Mix Examples - Triangles Questions in English

(A) In $\triangle OAC$ and $\triangle ODB$:
$1$. $\angle AOC = \angle DOB$ (Vertically opposite angles).
$2$. $\angle OAC = \angle ODB$ (Angles subtended by the same arc $CB$ at the circumference are equal).
$3$. $\angle OCA = \angle OBD$ (Angles subtended by the same arc $AD$ at the circumference are equal).
Thus, $\triangle OAC \sim \triangle ODB$ by $AAA$ similarity criterion.
Given $OB = OD$, in $\triangle ODB$, the angles opposite to these sides must be equal, i.e., $\angle ODB = \angle OBD$. Since $\angle OAC = \angle ODB$ and $\angle OCA = \angle OBD$, it follows that $\angle OAC = \angle OCA$. Therefore, $\triangle OAC$ is an isosceles triangle with $OA = OC$.
Since $\triangle OAC \sim \triangle ODB$ and $\triangle OAC$ is isosceles, $\triangle ODB$ must also be isosceles.
Hence, the triangles are isosceles and similar.

(D) We know that the diagonals of a rhombus are perpendicular bisectors of each other.
Let the rhombus be $ABCD$ with diagonals $AC = 16 \, cm$ and $BD = 12 \, cm$ intersecting at $O$.
Since the diagonals bisect each other at right angles,we have:
$AO = \frac{AC}{2} = \frac{16}{2} = 8 \, cm$
$BO = \frac{BD}{2} = \frac{12}{2} = 6 \, cm$
In the right-angled triangle $\triangle AOB$,by Pythagoras theorem:
$AB^2 = AO^2 + BO^2$
$AB^2 = 8^2 + 6^2 = 64 + 36 = 100$
$AB = \sqrt{100} = 10 \, cm$
Thus,the length of the side of the rhombus is $10 \, cm$.

(C) In $\triangle APB$ and $\triangle DPC$,we have:
$\angle APB = \angle DPC = 50^{\circ}$ [Vertically opposite angles]
Now,consider the ratios of the sides including the equal angles:
$\frac{PA}{PD} = \frac{6}{5} = 1.2$
$\frac{PB}{PC} = \frac{3}{2.5} = \frac{30}{25} = 1.2$
Since $\frac{PA}{PD} = \frac{PB}{PC}$ and the included angles are equal,by the $SAS$ similarity criterion:
$\triangle APB \sim \triangle DPC$
Since the triangles are similar,their corresponding angles are equal:
$\angle PAB = \angle PDC = 30^{\circ}$
In $\triangle APB$,the sum of angles is $180^{\circ}$:
$\angle PAB + \angle APB + \angle PBA = 180^{\circ}$
$30^{\circ} + 50^{\circ} + \angle PBA = 180^{\circ}$
$80^{\circ} + \angle PBA = 180^{\circ}$
$\angle PBA = 180^{\circ} - 80^{\circ} = 100^{\circ}$

(C) Given that $\triangle ABC \sim \triangle DFE$,the corresponding angles are equal and the corresponding sides are proportional.
Therefore,$\angle A = \angle D = 30^{\circ}$,$\angle B = \angle F$,and $\angle C = \angle E = 50^{\circ}$.
In $\triangle ABC$,the sum of angles is $180^{\circ}$,so $\angle B = 180^{\circ} - (30^{\circ} + 50^{\circ}) = 100^{\circ}$.
Thus,$\angle F = 100^{\circ}$.
From the similarity $\triangle ABC \sim \triangle DFE$,we have the ratio of corresponding sides: $\frac{AB}{DF} = \frac{AC}{DE} = \frac{BC}{FE}$.
Substituting the given values: $\frac{5}{7.5} = \frac{8}{DE}$.
Solving for $DE$: $DE = \frac{8 \times 7.5}{5} = \frac{60}{5} = 12 \, cm$.
Hence,$DE = 12 \, cm$ and $\angle F = 100^{\circ}$.

(A) Given,$\triangle ABC \sim \triangle QRP$.
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Therefore,$\frac{\operatorname{ar}(\triangle ABC)}{\operatorname{ar}(\triangle QRP)} = \frac{BC^2}{RP^2}$.
Given that $\frac{\operatorname{ar}(\triangle ABC)}{\operatorname{ar}(\triangle QRP)} = \frac{9}{4}$ and $BC = 15 \, cm$.
Substituting the values,we get $\frac{9}{4} = \frac{(15)^2}{RP^2}$.
$\frac{9}{4} = \frac{225}{RP^2}$.
$RP^2 = \frac{225 \times 4}{9} = 25 \times 4 = 100$.
$RP = \sqrt{100} = 10 \, cm$.
Since $\triangle ABC \sim \triangle QRP$,the corresponding side to $PR$ is $AC$. However,the question asks for $PR$ based on the similarity ratio. From the similarity $\triangle ABC \sim \triangle QRP$,we have $\frac{BC}{RP} = \frac{AC}{QP} = \frac{AB}{QR}$.
Thus,$RP = 10 \, cm$.

(B) Given,in $\triangle PQR$,$PS = QS = RS$ ... $(1)$
In $\triangle PSR$,$PS = RS$ (from Eq. $1$),therefore $\angle 1 = \angle 2$ ... $(2)$ (angles opposite to equal sides are equal).
Similarly,in $\triangle RSQ$,$RS = QS$ (from Eq. $1$),therefore $\angle 3 = \angle 4$ ... $(3)$ (angles opposite to equal sides are equal).
In $\triangle PQR$,the sum of angles is $180^{\circ}$.
$\angle P + \angle Q + \angle R = 180^{\circ}$
$\angle 2 + \angle 4 + (\angle 1 + \angle 3) = 180^{\circ}$
Substituting $\angle 2 = \angle 1$ and $\angle 4 = \angle 3$ from $(2)$ and $(3)$:
$\angle 1 + \angle 3 + \angle 1 + \angle 3 = 180^{\circ}$
$2(\angle 1 + \angle 3) = 180^{\circ}$
$\angle 1 + \angle 3 = 90^{\circ}$
Thus,$\angle R = 90^{\circ}$.
In $\triangle PQR$,by Pythagoras theorem,$PR^2 + QR^2 = PQ^2$.

