Mix Examples - Triangles Questions in English

(A) In a right-angled triangle $\Delta PQR$,where $\angle Q = 90^{\circ}$,the side $PR$ is the hypotenuse.
According to the Pythagorean theorem: $(PR)^2 = (PQ)^2 + (QR)^2$.
Given values are $PQ = 33$ and $PR = 65$.
Substituting the values: $(65)^2 = (33)^2 + (QR)^2$.
$4225 = 1089 + (QR)^2$.
$(QR)^2 = 4225 - 1089$.
$(QR)^2 = 3136$.
Taking the square root of both sides: $QR = \sqrt{3136} = 56$.
Therefore,the length of $QR$ is $56$.

(A) Let the diagonals $PR$ and $QS$ of the quadrilateral $PQRS$ intersect at point $O$ at right angles.
Thus,$\angle POQ = \angle QOR = \angle ROS = \angle SOP = 90^{\circ}$.
In right-angled $\triangle POQ$,by Pythagoras theorem: $PQ^{2} = PO^{2} + OQ^{2}$.
In right-angled $\triangle ROS$,by Pythagoras theorem: $RS^{2} = RO^{2} + OS^{2}$.
Adding these two equations: $PQ^{2} + RS^{2} = PO^{2} + OQ^{2} + RO^{2} + OS^{2}$ --- (Equation $1$).
In right-angled $\triangle SOP$,by Pythagoras theorem: $PS^{2} = SO^{2} + OP^{2}$.
In right-angled $\triangle QOR$,by Pythagoras theorem: $QR^{2} = OQ^{2} + OR^{2}$.
Adding these two equations: $PS^{2} + QR^{2} = SO^{2} + OP^{2} + OQ^{2} + OR^{2}$ --- (Equation $2$).
Comparing Equation $1$ and Equation $2$,we see that the right-hand sides are identical.
Therefore,$PQ^{2} + RS^{2} = PS^{2} + QR^{2}$.

(N/A) Given: In $\Delta ABC$,$\angle B = 90^{\circ}$. $N$ is a point on $AB$ and $M$ is a point on $BC$.
Step $1$: Apply the Pythagorean theorem in $\Delta ABM$ and $\Delta CBN$.
In $\Delta ABM$,since $\angle B = 90^{\circ}$,we have $AM^{2} = AB^{2} + BM^{2}$.
In $\Delta CBN$,since $\angle B = 90^{\circ}$,we have $CN^{2} = CB^{2} + BN^{2}$.
Step $2$: Add the two equations.
$AM^{2} + CN^{2} = (AB^{2} + BM^{2}) + (CB^{2} + BN^{2})$.
Step $3$: Rearrange the terms.
$AM^{2} + CN^{2} = (AB^{2} + CB^{2}) + (BM^{2} + BN^{2})$.
Step $4$: Use the Pythagorean theorem for $\Delta ABC$ and $\Delta MBN$.
In $\Delta ABC$,$AC^{2} = AB^{2} + CB^{2}$.
In $\Delta MBN$,$MN^{2} = BM^{2} + BN^{2}$.
Step $5$: Substitute these into the equation from Step $3$.
$AM^{2} + CN^{2} = AC^{2} + MN^{2}$.
Hence,the result is proved.

(N/A) Let $ABCD$ be a rectangle with sides $AB = CD = l$ and $BC = DA = b$.
Let the diagonals be $AC$ and $BD$.
In a rectangle,all interior angles are $90^{\circ}$.
Consider the right-angled triangle $\triangle ABC$. By the Pythagoras theorem:
$AC^2 = AB^2 + BC^2 = l^2 + b^2$.
Similarly,in the right-angled triangle $\triangle BCD$:
$BD^2 = BC^2 + CD^2 = b^2 + l^2$.
Sum of the squares of the diagonals = $AC^2 + BD^2 = (l^2 + b^2) + (b^2 + l^2) = 2l^2 + 2b^2$.
Sum of the squares of the sides = $AB^2 + BC^2 + CD^2 + DA^2 = l^2 + b^2 + l^2 + b^2 = 2l^2 + 2b^2$.
Thus,the sum of the squares of the sides is equal to the sum of the squares of the diagonals.

