ Area of $\Delta ABC$. (Statement $A$ is true,Statement $D$ is false).
$3$. Since the sides are proportional,$\Delta PQR \sim \Delta CBA$. The correspondence $ABC \leftrightarrow QRP$ implies $\frac{AB}{QR} = \frac{BC}{RP} = \frac{AC}{PQ} = 2$,which is a similarity. (Statement $B$ is true).
Therefore,the statement that is not true is $D$.

(A) Given the similarity condition for triangles based on the Side-Side-Side $(SSS)$ criterion,the ratios of the corresponding sides must be equal.
We are given $\frac{AB}{PQ} = \frac{BC}{PR} = \frac{CA}{QR}$.
By observing the vertices in the ratios:
- Side $AB$ corresponds to side $PQ$.
- Side $BC$ corresponds to side $PR$.
- Side $CA$ corresponds to side $QR$.
Matching the vertices accordingly: $A$ corresponds to $P$,$B$ corresponds to $Q$,and $C$ corresponds to $R$.
Therefore,the correspondence is $ABC \leftrightarrow PQR$.

(C) Given that in $\Delta ABC$ and $\Delta DEF$,we have $\frac{AB}{DF} = \frac{BC}{EF}$ and $\angle B = \angle F$.
According to the $SAS$ (Side-Angle-Side) similarity criterion,if one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional,then the two triangles are similar.
Here,the sides $AB$ and $BC$ include $\angle B$,and the sides $DF$ and $EF$ include $\angle F$.
Since the ratios of the sides are equal and the included angles are congruent,the correspondence $ABC \leftrightarrow DFE$ is a similarity by the $SAS$ condition.

(B) The criteria for the similarity of two triangles are $AAA$ (Angle-Angle-Angle),$AA$ (Angle-Angle),$SAS$ (Side-Angle-Side),and $SSS$ (Side-Side-Side).
$ASA$ (Angle-Side-Angle) is a criterion for the congruence of triangles,not for their similarity.
Therefore,$ASA$ is not a condition for the similarity of two triangles.

(C) In $\triangle OPQ$ and $\triangle OSR$,we are given $OQ = OR$.
Also,$\angle POQ = \angle SOR$ because they are vertically opposite angles.
To prove the triangles are congruent by the $SAS$ (Side-Angle-Side) congruence criterion,we need another side adjacent to the angle $\angle POQ$ and $\angle SOR$ to be equal.
The sides adjacent to the angles are $OP$ and $OQ$ in $\triangle OPQ$,and $OS$ and $OR$ in $\triangle OSR$.
Since we already have $OQ = OR$,we need $OP = OS$ to satisfy the $SAS$ condition.
Therefore,the condition $OP = OS$ is sufficient.

(B) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta XYZ$,$YM$ is the bisector of $\angle Y$,which intersects $\overline{XZ}$ at $M$.
Therefore,by the Angle Bisector Theorem,the ratio of the segments of the side $\overline{XZ}$ is equal to the ratio of the other two sides of the triangle:
$\frac{XM}{MZ} = \frac{XY}{YZ}$.
Thus,option $B$ is correct.

(D) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta ABC$,since $AD$ is the bisector of $\angle A$,it divides the side $BC$ such that:
$\frac{BD}{DC} = \frac{AB}{AC}$
By cross-multiplying the terms,we get:
$BD \times AC = DC \times AB$
Therefore,the correct option is $D$.

(B) $1$. In $\Delta ABC$ and $\Delta PQR$,if $\angle A = \angle P$ and $\angle C = \angle Q$,then by $AA$ similarity,$\Delta ABC \sim \Delta PRQ$. Thus,$(1-d)$.
$2$. In $\Delta ABC$ and $\Delta PQR$,if $\frac{AB}{QR} = \frac{BC}{PQ}$ and $\angle B = \angle Q$,then by $SAS$ similarity,$\Delta ABC \sim \Delta RQP$. Thus,$(2-a)$.
$3$. In $\Delta ABC$ and $\Delta PQR$,if $\frac{AB}{PQ} = \frac{BC}{PR} = \frac{CA}{QR}$,then by $SSS$ similarity,$\Delta ABC \sim \Delta PQR$. Thus,$(3-c)$.
$4$. In $\Delta ABC$ and $\Delta PQR$,if $\frac{AB}{PQ} = \frac{CA}{PR}$ and $\angle A = \angle P$,then by $SAS$ similarity,$\Delta ABC \sim \Delta QPR$. Thus,$(4-b)$.
Therefore,the correct matching is $(1-d), (2-a), (3-c), (4-b)$. Note: The provided options seem to have a slight mismatch in the sequence,but based on standard similarity criteria,the correct pairing is $(1-d), (2-a), (3-c), (4-b)$.

$(A)$ To determine the correct matching, we analyze the similarity criteria for triangles:
$1.$ Given $\frac{AB}{XY} = \frac{BC}{YZ}$ and $\angle B \cong \angle Y$. This satisfies the $SAS$ (Side-Angle-Side) similarity criterion. Thus, $1-b$.
$2.$ Given $\angle A \cong \angle X, \angle B \cong \angle Y$, and $\angle C \cong \angle Z$. This satisfies the $AAA$ (Angle-Angle-Angle) similarity criterion. Thus, $2-c$.
$3.$ Given $\frac{AB}{XY} = \frac{BC}{YZ} = \frac{CA}{ZX}$. This satisfies the $SSS$ (Side-Side-Side) similarity criterion. Since $SSS$ is not listed as an option in Part $II$, this corresponds to $d$ (None of the conditions apply).
Therefore, the correct matching is $1-b, 2-c, 3-d$.

(B) In $\Delta ABC$,$\angle B = 90^{\circ}$.
According to Pythagoras' theorem,in a right-angled triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Here,the side opposite to $\angle B$ is $\overline{AC}$,which is the hypotenuse.
Therefore,$AC^{2} = AB^{2} + BC^{2}$.