(N/A) No,$\triangle QPR$ is not similar to $\triangle TSM$.
We know that the sum of the three angles of a triangle is $180^{\circ}$.
In $\triangle PQR$,$\angle P + \angle Q + \angle R = 180^{\circ}$.
$55^{\circ} + 25^{\circ} + \angle R = 180^{\circ}$.
$\angle R = 180^{\circ} - 80^{\circ} = 100^{\circ}$.
In $\triangle TSM$,$\angle T + \angle S + \angle M = 180^{\circ}$.
$\angle T + 25^{\circ} + 100^{\circ} = 180^{\circ}$.
$\angle T = 180^{\circ} - 125^{\circ} = 55^{\circ}$.
Comparing the angles of $\triangle PQR$ and $\triangle TSM$:
$\angle P = 55^{\circ} = \angle T$
$\angle Q = 25^{\circ} = \angle S$
$\angle R = 100^{\circ} = \angle M$
Since all corresponding angles are equal,the triangles are similar,but the correct correspondence is $\triangle PQR \sim \triangle TSM$.
Therefore,$\triangle QPR$ is not similar to $\triangle TSM$ because the order of vertices does not match the correspondence of equal angles.

(N/A) No,it is not correct to say that $\triangle PQD \sim \triangle RPD$ in general.
In $\triangle PQD$ and $\triangle RPD$:
$1$. $\angle PDQ = \angle PDR = 90^{\circ}$ (Given that $PD \perp QR$)
$2$. $PD = PD$ (Common side)
For two triangles to be similar,we need either $AA$,$SAS$,or $SSS$ similarity criteria. Here,we only have one angle and one side equal. We do not have information about the equality of other angles or the proportionality of other sides.
Therefore,$\triangle PQD$ is not necessarily similar to $\triangle RPD$ unless $\triangle PQR$ is a specific type of triangle (e.g.,if $\angle P = 90^{\circ}$ and $PD$ is the altitude to the hypotenuse,then $\triangle PQD \sim \triangle RPD$ by $AA$ similarity).

(B) Let $ABC$ be a right triangle right-angled at $B$ with $AB = 16 \,cm$ and $BC = 8 \,cm$. The largest square $BRSP$ that can be inscribed in this triangle is shown in the figure.
Let the side of the square be $x \,cm$. So,$PB = x \,cm$ and $BR = x \,cm$.
Then $AP = AB - PB = (16 - x) \,cm$.
In $\triangle APS$ and $\triangle ABC$,$\angle A = \angle A$ (common) and $\angle APS = \angle ABC = 90^{\circ}$.
Therefore,$\triangle APS \sim \triangle ABC$ by $AA$ similarity.
This implies $\frac{AP}{AB} = \frac{PS}{BC}$.
Substituting the values,$\frac{16 - x}{16} = \frac{x}{8}$.
Cross-multiplying,$8(16 - x) = 16x$.
$128 - 8x = 16x$.
$128 = 24x$.
$x = \frac{128}{24} = \frac{16}{3} \,cm$.

(N/A) Given: $\triangle NSQ \cong \triangle MTR$ and $\angle 1 = \angle 2.$
To prove: $\triangle PTS \sim \triangle PRQ.$
Proof: Since $\triangle NSQ \cong \triangle MTR,$ by $CPCT$,we have $SQ = TR$ and $NS = MT.$
Also,given $\angle 1 = \angle 2,$ in $\triangle PST,$ sides opposite to equal angles are equal,so $PT = PS.$
From the congruence $\triangle NSQ \cong \triangle MTR,$ we have $NQ = MR.$ Subtracting $SQ = TR$ from both sides,we get $NQ - SQ = MR - TR,$ which implies $NS = MT$ (already known) or more simply,from $\triangle NSQ \cong \triangle MTR,$ we have $SQ = TR.$
Since $PT = PS$ and $PQ = PT + TQ$ and $PR = PS + SR,$ and given the symmetry,we can show $\frac{PS}{PQ} = \frac{PT}{PR}.$
Since $\angle P = \angle P$ (common) and $\frac{PS}{PQ} = \frac{PT}{PR},$ by $SAS$ similarity criterion,$\triangle PTS \sim \triangle PRQ.$

(B) Let $ABC$ be an equilateral triangle with side length $a = 8 \, cm$.
In an equilateral triangle,the altitude $AD$ drawn from vertex $A$ to the base $BC$ bisects the base.
Therefore,$BD = CD = \frac{BC}{2} = \frac{8}{2} = 4 \, cm$.
In the right-angled triangle $ABD$,by Pythagoras theorem:
$AB^2 = AD^2 + BD^2$
$8^2 = AD^2 + 4^2$
$64 = AD^2 + 16$
$AD^2 = 64 - 16 = 48$
$AD = \sqrt{48} = \sqrt{16 \times 3} = 4 \sqrt{3} \, cm$.
Alternatively,the altitude of an equilateral triangle is given by the formula $h = \frac{\sqrt{3}}{2} \times \text{side}$.
$h = \frac{\sqrt{3}}{2} \times 8 = 4 \sqrt{3} \, cm$.

(TRUE) True.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
For $\triangle ABC$:
$AB = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2 + 0^2} = 4$
$BC = \sqrt{(0 - 2)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$
$AC = \sqrt{(0 - (-2))^2 + (2 - 0)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$
For $\triangle DEF$:
$DE = \sqrt{(4 - (-4))^2 + (0 - 0)^2} = \sqrt{8^2 + 0^2} = 8$
$EF = \sqrt{(0 - 4)^2 + (4 - 0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
$DF = \sqrt{(0 - (-4))^2 + (4 - 0)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
Comparing the ratios of corresponding sides:
$\frac{AB}{DE} = \frac{4}{8} = \frac{1}{2}$
$\frac{BC}{EF} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2}$
$\frac{AC}{DF} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2}$
Since $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = \frac{1}{2}$,by the $SSS$ similarity criterion,$\triangle ABC \sim \triangle DEF$.

(N/A) Join $OA$,$OB$,and $OC$.
Applying the Pythagoras theorem in $\triangle AOF$,we obtain:
$OA^{2} = OF^{2} + AF^{2}$ --- $(i)$
Similarly,in $\triangle BOD$:
$OB^{2} = OD^{2} + BD^{2}$ --- $(ii)$
Similarly,in $\triangle COE$:
$OC^{2} = OE^{2} + CE^{2}$ --- $(iii)$
Adding equations $(i)$,$(ii)$,and $(iii)$,we get:
$OA^{2} + OB^{2} + OC^{2} = OF^{2} + AF^{2} + OD^{2} + BD^{2} + OE^{2} + CE^{2}$
Rearranging the terms,we get:
$OA^{2} + OB^{2} + OC^{2} - OD^{2} - OE^{2} - OF^{2} = AF^{2} + BD^{2} + CE^{2}$
Hence,the result is proved.

