Criteria for Similarity of Triangles Questions in English

(B) It is given that $\Delta ABC \sim \Delta DEF$.
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.
$\therefore \frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta DEF)} = \left(\frac{BC}{EF}\right)^2$
Given that $\text{ar}(\Delta ABC) = 64 \text{ cm}^2$,$\text{ar}(\Delta DEF) = 121 \text{ cm}^2$,and $EF = 15.4 \text{ cm}$.
$\frac{64}{121} = \left(\frac{BC}{15.4}\right)^2$
Taking the square root on both sides:
$\sqrt{\frac{64}{121}} = \frac{BC}{15.4}$
$\frac{8}{11} = \frac{BC}{15.4}$
$BC = \frac{8 \times 15.4}{11}$
$BC = 8 \times 1.4 = 11.2 \text{ cm}$.

(4:1) Since $AB \parallel CD,$
$\therefore \angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$ (Alternate interior angles).
In $\triangle AOB$ and $\triangle COD,$
$\angle AOB = \angle COD$ (Vertically opposite angles).
$\angle OAB = \angle OCD$ (Alternate interior angles).
$\angle OBA = \angle ODC$ (Alternate interior angles).
$\therefore \triangle AOB \sim \triangle COD$ (By $AAA$ similarity criterion).
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\triangle AOB)}{\operatorname{ar}(\triangle COD)} = \left(\frac{AB}{CD}\right)^2$.
Given $AB = 2 CD,$
$\therefore \frac{\operatorname{ar}(\triangle AOB)}{\operatorname{ar}(\triangle COD)} = \left(\frac{2 CD}{CD}\right)^2 = \left(\frac{2}{1}\right)^2 = \frac{4}{1} = 4:1$.
Thus,the ratio of the areas of triangles $AOB$ and $COD$ is $4:1$.

(N/A) Let us draw two perpendiculars $AP$ and $DM$ on line $BC$.
We know that the area of a triangle $= \frac{1}{2} \times \text{Base} \times \text{Height}$.
Therefore,$\frac{\operatorname{ar}(\Delta ABC)}{\operatorname{ar}(\Delta DBC)} = \frac{\frac{1}{2} \times BC \times AP}{\frac{1}{2} \times BC \times DM} = \frac{AP}{DM}$.
In $\triangle APO$ and $\triangle DMO$:
$\angle APO = \angle DMO = 90^{\circ}$ (By construction).
$\angle AOP = \angle DOM$ (Vertically opposite angles).
Therefore,$\triangle APO \sim \triangle DMO$ (By $AA$ similarity criterion).
Since the triangles are similar,the ratio of their corresponding sides is equal:
$\frac{AP}{DM} = \frac{AO}{DO}$.
Substituting this into our area ratio equation:
$\frac{\operatorname{ar}(\Delta ABC)}{\operatorname{ar}(\Delta DBC)} = \frac{AO}{DO}$.

(N/A) Let us assume two similar triangles as $\triangle ABC \sim \triangle PQR$.
According to the theorem of areas of similar triangles:
$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{AC}{PR}\right)^2$ $...(1)$
Given that,$\text{ar}(\triangle ABC) = \text{ar}(\triangle PQR)$.
Therefore,$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)} = 1$.
Substituting this value in equation $(1)$,we get:
$1 = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{AC}{PR}\right)^2$.
This implies:
$\left(\frac{AB}{PQ}\right)^2 = 1 \Rightarrow AB^2 = PQ^2 \Rightarrow AB = PQ$
$\left(\frac{BC}{QR}\right)^2 = 1 \Rightarrow BC^2 = QR^2 \Rightarrow BC = QR$
$\left(\frac{AC}{PR}\right)^2 = 1 \Rightarrow AC^2 = PR^2 \Rightarrow AC = PR$
Since all three corresponding sides are equal,by the $SSS$ (Side-Side-Side) congruence criterion,$\triangle ABC \cong \triangle PQR$.

(1:4) Given that $D, E$ and $F$ are the mid-points of sides $AB, BC$ and $CA$ respectively in $\Delta ABC$.
By the Mid-point Theorem,the line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.
Therefore,$DE \parallel AC$ and $DE = \frac{1}{2} AC$.
Similarly,$EF \parallel AB$ and $EF = \frac{1}{2} AB$,and $DF \parallel BC$ and $DF = \frac{1}{2} BC$.
In $\Delta DEF$ and $\Delta ABC$:
$\frac{DE}{AC} = \frac{EF}{AB} = \frac{DF}{BC} = \frac{1}{2}$.
By $SSS$ similarity criterion,$\Delta DEF \sim \Delta ABC$.
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\operatorname{ar}(\Delta DEF)}{\operatorname{ar}(\Delta ABC)} = \left(\frac{DE}{AC}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
Thus,the ratio of the areas of $\Delta DEF$ and $\Delta ABC$ is $1:4$.

(N/A) Let us assume two similar triangles $\Delta ABC \sim \Delta PQR$. Let $AD$ and $PS$ be the medians of these triangles.
Since $\Delta ABC \sim \Delta PQR$,we have:
$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} \quad ...(1)$
And $\angle B = \angle Q \quad ...(2)$
Since $AD$ and $PS$ are medians,$D$ and $S$ are midpoints of $BC$ and $QR$ respectively.
Therefore,$BD = \frac{BC}{2}$ and $QS = \frac{QR}{2}$.
Substituting these into equation $(1)$:
$\frac{AB}{PQ} = \frac{2BD}{2QS} = \frac{BD}{QS} \quad ...(3)$
In $\triangle ABD$ and $\triangle PQS$:
$\angle B = \angle Q$ [From $(2)$]
$\frac{AB}{PQ} = \frac{BD}{QS}$ [From $(3)$]
Therefore,$\triangle ABD \sim \triangle PQS$ by $SAS$ similarity criterion.
This implies that $\frac{AB}{PQ} = \frac{AD}{PS} \quad ...(4)$
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
$\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta PQR)} = \left(\frac{AB}{PQ}\right)^2$
From equation $(4)$,substituting $\frac{AB}{PQ} = \frac{AD}{PS}$:
$\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta PQR)} = \left(\frac{AD}{PS}\right)^2$
Hence proved.