(N/A) Let the sides of the triangle be $a = m^{2}-n^{2}$,$b = 2mn$,and $c = m^{2}+n^{2}$.
To prove that the triangle is a right-angled triangle,we need to check if the square of the longest side is equal to the sum of the squares of the other two sides (Pythagoras Theorem).
Here,$c = m^{2}+n^{2}$ is the longest side since $m > n > 0$.
Calculate $a^{2} + b^{2}$:
$a^{2} + b^{2} = (m^{2}-n^{2})^{2} + (2mn)^{2}$
$= (m^{4} - 2m^{2}n^{2} + n^{4}) + 4m^{2}n^{2}$
$= m^{4} + 2m^{2}n^{2} + n^{4}$
$= (m^{2}+n^{2})^{2}$
$= c^{2}$
Since $a^{2} + b^{2} = c^{2}$,the triangle satisfies the Pythagoras Theorem.
Therefore,the triangle is a right-angled triangle.

(N/A) Given: In $\Delta ABC$,$\angle B = 90^{\circ}$ (since $\overline{AC}$ is the hypotenuse),and $\overline{BE}$ is a median to the hypotenuse $\overline{AC}$.
Since $\overline{BE}$ is a median,$E$ is the midpoint of $\overline{AC}$. Thus,$AE = EC = \frac{1}{2} AC$,which implies $AC = 2AE$.
In right-angled $\Delta ABC$,by Pythagoras theorem: $AB^{2} + BC^{2} = AC^{2}$.
We need to prove: $AB^{2} + BC^{2} + AC^{2} = 8AE^{2}$.
Substitute $AB^{2} + BC^{2} = AC^{2}$ into the expression:
$AC^{2} + AC^{2} = 2AC^{2}$.
Since $AC = 2AE$,substitute this into the expression:
$2(2AE)^{2} = 2(4AE^{2}) = 8AE^{2}$.
Thus,$AB^{2} + BC^{2} + AC^{2} = 8AE^{2}$ is proved.

(A) In $\Delta GBS$,$\angle B = 90^{\circ}$ and $BM \perp GS$.
By the property of similar triangles in a right-angled triangle,when an altitude is drawn to the hypotenuse,the triangles formed on both sides of the altitude are similar to the original triangle and to each other.
Specifically,$\Delta GMB \sim \Delta GBS$ and $\Delta B M S \sim \Delta GBS$.
From $\Delta GMB \sim \Delta GBS$,we have $\frac{GB}{GS} = \frac{GM}{GB}$,which implies $GB^{2} = GM \cdot GS$ (Equation $1$).
From $\Delta BMS \sim \Delta GBS$,we have $\frac{BS}{GS} = \frac{SM}{BS}$,which implies $BS^{2} = SM \cdot GS$ (Equation $2$).
Dividing Equation $1$ by Equation $2$:
$\frac{GB^{2}}{BS^{2}} = \frac{GM \cdot GS}{SM \cdot GS} = \frac{GM}{SM}$.
Hence,the result is proved.