(N/A) Join $OA$,$OB$,and $OC$.
In right-angled triangles $\triangle AFO$,$\triangle BDO$,and $\triangle CEO$,by applying the Pythagoras theorem:
$OA^{2} = AF^{2} + OF^{2} \implies AF^{2} = OA^{2} - OF^{2}$ ... $(i)$
$OB^{2} = BD^{2} + OD^{2} \implies BD^{2} = OB^{2} - OD^{2}$ ... $(ii)$
$OC^{2} = CE^{2} + OE^{2} \implies CE^{2} = OC^{2} - OE^{2}$ ... $(iii)$
Adding $(i)$,$(ii)$,and $(iii)$,we get:
$AF^{2} + BD^{2} + CE^{2} = (OA^{2} - OF^{2}) + (OB^{2} - OD^{2}) + (OC^{2} - OE^{2})$ ... $(iv)$
Similarly,in right-angled triangles $\triangle AEO$,$\triangle CDO$,and $\triangle BFO$:
$OA^{2} = AE^{2} + OE^{2} \implies AE^{2} = OA^{2} - OE^{2}$
$OC^{2} = CD^{2} + OD^{2} \implies CD^{2} = OC^{2} - OD^{2}$
$OB^{2} = BF^{2} + OF^{2} \implies BF^{2} = OB^{2} - OF^{2}$
Adding these,we get:
$AE^{2} + CD^{2} + BF^{2} = (OA^{2} - OE^{2}) + (OC^{2} - OD^{2}) + (OB^{2} - OF^{2})$
Rearranging the terms,this is equal to the right-hand side of equation $(iv)$.
Thus,$AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}$.

(N/A) Given: $AD$ is the median of $\Delta ABC$,so $BD = DC = \frac{BC}{2}$.
Also,$AM \perp BC$.
In right-angled $\Delta ABM$,by Pythagoras theorem:
$AB^{2} = AM^{2} + BM^{2}$
In right-angled $\Delta ADM$,by Pythagoras theorem:
$AD^{2} = AM^{2} + DM^{2} \implies AM^{2} = AD^{2} - DM^{2}$
Substituting $AM^{2}$ in the first equation:
$AB^{2} = (AD^{2} - DM^{2}) + BM^{2}$
Since $BM = BD - DM$,we have:
$AB^{2} = AD^{2} - DM^{2} + (BD - DM)^{2}$
$AB^{2} = AD^{2} - DM^{2} + BD^{2} + DM^{2} - 2 \cdot BD \cdot DM$
$AB^{2} = AD^{2} + BD^{2} - 2 \cdot BD \cdot DM$
Since $BD = \frac{BC}{2}$,substituting this value:
$AB^{2} = AD^{2} + \left(\frac{BC}{2}\right)^{2} - 2 \cdot \left(\frac{BC}{2}\right) \cdot DM$
$AB^{2} = AD^{2} + \left(\frac{BC}{2}\right)^{2} - BC \cdot DM$

(N/A) Applying Pythagoras theorem in $\Delta ABM$,we obtain:
$AM^2 + MB^2 = AB^2 \quad \dots(1)$
Applying Pythagoras theorem in $\Delta AMC$,we obtain:
$AM^2 + MC^2 = AC^2 \quad \dots(2)$
Adding equations $(1)$ and $(2)$,we obtain:
$2AM^2 + MB^2 + MC^2 = AB^2 + AC^2$
Since $AD$ is a median,$BD = DC = \frac{BC}{2}$.
We can write $MB = BD - DM$ and $MC = DC + DM = BD + DM$.
Substituting these in the equation:
$2AM^2 + (BD - DM)^2 + (BD + DM)^2 = AB^2 + AC^2$
$2AM^2 + (BD^2 + DM^2 - 2BD \cdot DM) + (BD^2 + DM^2 + 2BD \cdot DM) = AB^2 + AC^2$
$2AM^2 + 2BD^2 + 2DM^2 = AB^2 + AC^2$
$2(AM^2 + DM^2) + 2BD^2 = AB^2 + AC^2$
In $\Delta ADM$,by Pythagoras theorem,$AM^2 + DM^2 = AD^2$.
So,$2AD^2 + 2(\frac{BC}{2})^2 = AB^2 + AC^2$
$2AD^2 + 2(\frac{BC^2}{4}) = AB^2 + AC^2$
$2AD^2 + \frac{BC^2}{2} = AB^2 + AC^2$

(N/A) Let $A$ be the tip of the fishing rod and $B$ be the point on the water surface directly below $A$. Let $C$ be the initial position of the fly. Given $AB = 1.8 \, m$ and $BC = 2.4 \, m$.
In the right-angled triangle $\triangle ABC$,by Pythagoras theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = (1.8)^2 + (2.4)^2 = 3.24 + 5.76 = 9.00$
$AC = \sqrt{9} = 3 \, m$.
So,the length of the string out is $3 \, m$.
She pulls the string at a rate of $5 \, cm/s = 0.05 \, m/s$.
In $12 \, seconds$,the length of the string pulled in is $12 \times 0.05 = 0.6 \, m$.
Let the new position of the fly be $D$ after $12 \, seconds$. The new length of the string is $AD = AC - 0.6 = 3 - 0.6 = 2.4 \, m$.
In the right-angled triangle $\triangle ADB$:
$AD^2 = AB^2 + BD^2$
$(2.4)^2 = (1.8)^2 + BD^2$
$5.76 = 3.24 + BD^2$
$BD^2 = 5.76 - 3.24 = 2.52$
$BD = \sqrt{2.52} \approx 1.587 \, m$.
The total horizontal distance of the fly from Nazima is $BD + 1.2 \, m = 1.587 + 1.2 = 2.787 \, m \approx 2.79 \, m$.

(C) Let $AB$ be the vertical wall and $AC = 10\,m$ be the length of the ladder.
The distance of the foot of the ladder from the base of the wall is $BC = 6\,m$.
In the right-angled $\triangle ABC$,by the Pythagoras theorem:
$AC^2 = AB^2 + BC^2$
$(10)^2 = AB^2 + (6)^2$
$100 = AB^2 + 36$
$AB^2 = 100 - 36 = 64$
$AB = \sqrt{64} = 8\,m$.
Thus,the height of the point on the wall where the top of the ladder reaches is $8\,m$.