(N/A) In $\Delta ACD$ and $\Delta ABC$:
$\angle ADC = \angle ACB = 90^{\circ}$
$\angle CAD = \angle CAB$ (Common angle)
Therefore,$\Delta ACD \sim \Delta ABC$ (by $AA$ similarity criterion).
This implies $\frac{AC}{AB} = \frac{AD}{AC}$,so $AC^{2} = AB \cdot AD$ $...(1)$
In $\Delta BCD$ and $\Delta BAC$:
$\angle BDC = \angle BCA = 90^{\circ}$
$\angle CBD = \angle ABC$ (Common angle)
Therefore,$\Delta BCD \sim \Delta BAC$ (by $AA$ similarity criterion).
This implies $\frac{BC}{BA} = \frac{BD}{BC}$,so $BC^{2} = BA \cdot BD$ $...(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{BC^{2}}{AC^{2}} = \frac{BA \cdot BD}{AB \cdot AD}$
Since $BA = AB$,we get:
$\frac{BC^{2}}{AC^{2}} = \frac{BD}{AD}$

(C) If two triangles are similar to each other,then the ratio of the areas of these triangles is equal to the square of the ratio of their corresponding sides.
Given that the ratio of the sides is $4:9$.
Therefore,the ratio of the areas of these triangles $= (4/9)^2 = 16/81$.
Hence,the correct answer is $(C)$.

(N/A) Let $\angle MPR = x$.
In $\Delta MPR$,
$\angle MRP = 180^{\circ} - 90^{\circ} - x = 90^{\circ} - x$.
Similarly,in $\Delta MPQ$,
$\angle MPQ = 90^{\circ} - \angle MPR = 90^{\circ} - x$.
In $\Delta MPQ$,
$\angle MQP = 180^{\circ} - 90^{\circ} - (90^{\circ} - x) = x$.
Now,in $\Delta QMP$ and $\Delta PMR$:
$\angle MPQ = \angle MRP = 90^{\circ} - x$
$\angle MQP = \angle MPR = x$
$\angle PMQ = \angle RMP = 90^{\circ}$
Therefore,$\Delta QMP \sim \Delta PMR$ (by $AA$ similarity criterion).
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{QM}{PM} = \frac{PM}{MR}$
$\Rightarrow PM^{2} = QM \cdot MR$.

(N/A) Given: In $\triangle ABD$,$\angle BAD = 90^\circ$ and $AC \perp BD$.
To prove: $AB^2 = BC \cdot BD$.
Proof: Consider $\triangle BCA$ and $\triangle BAD$.
In $\triangle BCA$ and $\triangle BAD$:
$\angle BCA = \angle BAD = 90^\circ$ (Given $AC \perp BD$ and $\angle BAD = 90^\circ$)
$\angle B = \angle B$ (Common angle)
Therefore,by $AA$ similarity criterion,$\triangle BCA \sim \triangle BAD$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{BC}{BA} = \frac{BA}{BD}$
Cross-multiplying gives:
$BA \cdot BA = BC \cdot BD$
$AB^2 = BC \cdot BD$.
Hence proved.

(N/A) Given: In $\triangle ABD$,$\angle A = 90^{\circ}$ and $AC \perp BD$.
To prove: $AC^{2} = BC \cdot DC$.
Proof:
In $\triangle BCA$ and $\triangle ACD$:
$\angle BCA = \angle ACD = 90^{\circ}$ (Since $AC \perp BD$)
$\angle CBA = 90^{\circ} - \angle CAB$ (In $\triangle ABC$,$\angle B + \angle CAB = 90^{\circ}$)
Also,$\angle CAD = 90^{\circ} - \angle CAB$ (Since $\angle A = 90^{\circ}$,$\angle CAD + \angle CAB = 90^{\circ}$)
Therefore,$\angle CBA = \angle CAD$.
By $AA$ similarity criterion,$\triangle BCA \sim \triangle ACD$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{BC}{AC} = \frac{AC}{DC}$
Cross-multiplying,we get:
$AC^{2} = BC \cdot DC$.
Hence proved.

(N/A) Given: In $\triangle ABD$,$\angle A = 90^{\circ}$ and $AC \perp BD$.
To prove: $AD^{2} = BD \cdot CD$.
Proof: Consider $\triangle ACD$ and $\triangle BCA$.
In $\triangle ACD$ and $\triangle BCA$:
$\angle ACD = \angle BCA = 90^{\circ}$ (Since $AC \perp BD$ and $\angle A = 90^{\circ}$).
$\angle CAD = \angle CBA$ (Since both are complementary to $\angle BAD$ or $\angle BDA$).
Therefore,by $AA$ similarity criterion,$\triangle ACD \sim \triangle BCA$.
Since the triangles are similar,the ratio of their corresponding sides is equal:
$\frac{CD}{AC} = \frac{AC}{BC} = \frac{AD}{AB}$.
Wait,let us consider $\triangle ACD$ and $\triangle BAD$.
In $\triangle ACD$ and $\triangle BAD$:
$\angle ACD = \angle BAD = 90^{\circ}$.
$\angle D = \angle D$ (Common angle).
Therefore,by $AA$ similarity criterion,$\triangle ACD \sim \triangle BAD$.
Since the triangles are similar,the ratio of their corresponding sides is equal:
$\frac{AC}{BA} = \frac{CD}{AD} = \frac{AD}{BD}$.
Taking the last two parts of the ratio:
$\frac{CD}{AD} = \frac{AD}{BD}$.
Cross-multiplying gives:
$AD^{2} = BD \cdot CD$.
Hence proved.