(N/A) $1$. In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $\overline{QD} \perp \overline{PR}$.
$2$. By the property of similarity in right-angled triangles,$\Delta PDQ \sim \Delta PQR$ and $\Delta RDQ \sim \Delta RQP$.
$3$. Specifically,$\Delta PDQ \sim \Delta QDR$ is not directly applicable,but we use the property that in a right triangle,the altitude to the hypotenuse divides the triangle into two triangles similar to the original and to each other.
$4$. Thus,$\Delta PDQ \sim \Delta QDR$.
$5$. From the similarity $\Delta PDQ \sim \Delta QDR$,we have the ratio of corresponding sides: $\frac{PD}{QD} = \frac{QD}{RD} = \frac{PQ}{QR}$.
$6$. We are given $PQ = 3QR$,which implies $\frac{PQ}{QR} = 3$.
$7$. From $\frac{PD}{QD} = \frac{PQ}{QR}$,we get $PD = 3QD$.
$8$. From $\frac{QD}{RD} = \frac{PQ}{QR}$,we get $QD = 3RD$.
$9$. Substituting $QD = 3RD$ into $PD = 3QD$,we get $PD = 3(3RD) = 9RD$.
$10$. Hence,$PD = 9RD$ is proved.

(N/A) In $\Delta XYZ$,$\angle Y = 90^{\circ}$ and $\overline{YM} \perp \overline{XZ}$.
According to the Geometric Mean Theorem (or properties of similar triangles),in a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other.
Specifically,$\Delta XMY \sim \Delta YMZ$.
From the similarity $\Delta XMY \sim \Delta YMZ$,we have the ratio of corresponding sides:
$\frac{XY}{YZ} = \frac{XM}{YM} = \frac{YM}{ZM}$.
From $\frac{XY}{YZ} = \frac{YM}{ZM}$,we get $YM^2 = XY \cdot ZM$ (not directly useful here).
Alternatively,using the property of the altitude in a right triangle:
$XY^2 = XM \cdot XZ$ and $YZ^2 = ZM \cdot XZ$.
Dividing these two equations:
$\frac{XY^2}{YZ^2} = \frac{XM \cdot XZ}{ZM \cdot XZ} = \frac{XM}{ZM}$.
Given $XM = 16ZM$,we substitute this into the ratio:
$\frac{XY^2}{YZ^2} = \frac{16ZM}{ZM} = 16$.
Taking the square root of both sides:
$\frac{XY}{YZ} = \sqrt{16} = 4$.
Therefore,$XY = 4YZ$.

(N/A) In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.
By the area of a right-angled triangle,the area of $\Delta ABC$ can be expressed in two ways:
Area $= \frac{1}{2} \times AB \times BC$ (using legs as base and height)
Area $= \frac{1}{2} \times AC \times BM$ (using hypotenuse as base and altitude as height)
Equating the two expressions: $\frac{1}{2} \times AB \times BC = \frac{1}{2} \times AC \times BM \implies AB \times BC = AC \times BM$.
Squaring both sides: $AB^2 \times BC^2 = AC^2 \times BM^2$.
Since $\Delta ABC$ is a right-angled triangle,by the Pythagorean theorem: $AC^2 = AB^2 + BC^2$.
Substituting $AC^2$ into the equation: $AB^2 \times BC^2 = (AB^2 + BC^2) \times BM^2$.
Rearranging for $BM^2$: $BM^2 = \frac{AB^2 \times BC^2}{AB^2 + BC^2}$.
Taking the reciprocal of both sides: $\frac{1}{BM^2} = \frac{AB^2 + BC^2}{AB^2 \times BC^2}$.
Splitting the fraction: $\frac{1}{BM^2} = \frac{AB^2}{AB^2 \times BC^2} + \frac{BC^2}{AB^2 \times BC^2}$.
Simplifying: $\frac{1}{BM^2} = \frac{1}{BC^2} + \frac{1}{AB^2}$.

(N/A) $1$. In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BD} \perp \overline{AC}$.
$2$. By the property of geometric mean in a right-angled triangle,we have $BD^2 = AD \cdot CD$.
$3$. Given $BD = 2CD$,substitute this into the equation: $(2CD)^2 = AD \cdot CD$.
$4$. This simplifies to $4CD^2 = AD \cdot CD$.
$5$. Dividing both sides by $CD$ (since $CD \neq 0$),we get $AD = 4CD$.
$6$. Since $AC = AD + CD$,substitute $AD = 4CD$ into the equation: $AC = 4CD + CD$.
$7$. Therefore,$AC = 5CD$.