(N/A) In $\triangle AOF$ and $\triangle BOD$:
$\angle O = \angle O$ (Common angle) and $\angle A = \angle B = 90^{\circ}$.
Therefore,$\triangle AOF \sim \triangle BOD$ ($AA$ similarity).
So,$\frac{OA}{OB} = \frac{FA}{DB}$ .........$(1)$
Also,in $\triangle FAC$ and $\triangle EBC$:
$\angle A = \angle B = 90^{\circ}$ and $\angle FCA = \angle ECB$ (Vertically opposite angles).
Therefore,$\triangle FAC \sim \triangle EBC$ ($AA$ similarity).
So,$\frac{FA}{EB} = \frac{AC}{BC}$.
Since $OB$ is the perpendicular bisector of $DE$,$EB = DB$.
So,$\frac{FA}{DB} = \frac{AC}{BC}$ ......$(2)$
From $(1)$ and $(2)$,we have:
$\frac{AC}{BC} = \frac{OA}{OB}$.
Since $AC = OC - OA$ and $BC = OB - OC$,we get:
$\frac{OC - OA}{OB - OC} = \frac{OA}{OB}$.
$OB(OC - OA) = OA(OB - OC)$.
$OB \cdot OC - OA \cdot OB = OA \cdot OB - OA \cdot OC$.
$OB \cdot OC + OA \cdot OC = 2(OA \cdot OB)$.
Dividing both sides by $(OA \cdot OB \cdot OC)$:
$\frac{OB \cdot OC}{OA \cdot OB \cdot OC} + \frac{OA \cdot OC}{OA \cdot OB \cdot OC} = \frac{2(OA \cdot OB)}{OA \cdot OB \cdot OC}$.
$\frac{1}{OA} + \frac{1}{OB} = \frac{2}{OC}$.

(B) Distance travelled by the first aeroplane in $1 \frac{1}{2}$ hours (or $1.5$ hours) $= 300 \times 1.5 = 450 \, km$.
Distance travelled by the second aeroplane in $1.5$ hours $= 400 \times 1.5 = 600 \, km$.
Since one aeroplane flies North and the other flies West,the paths are perpendicular to each other,forming a right-angled triangle with the airport at the vertex of the right angle.
Let $O$ be the airport,$A$ be the position of the first aeroplane,and $B$ be the position of the second aeroplane.
Then $OA = 450 \, km$ and $OB = 600 \, km$.
Using the Pythagoras theorem in $\triangle AOB$:
$AB^2 = OA^2 + OB^2$
$AB^2 = (450)^2 + (600)^2$
$AB^2 = 202500 + 360000$
$AB^2 = 562500$
$AB = \sqrt{562500} = 750 \, km$.
Thus,the two aeroplanes will be $750 \, km$ apart.

(N/A) Given that $\triangle ABC \sim \triangle DEF$.
Therefore,the ratios of their corresponding sides are equal:
$\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD}$
Substituting the given expressions:
$\frac{2x - 1}{18} = \frac{2x + 2}{3x + 9} = \frac{3x}{6x}$
Taking the first and third ratios:
$\frac{2x - 1}{18} = \frac{3x}{6x}$
$\frac{2x - 1}{18} = \frac{1}{2}$
Cross-multiplying:
$2(2x - 1) = 18$
$4x - 2 = 18$
$4x = 20$
$x = 5$
Now,calculating the side lengths:
For $\triangle ABC$:
$AB = 2(5) - 1 = 9 \, cm$
$BC = 2(5) + 2 = 12 \, cm$
$CA = 3(5) = 15 \, cm$
For $\triangle DEF$:
$DE = 18 \, cm$
$EF = 3(5) + 9 = 24 \, cm$
$FD = 6(5) = 30 \, cm$

(D) Let $AC$ be the ladder of length $5\,m$ and $BC = 4\,m$ be the height of the wall where the ladder is placed. If the foot of the ladder is moved $1.6\,m$ towards the wall,i.e.,$AD = 1.6\,m$,then the ladder slides upward to a new position $ED$.
In right-angled $\triangle ABC$,by Pythagoras theorem:
$AC^2 = AB^2 + BC^2$
$5^2 = AB^2 + 4^2$
$25 = AB^2 + 16$
$AB^2 = 9 \Rightarrow AB = 3\,m$.
Now,the new distance of the foot from the wall is $DB = AB - AD = 3 - 1.6 = 1.4\,m$.
In right-angled $\triangle EBD$,by Pythagoras theorem:
$ED^2 = EB^2 + DB^2$
$5^2 = EB^2 + (1.4)^2$
$25 = EB^2 + 1.96$
$EB^2 = 25 - 1.96 = 23.04$
$EB = \sqrt{23.04} = 4.8\,m$.
The distance by which the top of the ladder slides upwards is $EC = EB - BC = 4.8 - 4 = 0.8\,m$.

(B) Let $BC = 18 \, m$ be the flag pole and its shadow be $AB = 9.6 \, m$. The distance of the top of the pole $(C)$ from the far end of the shadow $(A)$ is $AC$.
In right-angled $\triangle ABC$,by Pythagoras theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = (9.6)^2 + (18)^2$
$AC^2 = 92.16 + 324$
$AC^2 = 416.16$
$AC = \sqrt{416.16} = 20.4 \, m$
Hence,the required distance is $20.4 \, m$.

(C) Let $A$ be the position of the street bulb fixed on a pole $AB = 6 \, m$ and $CD = 1.5 \, m$ be the height of a woman and her shadow be $ED = 3 \, m$. Let the distance between the pole and the woman be $x \, m$.
Here,the woman and the pole are both standing vertically.
So,$CD \parallel AB$.
In $\triangle CDE$ and $\triangle ABE$,$\angle E = \angle E$ [common angle].
$\angle ABE = \angle CDE = 90^{\circ}$ [each equal to $90^{\circ}$].
Therefore,$\triangle CDE \sim \triangle ABE$ [by $AAA$ similarity criterion].
Then,$\frac{ED}{EB} = \frac{CD}{AB}$.
$\frac{3}{3 + x} = \frac{1.5}{6}$.
$3 \times 6 = 1.5(3 + x)$.
$18 = 4.5 + 1.5x$.
$1.5x = 18 - 4.5$.
$1.5x = 13.5$.
$x = \frac{13.5}{1.5} = 9 \, m$.
Hence,she is at a distance of $9 \, m$ from the base of the pole.