(N/A) Join $DB.$
We have $DN \parallel CB,$ $DM \parallel AB,$ and $\angle B = 90^{\circ}.$
Therefore,$DMBN$ is a rectangle.
Thus,$DN = MB$ and $DM = NB.$
Since $D$ is the foot of the perpendicular drawn from $B$ to $AC,$ we have $\angle CDB = 90^{\circ}.$
$\angle 2 + \angle 3 = 90^{\circ} \dots(1)$
In $\Delta CDM, \angle 1 + \angle 2 + \angle DMC = 180^{\circ} \Rightarrow \angle 1 + \angle 2 = 90^{\circ} \dots(2)$
In $\Delta DMB, \angle 3 + \angle DMB + \angle 4 = 180^{\circ} \Rightarrow \angle 3 + \angle 4 = 90^{\circ} \dots(3)$
From $(1)$ and $(2),$ we get $\angle 1 = \angle 3.$
From $(1)$ and $(3),$ we get $\angle 2 = \angle 4.$
In $\Delta DCM$ and $\Delta BDM,$
$\angle 1 = \angle 3$ (Proved above)
$\angle 2 = \angle 4$ (Proved above)
Therefore,$\Delta DCM \sim \Delta BDM$ ($AA$ similarity criterion).
Thus,$\frac{BM}{DM} = \frac{DM}{MC}.$
Since $BM = DN,$ we have $\frac{DN}{DM} = \frac{DM}{MC}.$
Therefore,$DM^{2} = DN \cdot MC.$

(N/A) In quadrilateral $DMBN,$ since $\angle DMB = 90^{\circ},$ $\angle DNB = 90^{\circ},$ and $\angle MBN = 90^{\circ}$ (given $\angle ABC = 90^{\circ}$),the fourth angle $\angle MDN$ must also be $90^{\circ}.$ Thus,$DMBN$ is a rectangle.
Therefore,$DM = NB$ and $DN = MB.$
In $\Delta ABC,$ $BD \perp AC.$ By the property of similar triangles in a right triangle,$\Delta ADN \sim \Delta ABD$ and $\Delta BDC \sim \Delta ABD.$
More directly,in $\Delta DNB$ and $\Delta ADN:$
$\angle DNB = \angle DNA = 90^{\circ}.$
$\angle ND B = \angle DAN$ (since both are complementary to $\angle ADN$).
Thus,$\Delta DNB \sim \Delta ADN$ by $AA$ similarity.
Therefore,$\frac{DN}{AN} = \frac{NB}{DN}.$
$DN^{2} = AN \cdot NB.$
Since $NB = DM,$ we have $DN^{2} = AN \cdot DM.$

(N/A) To prove that $\Delta APC \sim \Delta DPB$,we follow these steps:
$1$. Join $CB$ and $AD$.
$2$. Consider $\Delta APC$ and $\Delta DPB$.
$3$. $\angle APC = \angle DPB$ (Vertically opposite angles).
$4$. $\angle CAP = \angle BDP$ (Angles subtended by the same arc $CB$ in the same segment of the circle are equal).
$5$. Therefore,by the $AA$ (Angle-Angle) similarity criterion,$\Delta APC \sim \Delta DPB$.

(N/A) To prove $AP \cdot PB = CP \cdot DP$,we consider the triangles $\Delta APC$ and $\Delta DPB$.
$1$. $\angle APC = \angle DPB$ (Vertically opposite angles).
$2$. $\angle CAP = \angle BDP$ (Angles in the same segment of a circle are equal).
By the $AA$ (Angle-Angle) similarity criterion,$\Delta APC \sim \Delta DPB$.
Since the corresponding sides of similar triangles are proportional,we have:
$\frac{AP}{DP} = \frac{PC}{PB} = \frac{AC}{DB}$
Taking the first two ratios:
$\frac{AP}{DP} = \frac{PC}{PB}$
By cross-multiplying,we get:
$AP \cdot PB = PC \cdot DP$ (or $AP \cdot PB = CP \cdot DP$).

(N/A) Consider $\Delta PAC$ and $\Delta PDB$.
$1$. $\angle P = \angle P$ (Common angle).
$2$. In a cyclic quadrilateral $ABDC$,the exterior angle is equal to the interior opposite angle. Therefore,$\angle PAC = \angle PDB$ (Since $\angle PAC + \angle CAB = 180^{\circ}$ and $\angle PDB + \angle CAB = 180^{\circ}$).
$3$. By the $AA$ (Angle-Angle) similarity criterion,$\Delta PAC \sim \Delta PDB$.

(N/A) To prove $PA \cdot PB = PC \cdot PD$,consider $\triangle PAD$ and $\triangle PCB$.
$1$. $\angle P = \angle P$ (Common angle).
$2$. $\angle PDA = \angle PBC$ (Exterior angle of a cyclic quadrilateral is equal to the interior opposite angle,or since $\angle PAD + \angle DAB = 180^{\circ}$ and $\angle BCD + \angle DAB = 180^{\circ}$,we have $\angle PAD = \angle BCD$,and thus $\triangle PAD \sim \triangle PCB$ by $AA$ similarity).
Since $\triangle PAD \sim \triangle PCB$,the ratios of their corresponding sides are equal:
$\frac{PA}{PC} = \frac{PD}{PB} = \frac{AD}{CB}$.
Taking the first two parts:
$\frac{PA}{PC} = \frac{PD}{PB}$
By cross-multiplying,we get:
$PA \cdot PB = PC \cdot PD$.

