Criteria for Similarity of 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) 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, $\text{Area}(\Delta ABC) = 12 \sqrt{3} \times 4 = 48 \sqrt{3}$.

(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) 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 $= \left(\frac{4}{9}\right)^2 = \frac{16}{81} = 16:81$.

(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 $\Delta XYZ \sim \Delta MNO$.

(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.

(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.

(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 $1-b, 2-c, 3-d$.

(C) In $\Delta ABC$ and $\Delta ADB$:
$1$. $m \angle ABC = m \angle ADB = 90^{\circ}$ (Given).
$2$. $m \angle BAC = m \angle DAB$ (Common angle).
By the $AA$ (Angle-Angle) similarity criterion,the triangles are similar.
Therefore,the correspondence $ABC \leftrightarrow ADB$ is a similarity.

(A) In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$.
According to the geometric mean theorem for right-angled triangles,the altitude to the hypotenuse divides the triangle into two triangles similar to the original and to each other.
Thus,$BM^2 = AM \cdot CM$.
Given $AC = 13$ and $CM = 9$,we have $AM = AC - CM = 13 - 9 = 4$.
Substituting these values into the formula: $BM^2 = 4 \cdot 9 = 36$.
Therefore,$BM = \sqrt{36} = 6$.

(B) In $\Delta ABC$,$m \angle B = 90^{\circ}$.
Using the Pythagorean theorem,$AC^2 = AB^2 + BC^2 = 8^2 + 6^2 = 64 + 36 = 100$.
Therefore,$AC = \sqrt{100} = 10$.
The area of $\Delta ABC$ can be calculated in two ways: $\text{Area} = \frac{1}{2} \times AB \times BC$ or $\text{Area} = \frac{1}{2} \times AC \times BM$.
Equating the two expressions: $\frac{1}{2} \times AB \times BC = \frac{1}{2} \times AC \times BM$.
$AB \times BC = AC \times BM$.
$8 \times 6 = 10 \times BM$.
$48 = 10 \times BM$.
$BM = \frac{48}{10} = 4.8$.