(N/A) Given,$\Delta PQR$ in which $\angle Q = 90^{\circ}$,$QS \perp PR$,$PQ = 6 \, cm$,and $PS = 4 \, cm$.
In right-angled $\Delta PSQ$,by Pythagoras theorem:
$PQ^2 = PS^2 + QS^2$
$6^2 = 4^2 + QS^2$
$36 = 16 + QS^2$
$QS^2 = 20$
$QS = \sqrt{20} = 2\sqrt{5} \, cm$.
Since $\Delta PSQ \sim \Delta QSR$,we have:
$\frac{PS}{QS} = \frac{QS}{RS}$
$QS^2 = PS \times RS$
$20 = 4 \times RS$
$RS = 5 \, cm$.
In right-angled $\Delta QSR$,by Pythagoras theorem:
$QR^2 = QS^2 + RS^2$
$QR^2 = 20 + 5^2$
$QR^2 = 20 + 25 = 45$
$QR = \sqrt{45} = 3\sqrt{5} \, cm$.
Thus,$QS = 2\sqrt{5} \, cm$,$RS = 5 \, cm$,and $QR = 3\sqrt{5} \, cm$.

(N/A) Given: In $\triangle PQR$,$PD \perp QR$,$PQ = a$,$PR = b$,$QD = c$,and $DR = d$.
To prove: $(a+b)(a-b) = (c+d)(c-d)$.
Proof: In right-angled $\triangle PDQ$,by Pythagoras theorem:
$PQ^2 = PD^2 + QD^2$
$a^2 = PD^2 + c^2$
$PD^2 = a^2 - c^2$ ...... $(i)$
In right-angled $\triangle PDR$,by Pythagoras theorem:
$PR^2 = PD^2 + DR^2$
$b^2 = PD^2 + d^2$
$PD^2 = b^2 - d^2$ ...... $(ii)$
From equations $(i)$ and $(ii)$:
$a^2 - c^2 = b^2 - d^2$
$a^2 - b^2 = c^2 - d^2$
$(a - b)(a + b) = (c - d)(c + d)$
Hence,$(a + b)(a - b) = (c + d)(c - d)$ is proved.

(N/A) Given: $A$ quadrilateral $ABCD$ in which $\angle A + \angle D = 90^{\circ}$.
To prove: $AC^{2} + BD^{2} = AD^{2} + BC^{2}$.
Construction: Produce $AB$ and $DC$ to meet at point $E$.
Proof: In $\triangle AED$,$\angle A + \angle D = 90^{\circ}$ (given).
Since the sum of angles in a triangle is $180^{\circ}$,we have $\angle E = 180^{\circ} - (\angle A + \angle D) = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
In right-angled $\triangle AED$,by Pythagoras theorem: $AD^{2} = AE^{2} + DE^{2}$ ... $(i)$
In right-angled $\triangle BEC$,by Pythagoras theorem: $BC^{2} = BE^{2} + CE^{2}$ ... $(ii)$
Adding $(i)$ and $(ii)$,we get: $AD^{2} + BC^{2} = AE^{2} + DE^{2} + BE^{2} + CE^{2}$ ... $(iii)$
In right-angled $\triangle AEC$,by Pythagoras theorem: $AC^{2} = AE^{2} + CE^{2}$ ... $(iv)$
In right-angled $\triangle BED$,by Pythagoras theorem: $BD^{2} = BE^{2} + DE^{2}$ ... $(v)$
Adding $(iv)$ and $(v)$,we get: $AC^{2} + BD^{2} = AE^{2} + CE^{2} + BE^{2} + DE^{2}$ ... $(vi)$
Comparing $(iii)$ and $(vi)$,we get: $AC^{2} + BD^{2} = AD^{2} + BC^{2}$.
Hence proved.

(N/A) Given $l \parallel m$ and line segments $AB, CD$ and $EF$ are concurrent at point $P$.
To prove: $\frac{AE}{BF} = \frac{AC}{BD} = \frac{CE}{FD}$.
Proof:
In $\triangle APC$ and $\triangle BPD$,
$\angle APC = \angle BPD$ [Vertically opposite angles]
$\angle PAC = \angle PBD$ [Alternate interior angles,as $l \parallel m$]
Therefore,$\triangle APC \sim \triangle BPD$ [By $AA$ similarity criterion]
Then,$\frac{AP}{PB} = \frac{AC}{BD} = \frac{PC}{PD}$ ......$(i)$
In $\triangle APE$ and $\triangle BPF$,
$\angle APE = \angle BPF$ [Vertically opposite angles]
$\angle PAE = \angle PBF$ [Alternate interior angles]
Therefore,$\triangle APE \sim \triangle BPF$ [By $AA$ similarity criterion]
Then,$\frac{AP}{PB} = \frac{AE}{BF} = \frac{PE}{PF}$ ......$(ii)$
In $\triangle PEC$ and $\triangle PFD$,
$\angle EPC = \angle FPD$ [Vertically opposite angles]
$\angle PCE = \angle PDF$ [Alternate interior angles]
Therefore,$\triangle PEC \sim \triangle PFD$ [By $AA$ similarity criterion]
Then,$\frac{PE}{PF} = \frac{PC}{PD} = \frac{CE}{FD}$ ......$(iii)$
From equations $(i), (ii)$ and $(iii)$,
$\frac{AP}{PB} = \frac{AC}{BD} = \frac{AE}{BF} = \frac{PE}{PF} = \frac{CE}{FD}$
Thus,$\frac{AE}{BF} = \frac{AC}{BD} = \frac{CE}{FD}$. Hence proved.

(N/A) Given: In $\triangle ABC$,$E$ is the mid-point of $CA$ and $\angle AEF = \angle AFE$.
To prove: $\frac{BD}{CD} = \frac{BF}{CE}$.
Construction: Draw a line $CG$ parallel to $EF$ such that $G$ lies on $AB$.
Proof: Since $E$ is the mid-point of $CA$,we have $CE = AE$ $(i)$.
In $\triangle ACG$,since $CG \parallel EF$ and $E$ is the mid-point of $CA$,by the converse of the Mid-point Theorem,$F$ is the mid-point of $AG$. Thus,$GF = AF$ $(ii)$.
Also,in $\triangle ACG$,since $CG \parallel EF$,by the Mid-point Theorem,$EF = \frac{1}{2} CG$. However,this is not needed here. Instead,consider $\triangle BDF$ and $\triangle BCG$. Since $CG \parallel EF$,by the Basic Proportionality Theorem in $\triangle BDF$,we have $\frac{BD}{CD} = \frac{BF}{GF}$.
Since $\angle AEF = \angle AFE$,in $\triangle AEF$,we have $AE = AF$. From $(i)$ and $(ii)$,$CE = AE = AF = GF$. Therefore,$CE = GF$.
Substituting $GF = CE$ in the ratio $\frac{BD}{CD} = \frac{BF}{GF}$,we get $\frac{BD}{CD} = \frac{BF}{CE}$.
Hence proved.