(B) Given,the ratio of corresponding sides of two similar triangles is $2:3$.
Let the area of the smaller triangle be $A_1 = 48 \, cm^2$ and the area of the larger triangle be $A_2$.
According to the theorem of the area of similar triangles,the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\frac{A_1}{A_2} = \left(\frac{2}{3}\right)^2$
$\frac{48}{A_2} = \frac{4}{9}$
By cross-multiplying,we get:
$4 \times A_2 = 48 \times 9$
$A_2 = \frac{432}{4} = 108 \, cm^2$.
Therefore,the area of the larger triangle is $108 \, cm^2$.

(N/A) Given: In $\Delta PQR$,$N$ is a point on $PR$ such that $QN \perp PR$ and $PN \cdot NR = QN^2$.
To prove: $\angle PQR = 90^{\circ}$.
Proof: We have $PN \cdot NR = QN^2$.
This can be written as $\frac{PN}{QN} = \frac{QN}{NR}$.
In $\Delta QNP$ and $\Delta RNQ$:
$1$. $\frac{PN}{QN} = \frac{QN}{NR}$ (Given)
$2$. $\angle PNQ = \angle RNQ = 90^{\circ}$ (Given $QN \perp PR$)
By $SAS$ similarity criterion,$\Delta QNP \sim \Delta RNQ$.
Since the triangles are similar,their corresponding angles are equal:
$\angle PQN = \angle QRN$ (let this be $\alpha$)
$\angle RQN = \angle QPN$ (let this be $\beta$)
In $\Delta PQR$,the sum of angles is $180^{\circ}$:
$\angle P + \angle R + \angle PQR = 180^{\circ}$
$\angle QPN + \angle QRN + (\angle PQN + \angle RQN) = 180^{\circ}$
Substituting the equal angles:
$\beta + \alpha + (\alpha + \beta) = 180^{\circ}$
$2(\alpha + \beta) = 180^{\circ}$
$\alpha + \beta = 90^{\circ}$
Since $\angle PQR = \alpha + \beta$,we have $\angle PQR = 90^{\circ}$.
Hence proved.

(D) Given,area of smaller triangle $= 36\, cm^{2}$ and area of larger triangle $= 100\, cm^{2}$.
Also,length of a side of the larger triangle $= 20\, cm$.
Let the length of the corresponding side of the smaller triangle be $x\, cm$.
By the property 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.
$\frac{\text{Area of larger triangle}}{\text{Area of smaller triangle}} = \frac{(\text{Side of larger triangle})^{2}}{(\text{Side of smaller triangle})^{2}}$
$\Rightarrow \frac{100}{36} = \frac{(20)^{2}}{x^{2}}$
$\Rightarrow x^{2} = \frac{20^{2} \times 36}{100} = \frac{400 \times 36}{100} = 4 \times 36 = 144$
$x = \sqrt{144} = 12\, cm$.
Hence,the length of the corresponding side of the smaller triangle is $12\, cm$.

(A) Given: $AC = 8 \, cm$,$AD = 3 \, cm$ and $\angle ACB = \angle CDA = 90^{\circ}$ (from the figure).
In right-angled $\triangle ADC$,by Pythagoras theorem:
$AC^2 = AD^2 + CD^2$
$8^2 = 3^2 + CD^2$
$64 = 9 + CD^2$
$CD^2 = 55$
$CD = \sqrt{55} \, cm$.
Now,consider $\triangle CDB$ and $\triangle ADC$:
$1$. $\angle BDC = \angle ADC = 90^{\circ}$ (given).
$2$. $\angle DBC = \angle DCA$ (since both are equal to $90^{\circ} - \angle A$).
By $AA$ similarity criterion,$\triangle CDB \sim \triangle ADC$.
Therefore,the ratio of corresponding sides is equal:
$\frac{CD}{BD} = \frac{AD}{CD}$
$CD^2 = AD \times BD$
$55 = 3 \times BD$
$BD = \frac{55}{3} \, cm$.

(B) Let $BC = 15\, m$ be the height of the tower and $AB = 24\, m$ be the length of its shadow. At that time,the angle of elevation of the sun is $\theta$,so $\angle CAB = \theta$.
Let $EF = h$ be the height of the telephone pole and $DE = 16\, m$ be the length of its shadow. At the same time,the angle of elevation of the sun is the same,so $\angle EDF = \theta$.
In $\triangle ABC$ and $\triangle DEF$:
$\angle CAB = \angle EDF = \theta$ (Angle of elevation of the sun)
$\angle ABC = \angle DEF = 90^{\circ}$ (Both are vertical objects on level ground)
Therefore,$\triangle ABC \sim \triangle DEF$ by $AA$ similarity criterion.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AB}{DE} = \frac{BC}{EF}$
Substituting the given values:
$\frac{24}{16} = \frac{15}{h}$
Solving for $h$:
$h = \frac{15 \times 16}{24}$
$h = \frac{15 \times 2}{3}$
$h = 5 \times 2 = 10\, m$.
Hence,the height of the telephone pole is $10\, m$.