(N/A) In $\Delta ABC$,$\angle B = 90^\circ$ and $\overline{BM} \perp \overline{AC}$.
By the property of geometric mean in a right-angled triangle,we have $\Delta AMB \sim \Delta BMC$.
Therefore,the ratio of their corresponding sides is equal: $\frac{AB}{BC} = \frac{AM}{BM} = \frac{BM}{CM}$.
From $\frac{AB}{BC} = \frac{BM}{CM}$,we get $BM = \frac{AB \cdot CM}{BC}$.
Also,from $\Delta AMB \sim \Delta ABC$,we have $\frac{AB}{AC} = \frac{AM}{AB}$,which implies $AB^2 = AM \cdot AC$.
Similarly,from $\Delta BMC \sim \Delta ABC$,we have $\frac{BC}{AC} = \frac{CM}{BC}$,which implies $BC^2 = CM \cdot AC$.
Dividing the two equations: $\frac{AB^2}{BC^2} = \frac{AM \cdot AC}{CM \cdot AC} = \frac{AM}{CM}$.
Given $AM = 4CM$,we substitute this into the ratio: $\frac{AB^2}{BC^2} = \frac{4CM}{CM} = 4$.
Taking the square root of both sides: $\frac{AB}{BC} = \sqrt{4} = 2$.
Thus,$AB = 2BC$.

(N/A) In $\Delta ABC$,since $m \angle B = 90^{\circ}$,by the Pythagorean theorem,we have $AC^{2} = AB^{2} + BC^{2}$. (Equation $1$)
In $\Delta ABD$,since $m \angle B = 90^{\circ}$,by the Pythagorean theorem,we have $AD^{2} = AB^{2} + BD^{2}$. (Equation $2$)
From Equation $2$,we can write $AB^{2} = AD^{2} - BD^{2}$.
Substitute this value of $AB^{2}$ into Equation $1$:
$AC^{2} = (AD^{2} - BD^{2}) + BC^{2}$.
Rearranging the terms,we get $AC^{2} + BD^{2} = AD^{2} + BC^{2}$.
Hence,the statement is proved.

(A) Given: In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{AD}$ is a median to side $\overline{BC}$.
Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$. Therefore,$BD = DC = \frac{1}{2} BC$,which implies $BC = 2BD$.
In right-angled $\Delta ABC$,by Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2}$.
Substituting $BC = 2BD$,we get $AC^{2} = AB^{2} + (2BD)^{2} = AB^{2} + 4BD^{2}$. (Equation $1$)
In right-angled $\Delta ABD$,by Pythagoras theorem: $AD^{2} = AB^{2} + BD^{2}$,which implies $AB^{2} = AD^{2} - BD^{2}$. (Equation $2$)
Substitute Equation $2$ into Equation $1$:
$AC^{2} = (AD^{2} - BD^{2}) + 4BD^{2}$
$AC^{2} = AD^{2} + 3BD^{2}$.
Hence,the result is proved.

(C) In $\Delta PQR$,$\overline{PM}$ is a median.
By Apollonius' theorem:
$PQ^{2} + PR^{2} = 2(PM^{2} + QM^{2})$
Given $PQ^{2} + PR^{2} = 148$ and $PM = 7$:
$148 = 2(7^{2} + QM^{2})$
$74 = 49 + QM^{2}$
$QM^{2} = 74 - 49$
$QM^{2} = 25$
$QM = 5$
Since $\overline{PM}$ is a median,$M$ is the midpoint of $\overline{QR}$.
Therefore,$QR = 2 \times QM = 2 \times 5 = 10$.