(N/A) Let $ABC$ be a right triangle,right-angled at $B$ with $AB = y$ and $BC = x$.
Three semicircles are drawn on the sides $AB$,$BC$,and $AC$ respectively,with diameters $AB$,$BC$,and $AC$.
Let the areas of the semicircles with diameters $AB$,$BC$,and $AC$ be $A_1$,$A_2$,and $A_3$ respectively.
To prove: $A_3 = A_1 + A_2$.
Proof: In $\triangle ABC$,by the Pythagoras theorem,
$AC^2 = AB^2 + BC^2$
$AC^2 = y^2 + x^2$
$AC = \sqrt{y^2 + x^2}$
We know that the area of a semicircle with diameter $d$ is $\frac{\pi}{2} \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{8}$.
Therefore,the area of the semicircle drawn on $AC$ is:
$A_3 = \frac{\pi (AC)^2}{8} = \frac{\pi (y^2 + x^2)}{8}$ .....$(i)$
Now,the area of the semicircle drawn on $AB$ is:
$A_1 = \frac{\pi (AB)^2}{8} = \frac{\pi y^2}{8}$ .....$(ii)$
And the area of the semicircle drawn on $BC$ is:
$A_2 = \frac{\pi (BC)^2}{8} = \frac{\pi x^2}{8}$ .....$(iii)$
Adding equations $(ii)$ and $(iii)$:
$A_1 + A_2 = \frac{\pi y^2}{8} + \frac{\pi x^2}{8} = \frac{\pi (y^2 + x^2)}{8}$
From equation $(i)$,we see that $A_1 + A_2 = A_3$.
Hence,the area of the semicircle on the hypotenuse is equal to the sum of the areas of the semicircles on the other two sides.

(N/A) Let $\triangle ABC$ be a right-angled triangle with $\angle A = 90^\circ$,$AC = y$,and $AB = x$.
Three equilateral triangles $\triangle AEC$,$\triangle AFB$,and $\triangle CBD$ are drawn on sides $AC$,$AB$,and $BC$ respectively.
Let the areas of these equilateral triangles be $A_1$,$A_2$,and $A_3$ respectively.
We need to prove that $A_3 = A_1 + A_2$.
In $\triangle ABC$,by Pythagoras theorem: $BC^2 = AC^2 + AB^2 = y^2 + x^2$.
The area of an equilateral triangle with side $s$ is given by $\frac{\sqrt{3}}{4} s^2$.
$A_1 = \text{Area}(\triangle AEC) = \frac{\sqrt{3}}{4} AC^2 = \frac{\sqrt{3}}{4} y^2$ ... $(i)$
$A_2 = \text{Area}(\triangle AFB) = \frac{\sqrt{3}}{4} AB^2 = \frac{\sqrt{3}}{4} x^2$ ... $(ii)$
$A_3 = \text{Area}(\triangle CBD) = \frac{\sqrt{3}}{4} BC^2 = \frac{\sqrt{3}}{4} (y^2 + x^2)$ ... $(iii)$
From $(i)$,$(ii)$,and $(iii)$:
$A_3 = \frac{\sqrt{3}}{4} y^2 + \frac{\sqrt{3}}{4} x^2 = A_1 + A_2$.
Hence,the area of the equilateral triangle on the hypotenuse is equal to the sum of the areas of the equilateral triangles on the other two sides.

(B) In $\Delta XYZ$,$XM$ is the bisector of $\angle X$,which intersects $\overline{YZ}$ at $M$.
According to the Angle Bisector Theorem,the ratio of the sides forming the angle is equal to the ratio of the segments of the opposite side.
$\therefore \frac{XY}{XZ} = \frac{YM}{MZ}$
Substituting the given values:
$\therefore \frac{6}{8} = \frac{4.2}{MZ}$
$\therefore MZ = \frac{4.2 \times 8}{6}$
$\therefore MZ = 0.7 \times 8 = 5.6$
Now,$YZ = YM + MZ$ (since $Y-M-Z$ are collinear).
$YZ = 4.2 + 5.6 = 9.8$.

(B) In a right-angled triangle,the hypotenuse $AC$ can be calculated using the Pythagorean theorem: $AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$.
The radius $r$ of the incircle of a right-angled triangle is given by the formula $r = \frac{AB + BC - AC}{2}$.
Substituting the values: $r = \frac{8 + 6 - 10}{2} = \frac{14 - 10}{2} = \frac{4}{2} = 2$.
Therefore,the radius of the circle is $2$.

(B) $1$. In right-angled $\Delta PQR$,by the Pythagorean theorem,$PR^2 = PQ^2 + QR^2$.
$2$. $PR^2 = 40^2 + 30^2 = 1600 + 900 = 2500$,so $PR = 50$.
$3$. The area of $\Delta PQR = \frac{1}{2} \times PQ \times QR = \frac{1}{2} \times 40 \times 30 = 600$.
$4$. Also,the area of $\Delta PQR = \frac{1}{2} \times PR \times QM = 600$,so $\frac{1}{2} \times 50 \times QM = 600$,which gives $QM = 24$.
$5$. In $\Delta QMR$,$\angle QMR = 90^{\circ}$. By the Pythagorean theorem,$MR^2 = QR^2 - QM^2 = 30^2 - 24^2 = 900 - 576 = 324$,so $MR = 18$.
$6$. The area of $\Delta QMR = \frac{1}{2} \times QM \times MR = \frac{1}{2} \times 24 \times 18 = 12 \times 18 = 216$.