(D) Given,$\angle A = \angle C$,$AB = 6 \, cm$,$BP = 15 \, cm$,$AP = 12 \, cm$,and $CP = 4 \, cm$.
In $\triangle APB$ and $\triangle CPD$:
$\angle A = \angle C$ [Given]
$\angle APB = \angle CPD$ [Vertically opposite angles]
Therefore,$\triangle APB \sim \triangle CPD$ [By $AA$ similarity criterion]
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AP}{CP} = \frac{BP}{PD} = \frac{AB}{CD}$
Substituting the given values:
$\frac{12}{4} = \frac{15}{PD} = \frac{6}{CD}$
Taking the first two terms:
$\frac{12}{4} = \frac{15}{PD} \implies 3 = \frac{15}{PD} \implies PD = \frac{15}{3} = 5 \, cm$
Taking the first and last terms:
$\frac{12}{4} = \frac{6}{CD} \implies 3 = \frac{6}{CD} \implies CD = \frac{6}{3} = 2 \, cm$
Hence,the length of $PD = 5 \, cm$ and the length of $CD = 2 \, cm$.

(N/A) Given,$\triangle ABC \sim \triangle EDF$. Since the triangles are similar,their corresponding sides are in the same ratio.
i.e.,$\frac{AB}{ED} = \frac{AC}{EF} = \frac{BC}{DF}$ ...... $(i)$
Given values are $AB = 5 \, cm$,$AC = 7 \, cm$,$DF = 15 \, cm$,and $DE = 12 \, cm$.
Substituting these values in Eq. $(i)$,we get:
$\frac{5}{12} = \frac{7}{EF} = \frac{BC}{15}$
Taking the first and second terms:
$\frac{5}{12} = \frac{7}{EF}$
$\Rightarrow EF = \frac{7 \times 12}{5} = \frac{84}{5} = 16.8 \, cm$
Taking the first and third terms:
$\frac{5}{12} = \frac{BC}{15}$
$\Rightarrow BC = \frac{5 \times 15}{12} = \frac{75}{12} = 6.25 \, cm$
Hence,the lengths of the remaining sides are $EF = 16.8 \, cm$ and $BC = 6.25 \, cm$.

(N/A) Given,$\triangle ABC$ in which $\angle B = 90^{\circ}$ and $BD \perp AC$.
Also,$AD = 4 \, cm$ and $CD = 5 \, cm$.
In $\triangle ADB$ and $\triangle CDB$,$\angle ADB = \angle CDB = 90^{\circ}$.
Also,$\angle BAD = \angle DBC$ (since both are equal to $90^{\circ} - \angle C$).
Therefore,$\triangle ADB \sim \triangle CDB$ by $AA$ similarity criterion.
Thus,$\frac{BD}{AD} = \frac{CD}{BD}$.
$\Rightarrow BD^2 = AD \times CD = 4 \times 5 = 20$.
$\Rightarrow BD = \sqrt{20} = 2\sqrt{5} \, cm$.
Now,in right-angled $\triangle ADB$,by Pythagoras theorem:
$AB^2 = AD^2 + BD^2 = 4^2 + (2\sqrt{5})^2 = 16 + 20 = 36$.
$\Rightarrow AB = \sqrt{36} = 6 \, cm$.
Hence,$BD = 2\sqrt{5} \, cm$ and $AB = 6 \, cm$.

(N/A) In $\Delta ABC$,the ratio of the angles is $m \angle A : m \angle B : m \angle C = 15 : 8 : 7$.
Let the angles be $15t$,$8t$,and $7t$ respectively.
Since the sum of angles in a triangle is $180^\circ$,we have $15t + 8t + 7t = 180^\circ$.
$30t = 180^\circ$,which gives $t = 6^\circ$.
Therefore,$m \angle A = 15 \times 6^\circ = 90^\circ$,$m \angle B = 8 \times 6^\circ = 48^\circ$,and $m \angle C = 7 \times 6^\circ = 42^\circ$.
Given the similarity correspondence $ABC \leftrightarrow QPR$,the corresponding angles are equal.
Thus,$m \angle Q = m \angle A = 90^\circ$,$m \angle P = m \angle B = 48^\circ$,and $m \angle R = m \angle C = 42^\circ$.
Hence,the angles of $\Delta PQR$ are $m \angle P = 48^\circ$,$m \angle Q = 90^\circ$,and $m \angle R = 42^\circ$.

(C) Given that the correspondence $ABC \leftrightarrow XYZ$ between $\Delta ABC$ and $\Delta XYZ$ is a similarity.
Therefore,the ratios of their corresponding sides are equal:
$\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$
From the given equation $4AB = 5XY$,we can write:
$\frac{AB}{XY} = \frac{5}{4}$
Since the triangles are similar,we have:
$\frac{AB}{XY} = \frac{BC}{YZ}$
Substituting the known values:
$\frac{5}{4} = \frac{12}{YZ}$
Solving for $YZ$:
$YZ = \frac{12 \times 4}{5} = \frac{48}{5} = 9.6$
Thus,$YZ = 9.6$.

(D) Since $\Delta ABC \sim \Delta PQR$,the corresponding angles are equal: $m \angle A = m \angle P$,$m \angle B = m \angle Q$,and $m \angle C = m \angle R = 100^{\circ}$.
In $\Delta ABC$,the sum of angles is $180^{\circ}$,so $m \angle A + m \angle B + m \angle C = 180^{\circ}$.
Substituting $m \angle C = 100^{\circ}$,we get $m \angle A + m \angle B = 80^{\circ}$.
Given $2 m \angle P = 3 m \angle Q$,and since $m \angle P = m \angle A$ and $m \angle Q = m \angle B$,we have $2 m \angle A = 3 m \angle B$,which implies $m \angle A = 1.5 m \angle B$.
Substitute this into $m \angle A + m \angle B = 80^{\circ}$:
$1.5 m \angle B + m \angle B = 80^{\circ}$
$2.5 m \angle B = 80^{\circ}$
$m \angle B = 80 / 2.5 = 32^{\circ}$.