(D) In $\Delta XYZ$,$\overline{XN}$ is a median.
By Apollonius' theorem:
$XY^2 + XZ^2 = 2(XN^2 + YN^2)$
$200 = 2(8^2 + YN^2)$
$100 = 64 + YN^2$
$YN^2 = 36$
$YN = 6$
Since $\overline{XN}$ is a median,$N$ is the midpoint of $\overline{YZ}$.
Therefore,$YZ = 2 \times YN = 2 \times 6 = 12$.

(A) In $\Delta ABC$,$\overline{AD}$ is a median to side $\overline{BC}$.
Therefore,$D$ is the midpoint of $\overline{BC}$.
$BD = \frac{1}{2} BC = \frac{1}{2}(16) = 8$.
By Apollonius' theorem for $\Delta ABC$ with median $\overline{AD}$:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
Substituting the given values:
$10^2 + AC^2 = 2((\sqrt{58})^2 + 8^2)$
$100 + AC^2 = 2(58 + 64)$
$100 + AC^2 = 2(122)$
$100 + AC^2 = 244$
$AC^2 = 244 - 100$
$AC^2 = 144$
$AC = \sqrt{144} = 12$.

(B) In $\Delta ABC,$ $\overline{AD}$ is a median.
Since $D$ is the midpoint of $\overline{BC},$ we have $BC = 2BD = 2(4) = 8.$
By Apollonius' theorem for $\Delta ABC$ with median $\overline{AD},$
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
$AB^2 + 7^2 = 2(7^2 + 4^2)$
$AB^2 + 49 = 2(49 + 16)$
$AB^2 + 49 = 2(65)$
$AB^2 + 49 = 130$
$AB^2 = 130 - 49 = 81$
$AB = 9$
Perimeter of $\Delta ABC = AB + BC + AC = 9 + 8 + 7 = 24.$

(N/A) In $\Delta ABC$,$\overline{AD}$,$\overline{BE}$ and $\overline{CF}$ are medians. Therefore,$D$,$E$,and $F$ are the midpoints of $\overline{BC}$,$\overline{AC}$,and $\overline{AB}$ respectively.
By Apollonius' theorem in $\Delta ABC$ with median $\overline{AD}$:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
Since $BD = \frac{1}{2}BC$,we have:
$AB^2 + AC^2 = 2AD^2 + 2(\frac{BC}{2})^2 = 2AD^2 + \frac{BC^2}{2}$
$2AD^2 = AB^2 + AC^2 - \frac{BC^2}{2} \quad \dots(1)$
Similarly,for medians $\overline{BE}$ and $\overline{CF}$:
$2BE^2 = AB^2 + BC^2 - \frac{AC^2}{2} \quad \dots(2)$
$2CF^2 = AC^2 + BC^2 - \frac{AB^2}{2} \quad \dots(3)$
Adding equations $(1)$,$(2)$,and $(3)$:
$2(AD^2 + BE^2 + CF^2) = (AB^2 + AB^2 - \frac{AB^2}{2}) + (BC^2 + BC^2 - \frac{BC^2}{2}) + (AC^2 + AC^2 - \frac{AC^2}{2})$
$2(AD^2 + BE^2 + CF^2) = \frac{3}{2}AB^2 + \frac{3}{2}BC^2 + \frac{3}{2}AC^2$
$2(AD^2 + BE^2 + CF^2) = \frac{3}{2}(AB^2 + BC^2 + AC^2)$
Multiplying both sides by $2$:
$4(AD^2 + BE^2 + CF^2) = 3(AB^2 + BC^2 + AC^2)$

(D) According to Apollonius's Theorem,for any triangle $\Delta ABC$ with median $\overline{AD}$ to side $\overline{BC}$,the following relation holds:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
Given that $AB^2 + AC^2 = 122$ and $AD = 6$,we substitute these values into the formula:
$122 = 2(6^2 + BD^2)$
$122 = 2(36 + BD^2)$
Divide both sides by $2$:
$61 = 36 + BD^2$
$BD^2 = 61 - 36$
$BD^2 = 25$
$BD = \sqrt{25} = 5$
Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$,so $BC = 2 \times BD$.
$BC = 2 \times 5 = 10$.