(N/A) In $\Delta ABC$,$m\angle B = 90^\circ$.
By Pythagoras theorem,$AC^2 = AB^2 + BC^2$.
$AC^2 = 8^2 + 6^2 = 64 + 36 = 100$.
$AC = 10$.
The area of $\Delta ABC = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times AC \times BM$.
$\frac{1}{2} \times 8 \times 6 = \frac{1}{2} \times 10 \times BM$.
$48 = 10 \times BM \implies BM = 4.8$.
In $\Delta AMB$,$m\angle M = 90^\circ$. By Pythagoras theorem,$AB^2 = AM^2 + BM^2$.
$8^2 = AM^2 + (4.8)^2$.
$64 = AM^2 + 23.04$.
$AM^2 = 64 - 23.04 = 40.96$.
$AM = \sqrt{40.96} = 6.4$.
Since $AC = AM + CM$,$10 = 6.4 + CM$.
$CM = 10 - 6.4 = 3.6$.
Thus,$AM = 6.4$,$BM = 4.8$,and $CM = 3.6$.

(N/A) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
$m \angle A + m \angle C + m \angle B = 180^{\circ}$
Since $m \angle A + m \angle C = m \angle B$,we substitute:
$m \angle B + m \angle B = 180^{\circ}$
$2 m \angle B = 180^{\circ}$
$m \angle B = 90^{\circ}$
In a right-angled triangle $\Delta ABC$ with altitude $\overline{BM}$ to the hypotenuse $\overline{AC}$,by the geometric mean theorem:
$BM^2 = AM \times CM$
$BM^2 = 16 \times 9 = 144$
$BM = 12$
Now,using the properties of right triangles:
$AC = AM + CM = 16 + 9 = 25$
$AB^2 = AM \times AC = 16 \times 25 = 400 \implies AB = 20$
$BC^2 = CM \times AC = 9 \times 25 = 225 \implies BC = 15$
Thus,$BM = 12$,$AB = 20$,and $BC = 15$.

(D) To determine the correct matching,we analyze the properties of the diagonals of quadrilaterals:
$1$. For a rhombus $(1)$,the diagonals bisect each other at right angles $(b)$.
$2$. For a parallelogram $(2)$,the diagonals bisect each other $(a)$.
$3$. For a rectangle $(3)$,the diagonals are congruent and bisect each other $(d)$.
$4$. For a square $(4)$,the diagonals are congruent and bisect each other at right angles $(c)$.
Thus,the correct matching is $(1-b), (2-a), (3-d), (4-c)$.

(A) In $\Delta ABC$,$\overline{AD}$ is a median to the base $\overline{BC}$.
Since $\Delta ABC$ is an isosceles triangle with $\overline{AB} \cong \overline{AC}$,the median $\overline{AD}$ to the base is also the altitude to the base.
Therefore,$\overline{AD} \perp \overline{BC}$ and $\angle ADB = 90^{\circ}$.
Let $AB = AC = x$. The perimeter of $\Delta ABC$ is $AB + AC + BC = 36$.
$x + x + BC = 36 \implies BC = 36 - 2x$.
Since $D$ is the midpoint of $\overline{BC}$,$BD = \frac{BC}{2} = \frac{36 - 2x}{2} = 18 - x$.
In right-angled $\Delta ABD$,by the Pythagorean theorem:
$AB^2 = AD^2 + BD^2$
$x^2 = 12^2 + (18 - x)^2$
$x^2 = 144 + 324 - 36x + x^2$
$36x = 468$
$x = 13$.
Thus,$BD = 18 - 13 = 5$,so $BC = 2 \times 5 = 10$.
The area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AD$.
Area $= \frac{1}{2} \times 10 \times 12 = 60$.

(B) In $\Delta ABC$,we are given $AC - AB = 27$ and $AC - BC = 6$.
Therefore,$AB = AC - 27$ and $BC = AC - 6$.
Since $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^\circ$,by the Pythagorean theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = (AC - 27)^2 + (AC - 6)^2$
$AC^2 = (AC^2 - 54AC + 729) + (AC^2 - 12AC + 36)$
$AC^2 = 2AC^2 - 66AC + 765$
$0 = AC^2 - 66AC + 765$
Factoring the quadratic equation:
$(AC - 51)(AC - 15) = 0$
So,$AC = 51$ or $AC = 15$.
If $AC = 15$,then $AB = 15 - 27 = -12$,which is impossible as the length of a side cannot be negative.
Thus,$AC = 51$.
Now,$AB = 51 - 27 = 24$ and $BC = 51 - 6 = 45$.
The perimeter of $\Delta ABC = AB + BC + AC = 24 + 45 + 51 = 120$.

(B) Let $\overline{AB}$ represent the wall,$\overline{AC}$ represent the ladder,and $C$ be the lower end of the ladder on the ground.
Given: $AB = 12 \, m$,$BC = 9 \, m$,and $\angle B = 90^{\circ}$.
In right-angled $\Delta ABC$,by Pythagoras theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = 12^2 + 9^2$
$AC^2 = 144 + 81$
$AC^2 = 225$
$AC = \sqrt{225} = 15 \, m$.
Thus,the length of the ladder is $15 \, m$.

(C) The diagonals of a rhombus bisect each other at right angles.
Let the diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $M$.
Then,$m\angle AMB = 90^{\circ}$.
Since the diagonals bisect each other,$AM = \frac{1}{2} AC = \frac{1}{2}(12) = 6$ and $BM = \frac{1}{2} BD = \frac{1}{2}(16) = 8$.
In right-angled $\Delta AMB$,by the Pythagorean theorem:
$AB^2 = AM^2 + BM^2$
$AB^2 = 6^2 + 8^2$
$AB^2 = 36 + 64 = 100$
$AB = \sqrt{100} = 10$.
The perimeter of a rhombus is $4 \times \text{side length}$.
Perimeter $= 4 \times AB = 4 \times 10 = 40$.
Thus,the perimeter of rhombus $ABCD$ is $40$.

(D) The perimeter of a square is given by $4 \times \text{side}$.
Let the side length of the square be $s$.
$4s = 48$
$s = 12$
In square $ABCD$, all sides are equal to $12$ and each angle is $90^{\circ}$.
In right-angled triangle $\Delta ABC$, by Pythagoras theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = 12^2 + 12^2$
$AC^2 = 144 + 144 = 288$
$AC = \sqrt{288} = \sqrt{144 \times 2} = 12 \sqrt{2}$.
Thus, the length of the diagonal $\overline{AC}$ is $12 \sqrt{2}$.