(A) Given that the correspondence $PQR \leftrightarrow ZYX$ is a similarity,we have $\Delta PQR \sim \Delta ZYX$.
From the property of similar triangles,the ratios of corresponding sides are equal:
$\frac{PQ}{ZY} = \frac{QR}{YX} = \frac{PR}{ZX}$.
We are given $\frac{PQ}{ZY} = \frac{5}{3}$ and $PR = 10$.
Substituting these values into the ratio $\frac{PQ}{ZY} = \frac{PR}{ZX}$,we get:
$\frac{5}{3} = \frac{10}{ZX}$.
Cross-multiplying,we get $5 \times ZX = 3 \times 10$.
$5 \times ZX = 30$.
$ZX = \frac{30}{5} = 6$.
Therefore,the length of $ZX$ is $6$.

(A) Given that $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow QRP$.
This implies that the corresponding angles are equal:
$m \angle A = m \angle Q$
$m \angle B = m \angle R$
$m \angle C = m \angle P$
Given $m \angle A = 80^{\circ}, m \angle B = 40^{\circ}, m \angle C = 60^{\circ}$.
Substituting these values:
$m \angle Q = 80^{\circ}$
$m \angle R = 40^{\circ}$
$m \angle P = 60^{\circ}$
Thus,the angles of $\Delta PQR$ are $m \angle P = 60^{\circ}, m \angle Q = 80^{\circ}, m \angle R = 40^{\circ}$.

(C) Given that $\Delta XYZ \sim \Delta DEF$ with the correspondence $XYZ \leftrightarrow EDF$.
This implies the ratio of corresponding sides is equal: $\frac{XY}{ED} = \frac{YZ}{DF} = \frac{ZX}{FE} = k$.
Given $XY = 3, YZ = 4, ZX = 6$ and $DF = 12$.
From the ratio $\frac{YZ}{DF} = \frac{4}{12} = \frac{1}{3}$,the scale factor $k = \frac{1}{3}$.
Thus,$\frac{XY}{ED} = \frac{1}{3} \implies \frac{3}{ED} = \frac{1}{3} \implies ED = 9$.
And $\frac{ZX}{FE} = \frac{1}{3} \implies \frac{6}{FE} = \frac{1}{3} \implies FE = 18$.
The perimeter of $\Delta DEF = ED + DF + FE = 9 + 12 + 18 = 39$.

(D) Given that $\Delta ABC \sim \Delta XYZ$ under the correspondence $ABC \leftrightarrow ZYX$.
This implies that the ratios of corresponding sides are equal: $\frac{AB}{ZY} = \frac{BC}{YX} = \frac{CA}{XZ} = k$.
From the given values,$AB = 3, BC = 5, CA = 6$ and $XY = 6$.
Using the correspondence $ABC \leftrightarrow ZYX$,we have the ratios: $\frac{AB}{ZY} = \frac{BC}{YX} = \frac{CA}{XZ}$.
Substituting the known values: $\frac{3}{ZY} = \frac{5}{6} = \frac{6}{XZ}$.
From $\frac{5}{6} = \frac{3}{ZY}$,we get $ZY = \frac{3 \times 6}{5} = \frac{18}{5} = 3.6$.
From $\frac{5}{6} = \frac{6}{XZ}$,we get $XZ = \frac{6 \times 6}{5} = \frac{36}{5} = 7.2$.
The perimeter of $\Delta XYZ = XY + YZ + ZX = 6 + 3.6 + 7.2 = 16.8$.

(A) In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^{\circ}$.
Given $m \angle A + m \angle B = 130^{\circ}$,substituting this into the sum gives $130^{\circ} + m \angle C = 180^{\circ}$,so $m \angle C = 50^{\circ}$.
Given $m \angle B + m \angle C = 125^{\circ}$,substituting $m \angle C = 50^{\circ}$ gives $m \angle B + 50^{\circ} = 125^{\circ}$,so $m \angle B = 75^{\circ}$.
Since $m \angle A + m \angle B = 130^{\circ}$,we have $m \angle A + 75^{\circ} = 130^{\circ}$,so $m \angle A = 55^{\circ}$.
Given the correspondence $ABC \leftrightarrow QPR$,the corresponding angles are $\angle A \leftrightarrow \angle Q$,$\angle B \leftrightarrow \angle P$,and $\angle C \leftrightarrow \angle R$.
Therefore,$m \angle Q = m \angle A = 55^{\circ}$.

(N/A) Two triangles are said to be similar if:
$1$. Their corresponding angles are equal.
$2$. Their corresponding sides are in the same ratio (or proportion).

(C) Given that $\Delta ABC \sim \Delta RPQ$.
By the property of similar triangles,the corresponding angles are equal.
The correspondence is $A \leftrightarrow R$,$B \leftrightarrow P$,and $C \leftrightarrow Q$.
Therefore,$m\angle A = m\angle R = 35^{\circ}$ and $m\angle B = m\angle P = 50^{\circ}$.
In $\Delta ABC$,the sum of angles is $180^{\circ}$.
$m\angle C = 180^{\circ} - (m\angle A + m\angle B) = 180^{\circ} - (35^{\circ} + 50^{\circ}) = 180^{\circ} - 85^{\circ} = 95^{\circ}$.
Since $C \leftrightarrow Q$,we have $m\angle C = m\angle Q$.
Thus,$m\angle Q = 95^{\circ}$.