(A) According to Apollonius' Theorem for $\Delta ABC$ with median $\overline{AD}$:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
Substitute the given values: $18^2 + 14^2 = 2(14^2 + BD^2)$
$324 + 196 = 2(196 + BD^2)$
$520 = 2(196 + BD^2)$
$260 = 196 + BD^2$
$BD^2 = 260 - 196 = 64$
$BD = 8$
Since $\overline{AD}$ is a median,$BD = DC = 8$.
Therefore,$BC = BD + DC = 8 + 8 = 16$.
The perimeter of $\Delta ABC = AB + AC + BC = 18 + 14 + 16 = 48$.

(B) According to Apollonius's Theorem,for any triangle $\Delta ABC$ with median $\overline{AD}$ to side $\overline{BC}$,the following relation holds:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
Given that $AD = 7$ and $AB^2 + AC^2 = 148$,we substitute these values into the formula:
$148 = 2(7^2 + BD^2)$
$148 = 2(49 + BD^2)$
Divide both sides by $2$:
$74 = 49 + BD^2$
$BD^2 = 74 - 49 = 25$
$BD = \sqrt{25} = 5$
Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$,so $BC = 2 \times BD$.
$BC = 2 \times 5 = 10$.

(C) According to Apollonius's Theorem,for any triangle $\Delta ABC$ with median $\overline{AD}$ to side $\overline{BC}$:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
Given $AB = 8$,$AC = 15$,and $AD = 8.5$.
Substituting the values:
$8^2 + 15^2 = 2(8.5^2 + BD^2)$
$64 + 225 = 2(72.25 + BD^2)$
$289 = 2(72.25 + BD^2)$
$144.5 = 72.25 + BD^2$
$BD^2 = 144.5 - 72.25 = 72.25$
$BD = \sqrt{72.25} = 8.5$
Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$,so $BC = 2 \times BD$.
$BC = 2 \times 8.5 = 17$.

(D) According to Apollonius's Theorem for $\Delta XYZ$ with median $\overline{XM}$ to side $\overline{YZ}$:
$XY^2 + XZ^2 = 2(XM^2 + YM^2)$
Given $XY = 20$,$XZ = 21$,and $XM = 14.5$.
Substituting the values:
$20^2 + 21^2 = 2(14.5^2 + YM^2)$
$400 + 441 = 2(210.25 + YM^2)$
$841 = 2(210.25 + YM^2)$
$420.5 = 210.25 + YM^2$
$YM^2 = 420.5 - 210.25 = 210.25$
$YM = \sqrt{210.25} = 14.5$
Since $\overline{XM}$ is a median,$M$ is the midpoint of $\overline{YZ}$,so $YZ = 2 \times YM$.
$YZ = 2 \times 14.5 = 29$.

(A) According to Apollonius's Theorem for $\Delta PQR$ with median $\overline{PS}$:
$PQ^2 + PR^2 = 2(PS^2 + QS^2)$
Given $PQ = 9$,$PR = 40$,and $PS = 20.5$:
$9^2 + 40^2 = 2(20.5^2 + QS^2)$
$81 + 1600 = 2(420.25 + QS^2)$
$1681 = 2(420.25 + QS^2)$
$840.5 = 420.25 + QS^2$
$QS^2 = 840.5 - 420.25 = 420.25$
$QS = \sqrt{420.25} = 20.5$
Since $\overline{PS}$ is a median,$S$ is the midpoint of $\overline{QR}$,so $QR = 2 \times QS = 2 \times 20.5 = 41$.