(C) In $\Delta ABC$,$m \angle B = 90^{\circ}$.
By the Pythagorean theorem:
$AC^{2} = AB^{2} + BC^{2}$
$AC^{2} = 20^{2} + 21^{2}$
$AC^{2} = 400 + 441$
$AC^{2} = 841$
$AC = \sqrt{841} = 29$.
In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
Therefore,$BM = \frac{1}{2} AC = \frac{1}{2} \times 29 = 14.5$.
Thus,$BM = 14.5$.

(B) In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$.
According to the geometric mean theorem (or properties of right triangles with an altitude to the hypotenuse):
$AB^{2} = AM \times AC$
$BC^{2} = CM \times AC$
Subtracting the two equations:
$AB^{2} - BC^{2} = (AM \times AC) - (CM \times AC)$
$AB^{2} - BC^{2} = AC(AM - CM)$
Given $AB^{2} - BC^{2} = 260$ and $AM - CM = 10$:
$260 = AC(10)$
$AC = \frac{260}{10} = 26$
Thus,$AC = 26$.

(N/A) Given: $AB + BC = 23$ $(i)$,$BC + AC = 32$ $(ii)$,and $AB + AC = 25$ $(iii)$.
Adding $(i)$,$(ii)$,and $(iii)$,we get: $2(AB + BC + AC) = 23 + 32 + 25 = 80$.
Therefore,$AB + BC + AC = 40$ $(iv)$.
Subtracting $(ii)$ from $(iv)$: $AB = 40 - 32 = 8$.
Subtracting $(iii)$ from $(iv)$: $BC = 40 - 25 = 15$.
Subtracting $(i)$ from $(iv)$: $AC = 40 - 23 = 17$.
Now,check the sides: $AB^2 + BC^2 = 8^2 + 15^2 = 64 + 225 = 289$.
Also,$AC^2 = 17^2 = 289$.
Since $AB^2 + BC^2 = AC^2$,by the converse of the Pythagoras theorem,$\Delta ABC$ is a right-angled triangle with $\angle B = 90^\circ$.

(C) In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^{\circ}$.
Given $m \angle A = m \angle B + m \angle C$,substituting this into the sum gives $m \angle A + m \angle A = 180^{\circ}$,so $2m \angle A = 180^{\circ}$,which means $m \angle A = 90^{\circ}$.
Since $\overline{AM}$ is an altitude to $\overline{BC}$,$\Delta ABM$ and $\Delta ACM$ are right-angled triangles at $M$.
Let $BM = x$. Then $MC = 8 - x$.
In $\Delta ABM$,$AB^2 = AM^2 + BM^2 = 12 + x^2$.
In $\Delta ACM$,$AC^2 = AM^2 + MC^2 = 12 + (8 - x)^2$.
In $\Delta ABC$,by the Pythagorean theorem,$AB^2 + AC^2 = BC^2$,so $(12 + x^2) + (12 + (8 - x)^2) = 8^2$.
$24 + x^2 + 64 - 16x + x^2 = 64$.
$2x^2 - 16x + 24 = 0$.
$x^2 - 8x + 12 = 0$.
$(x - 6)(x - 2) = 0$.
Thus,$x = 6$ or $x = 2$.

(B) Given that in $\Delta PQR$,$m \angle P = m \angle Q + m \angle R$.
We know that the sum of all angles in a triangle is $180^\circ$,so $m \angle P + m \angle Q + m \angle R = 180^\circ$.
Substituting $m \angle Q + m \angle R = m \angle P$,we get $m \angle P + m \angle P = 180^\circ$,which implies $2(m \angle P) = 180^\circ$,so $m \angle P = 90^\circ$.
Thus,$\Delta PQR$ is a right-angled triangle with the hypotenuse $QR = 25$ and one side $PQ = 7$.
Using the Pythagorean theorem,$PQ^2 + PR^2 = QR^2$.
$7^2 + PR^2 = 25^2$.
$49 + PR^2 = 625$.
$PR^2 = 625 - 49 = 576$.
$PR = \sqrt{576} = 24$.
The perimeter of $\Delta PQR = PQ + QR + PR = 7 + 25 + 24 = 56$.

(D) $1$. The perimeter of a rhombus is given by $4 \times \text{side}$. Let the side length be $s$. So, $4s = 116$, which gives $s = 29$.
$2$. In a rhombus, the diagonals bisect each other at right angles $(90^{\circ})$.
$3$. Let the diagonals $AC$ and $BD$ intersect at point $O$. Then $AO = AC / 2 = 42 / 2 = 21$.
$4$. In the right-angled triangle $\triangle AOB$, by the Pythagorean theorem: $AO^2 + BO^2 = AB^2$.
$5$. Substituting the values: $21^2 + BO^2 = 29^2$.
$6$. $441 + BO^2 = 841$.
$7$. $BO^2 = 841 - 441 = 400$.
$8$. $BO = \sqrt{400} = 20$.
$9$. Since the diagonals bisect each other, $BD = 2 \times BO = 2 \times 20 = 40$.

(A) Given that $m \angle A + m \angle B = m \angle C$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $m \angle A + m \angle B + m \angle C = 180^{\circ}$.
Substituting $m \angle A + m \angle B = m \angle C$,we get $m \angle C + m \angle C = 180^{\circ}$,which implies $2m \angle C = 180^{\circ}$,so $m \angle C = 90^{\circ}$.
Thus,$\Delta ABC$ is a right-angled triangle with the hypotenuse $AB = 17.5$.
Given $AC : BC = 3 : 4$,let $AC = 3x$ and $BC = 4x$.
By the Pythagorean theorem,$AC^2 + BC^2 = AB^2$.
$(3x)^2 + (4x)^2 = (17.5)^2$.
$9x^2 + 16x^2 = 306.25$.
$25x^2 = 306.25$.
$x^2 = 12.25$,so $x = 3.5$.
Thus,$AC = 3(3.5) = 10.5$ and $BC = 4(3.5) = 14$.
The perimeter of $\Delta ABC = AC + BC + AB = 10.5 + 14 + 17.5 = 42$.

(D) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the diagonals $AC$ and $BD$ intersect at point $O$.
Given $AB = 25$ and $AC = 14$.
Since diagonals bisect each other,$AO = OC = AC / 2 = 14 / 2 = 7$.
In the right-angled triangle $\triangle AOB$,by the Pythagorean theorem:
$AO^2 + OB^2 = AB^2$
$7^2 + OB^2 = 25^2$
$49 + OB^2 = 625$
$OB^2 = 625 - 49 = 576$
$OB = \sqrt{576} = 24$.
Since $BD = 2 \times OB$,we have $BD = 2 \times 24 = 48$.