(D) Given that $\Delta ABC \sim \Delta XYZ$ under the correspondence $ABC \leftrightarrow XZY$.
This implies that the ratio of the corresponding sides is equal to the ratio of their perimeters.
Therefore,$\frac{AB}{XZ} = \frac{BC}{ZY} = \frac{AC}{XY} = \frac{\text{Perimeter of } \Delta ABC}{\text{Perimeter of } \Delta XYZ}$.
Given: $\text{Perimeter of } \Delta ABC = 45$,$\text{Perimeter of } \Delta XYZ = 30$,and $AB = 21$.
Substituting the values into the ratio: $\frac{21}{XZ} = \frac{45}{30}$.
Simplifying the fraction $\frac{45}{30}$ gives $\frac{3}{2}$.
So,$\frac{21}{XZ} = \frac{3}{2}$.
Cross-multiplying gives $3 \times XZ = 21 \times 2$.
$3 \times XZ = 42$.
$XZ = \frac{42}{3} = 14$.
Thus,the length of $XZ$ is $14$.

(A) Since $\Delta PQR \sim \Delta XYZ$,the ratios of their corresponding sides are equal:
$\frac{PQ}{XY} = \frac{QR}{YZ} = \frac{PR}{XZ}$.
From the given equation $\frac{PQ}{3} = \frac{XY}{5}$,we can rearrange it to get $\frac{PQ}{XY} = \frac{3}{5}$.
Substituting this ratio into the similarity condition $\frac{PQ}{XY} = \frac{QR}{YZ}$,we get:
$\frac{3}{5} = \frac{9}{YZ}$.
Cross-multiplying,we find $3 \times YZ = 5 \times 9$,which simplifies to $3 \times YZ = 45$.
Therefore,$YZ = \frac{45}{3} = 15$.

(A) Given that $\Delta ABC \sim \Delta PQR$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$.
Substituting the given values:
$\frac{3}{PQ} = \frac{4}{6} = \frac{5}{PR}$.
First,solve for $PQ$:
$\frac{3}{PQ} = \frac{4}{6} \implies \frac{3}{PQ} = \frac{2}{3} \implies 2 \cdot PQ = 9 \implies PQ = 4.5$.
Next,solve for $PR$:
$\frac{4}{6} = \frac{5}{PR} \implies \frac{2}{3} = \frac{5}{PR} \implies 2 \cdot PR = 15 \implies PR = 7.5$.
Thus,$PQ = 4.5$ and $PR = 7.5$.

(C) Since $\Delta ABC \sim \Delta DEF$,the ratios of their corresponding sides are equal:
$\frac{AB}{DE} = \frac{AC}{DF} = \frac{BC}{EF} = k$
Given $AB = 8$,$DE = 12$,$AC = 10$,and $EF = 18$.
The ratio of similarity $k = \frac{AB}{DE} = \frac{8}{12} = \frac{2}{3}$.
Now,find the remaining sides of $\Delta DEF$:
$\frac{AC}{DF} = \frac{2}{3} \implies \frac{10}{DF} = \frac{2}{3} \implies DF = \frac{10 \times 3}{2} = 15$.
$\frac{BC}{EF} = \frac{2}{3} \implies \frac{BC}{18} = \frac{2}{3} \implies BC = \frac{18 \times 2}{3} = 12$.
Now,find the perimeter of $\Delta DEF = DE + EF + DF = 12 + 18 + 15 = 45$.

(D) Given the correspondence $ABC \leftrightarrow XZY$,the triangles are similar,i.e.,$\Delta ABC \sim \Delta XZY$.
From the property of similar triangles,the ratio of corresponding sides is equal:
$\frac{AB}{XZ} = \frac{BC}{ZY} = \frac{AC}{XY}$.
Given $AB = 6$,$XY = 12$,$YZ = 6$,and $ZX = 9$.
Using the correspondence $ABC \leftrightarrow XZY$,the sides are proportional as follows:
$\frac{AB}{XZ} = \frac{BC}{ZY} = \frac{AC}{XY}$
$\frac{6}{9} = \frac{BC}{6} = \frac{AC}{12}$
First,find $BC$: $\frac{6}{9} = \frac{BC}{6} \implies BC = \frac{6 \times 6}{9} = \frac{36}{9} = 4$.
Next,find $AC$: $\frac{6}{9} = \frac{AC}{12} \implies AC = \frac{6 \times 12}{9} = \frac{72}{9} = 8$.
The perimeter of $\Delta ABC = AB + BC + AC = 6 + 4 + 8 = 18$.

(A) Given that $\Delta ABC \sim \Delta PQR.$
According to the properties of similar triangles,the ratio of their perimeters is equal to the ratio of their corresponding sides.
Therefore,$\frac{\text{Perimeter of } \Delta ABC}{\text{Perimeter of } \Delta PQR} = \frac{AB}{PQ}.$
Substituting the given values: $\frac{16}{24} = \frac{6}{PQ}.$
Simplifying the ratio $\frac{16}{24}$ gives $\frac{2}{3}.$
So,$\frac{2}{3} = \frac{6}{PQ}.$
Cross-multiplying gives $2 \times PQ = 18.$
Thus,$PQ = \frac{18}{2} = 9.$
The correct option is $A$.

(B) In $\Delta PQR$,the sum of angles is $180^{\circ}$.
$m \angle P + m \angle Q + m \angle R = 180^{\circ}$.
$80^{\circ} + 40^{\circ} + m \angle R = 180^{\circ}$.
$120^{\circ} + m \angle R = 180^{\circ} \implies m \angle R = 60^{\circ}$.
Given the correspondence $PQR \leftrightarrow YZX$,the corresponding angles are equal:
$m \angle P = m \angle Y = 80^{\circ}$,
$m \angle Q = m \angle Z = 40^{\circ}$,
$m \angle R = m \angle X = 60^{\circ}$.
Therefore,$m \angle Z = m \angle Q = 40^{\circ}$.