(B) According to Apollonius's Theorem,for any triangle $ABC$ with median $AD$ to side $BC$:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
Given $AB = 10$,$AC = 24$,and $AD = 13$.
Substituting the values:
$10^2 + 24^2 = 2(13^2 + BD^2)$
$100 + 576 = 2(169 + BD^2)$
$676 = 2(169 + BD^2)$
$338 = 169 + BD^2$
$BD^2 = 338 - 169 = 169$
$BD = \sqrt{169} = 13$
Since $AD$ is a median,$D$ is the midpoint of $BC$,so $BC = 2 \times BD$.
$BC = 2 \times 13 = 26$.

(C) Given that $\overline{AD}$ is a median to side $\overline{BC}$,so $D$ is the midpoint of $\overline{BC}$.
Since $AD = BD = 8.5$,in $\Delta ABD$,we have $AD = BD = 8.5$. This implies $\angle DAB = \angle DBA$.
In $\Delta ABD$,by the Law of Cosines or using the property of isosceles triangles,we can find $\angle ADB$. Since $AD = BD = 8.5$,$D$ is the center of a circle passing through $A, B,$ and $C$ if $CD = 8.5$ as well.
Actually,since $AD$ is a median,$BD = DC = 8.5$. Thus,$AD = BD = DC = 8.5$.
This means $D$ is the circumcenter of $\Delta ABC$,and $BC$ is the diameter of the circumcircle.
Therefore,$\angle BAC = 90^\circ$ (angle in a semicircle).
In right-angled $\Delta ABC$,by Pythagoras theorem,$AC^2 + AB^2 = BC^2$.
We know $AB = 8$ and $BC = BD + DC = 8.5 + 8.5 = 17$.
$AC^2 + 8^2 = 17^2$
$AC^2 + 64 = 289$
$AC^2 = 225$
$AC = 15$.

(D) In $\Delta XYZ$,$\overline{XM}$ is a median to the side $\overline{YZ}$.
By definition of a median,$M$ is the midpoint of $\overline{YZ}$,so $YM = MZ = 30.5$.
Thus,the length of the side $YZ = YM + MZ = 30.5 + 30.5 = 61$.
We use Apollonius's Theorem for $\Delta XYZ$ with median $\overline{XM}$:
$XY^2 + XZ^2 = 2(XM^2 + YM^2)$.
Substitute the given values: $11^2 + XZ^2 = 2(30.5^2 + 30.5^2)$.
$121 + XZ^2 = 2(930.25 + 930.25)$.
$121 + XZ^2 = 2(1860.5)$.
$121 + XZ^2 = 3721$.
$XZ^2 = 3721 - 121$.
$XZ^2 = 3600$.
$XZ = \sqrt{3600} = 60$.

(A) Given that $\overline{PS}$ is a median of $\Delta PQR,$ it divides the side $QR$ into two equal parts. Thus,$QS = SR = 18.5.$
Since $PS = QS = SR = 18.5,$ the point $S$ is the circumcenter of $\Delta PQR$ with respect to the side $QR,$ implying $\angle QPR = 90^\circ$ (by the property that if the median to a side is half the length of that side,the triangle is a right-angled triangle).
In the right-angled $\Delta PQR,$ by the Pythagorean theorem:
$PQ^2 + PR^2 = QR^2$
$QR = QS + SR = 18.5 + 18.5 = 37$
$12^2 + PR^2 = 37^2$
$144 + PR^2 = 1369$
$PR^2 = 1369 - 144 = 1225$
$PR = \sqrt{1225} = 35.$