(C) Given that $\Delta XYZ \sim \Delta DEF$,the corresponding angles are equal: $m \angle X = m \angle D$,$m \angle Y = m \angle E$,and $m \angle Z = m \angle F$.
In $\Delta XYZ$,the sum of angles is $180^{\circ}$,so $m \angle X + m \angle Y + m \angle Z = 180^{\circ}$.
Given $2 m \angle X = 3 m \angle Y$,we have $m \angle X = 1.5 m \angle Y$.
Substituting the values into the sum equation: $1.5 m \angle Y + m \angle Y + 30^{\circ} = 180^{\circ}$.
$2.5 m \angle Y = 150^{\circ}$.
$m \angle Y = 150^{\circ} / 2.5 = 60^{\circ}$.
Since $\Delta XYZ \sim \Delta DEF$,$m \angle E = m \angle Y = 60^{\circ}$.

(D) Given that $\Delta PQR \sim \Delta XYZ$ under the correspondence $PQR \leftrightarrow ZYX$,the ratios of their corresponding sides are equal.
This implies that $\frac{PQ}{ZY} = \frac{QR}{YX} = \frac{PR}{ZX}$.
We are given $PQ = 6$,$QR = 8$,and $XY = 12$.
From the correspondence $PQR \leftrightarrow ZYX$,the side $QR$ corresponds to $YX$.
Thus,$\frac{PQ}{ZY} = \frac{QR}{YX}$.
Substituting the given values: $\frac{6}{YZ} = \frac{8}{12}$.
Simplifying the fraction $\frac{8}{12}$ gives $\frac{2}{3}$.
So,$\frac{6}{YZ} = \frac{2}{3}$.
Cross-multiplying gives $2 \times YZ = 6 \times 3$,which means $2 \times YZ = 18$.
Therefore,$YZ = \frac{18}{2} = 9$.

(N/A) Given that $\Delta PQR \sim \Delta DEF$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{PQ}{DE} = \frac{QR}{EF} = \frac{PR}{DF}$.
From the given equation $3 PQ = 2 DE$,we get $\frac{PQ}{DE} = \frac{2}{3}$.
Now,using the ratio $\frac{PQ}{DE} = \frac{QR}{EF}$:
$\frac{2}{3} = \frac{QR}{6} \implies QR = \frac{2 \times 6}{3} = 4$.
Next,using the ratio $\frac{PQ}{DE} = \frac{PR}{DF}$:
$\frac{2}{3} = \frac{8}{DF} \implies DF = \frac{8 \times 3}{2} = 12$.
Thus,$QR = 4$ and $DF = 12$.

(B) Given that $\Delta ABC \sim \Delta PQR$.
According to the properties of similar triangles,the ratio of their perimeters is equal to the ratio of their corresponding sides.
Therefore,$\frac{\text{Perimeter of } \Delta ABC}{\text{Perimeter of } \Delta PQR} = \frac{AB}{PQ}$.
We are given that $\frac{AB}{PQ} = \frac{3}{4}$ and the perimeter of $\Delta PQR = 24$.
Let the perimeter of $\Delta ABC$ be $x$.
Substituting the values,we get $\frac{x}{24} = \frac{3}{4}$.
$x = \frac{3}{4} \times 24$.
$x = 3 \times 6 = 18$.
Thus,the perimeter of $\Delta ABC$ is $18$.

(C) Given that $\Delta PQR \sim \Delta XYZ$,the ratios of their corresponding sides are equal.
Therefore,$\frac{PQ}{XY} = \frac{QR}{YZ} = \frac{PR}{XZ}$.
From the given equation $3 QR = 4 YZ$,we can write the ratio $\frac{QR}{YZ} = \frac{4}{3}$.
Since the triangles are similar,the ratio of corresponding sides is constant,so $\frac{PR}{XZ} = \frac{QR}{YZ} = \frac{4}{3}$.
Substituting the given value $PR = 8$ into the equation,we get $\frac{8}{XZ} = \frac{4}{3}$.
Solving for $XZ$: $4 XZ = 8 \times 3$,which means $4 XZ = 24$.
Thus,$XZ = \frac{24}{4} = 6$.

(D) Since $\Delta ABC \sim \Delta XYZ$,the ratio of their corresponding sides is equal.
Therefore,$\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$.
Given that $\frac{AB}{XY} = \frac{4}{5}$ and $YZ = 20$.
Substituting the values into the ratio $\frac{AB}{XY} = \frac{BC}{YZ}$,we get:
$\frac{4}{5} = \frac{BC}{20}$.
Multiplying both sides by $20$,we get:
$BC = \frac{4}{5} \times 20$.
$BC = 4 \times 4 = 16$.
Thus,the length of $BC$ is $16$.

(A) Since $\Delta ABC \sim \Delta PQR$,the ratio of their corresponding sides is equal.
Therefore,$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = k$ (where $k$ is the scale factor).
From the properties of ratios,if $\frac{a}{b} = \frac{c}{d} = k$,then $\frac{a+c}{b+d} = k$.
Thus,$\frac{AB + BC}{PQ + QR} = \frac{AC}{PR}$.
Substituting the given values: $\frac{12}{15} = \frac{8}{PR}$.
Simplifying the fraction: $\frac{4}{5} = \frac{8}{PR}$.
Cross-multiplying gives: $4 \times PR = 5 \times 8$.
$4 \times PR = 40$.
$PR = \frac{40}{4} = 10$.
Hence,the length of $PR$ is $10$.