(N/A) Let $ABCD$ be a parallelogram with diagonals $AC$ and $BD$ intersecting at $O$.
By the Parallelogram Law,we know that the sum of the squares of the four sides is equal to the sum of the squares of the two diagonals.
Let the sides be $AB$,$BC$,$CD$,and $DA$. Since $ABCD$ is a parallelogram,$AB = CD$ and $BC = DA$.
According to Apollonius Theorem in $\triangle ABC$ and $\triangle ADC$ with median $BO$ (where $O$ is the midpoint of $AC$):
$AB^2 + BC^2 = 2(BO^2 + AO^2)$
$AD^2 + CD^2 = 2(DO^2 + AO^2)$
Since $O$ is the midpoint of $BD$,$BO = DO$.
Adding the two equations:
$AB^2 + BC^2 + CD^2 + DA^2 = 2(BO^2 + AO^2 + DO^2 + AO^2) = 2(2BO^2 + 2AO^2) = 4BO^2 + 4AO^2$.
Since $BD = 2BO$ and $AC = 2AO$,then $BD^2 = 4BO^2$ and $AC^2 = 4AO^2$.
Therefore,$AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2$.

(C) Given that $\Delta PQR$ is a right-angled triangle with $m \angle Q = 90^\circ$.
$PQ$ and $QR$ are the legs,and $PR$ is the hypotenuse.
Using the Pythagorean theorem: $PQ^2 + QR^2 = PR^2$.
Substitute the given values: $5^2 + QR^2 = 13^2$.
$25 + QR^2 = 169$.
$QR^2 = 169 - 25 = 144$.
$QR = \sqrt{144} = 12$.
The area of a right-angled triangle is given by $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
$\text{Area} = \frac{1}{2} \times PQ \times QR = \frac{1}{2} \times 5 \times 12$.
$\text{Area} = 5 \times 6 = 30$.

(D) Given that $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^{\circ}$.
Here,$AC$ is the hypotenuse and $AB$ is one of the legs.
Using the Pythagorean theorem: $AC^2 = AB^2 + BC^2$.
Substituting the given values: $10^2 = 6^2 + BC^2$.
$100 = 36 + BC^2$.
$BC^2 = 100 - 36 = 64$.
$BC = \sqrt{64} = 8$.
The area of a right-angled triangle is given by the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Here,base $BC = 8$ and height $AB = 6$.
$\text{Area} = \frac{1}{2} \times 8 \times 6 = 4 \times 6 = 24$.

(A) $1$. In $\Delta ABC$,since $m\angle B = 90^{\circ}$,we can find the length of $BC$ using the Pythagorean theorem: $AB^2 + BC^2 = AC^2$.
$2$. Substituting the given values: $3^2 + BC^2 = 5^2$,which gives $9 + BC^2 = 25$.
$3$. Therefore,$BC^2 = 25 - 9 = 16$,so $BC = 4$.
$4$. The area of $\Delta ABC$ can be calculated in two ways: $\text{Area} = \frac{1}{2} \times AB \times BC$ or $\text{Area} = \frac{1}{2} \times AC \times BM$.
$5$. Equating the two expressions: $\frac{1}{2} \times 3 \times 4 = \frac{1}{2} \times 5 \times BM$.
$6$. Simplifying: $12 = 5 \times BM$,which leads to $BM = \frac{12}{5} = 2.4$.

(B) Given that $\Delta ABC$ is a right-angled triangle where $m \angle B = 90^{\circ}$.
Since $\overline{AB} \cong \overline{BC}$,let $AB = BC = x$.
According to the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Substituting the values,$AC^2 = x^2 + x^2 = 2x^2$.
Therefore,$AC = \sqrt{2x^2} = x\sqrt{2}$.
We need to find the ratio $AB : AC$.
$AB : AC = x : x\sqrt{2} = 1 : \sqrt{2}$.

(C) In $\Delta ABC$,since $m \angle B = 90^{\circ}$,the triangle is a right-angled triangle.
By the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Given $AB = 20$ and $AC = 29$,we have $29^2 = 20^2 + BC^2$.
$841 = 400 + BC^2$.
$BC^2 = 841 - 400 = 441$.
$BC = \sqrt{441} = 21$.
The perimeter of $\Delta ABC = AB + BC + AC$.
Perimeter $= 20 + 21 + 29 = 70$.

(B) Let the side length of the square be $s$.
The perimeter of a square is given by $4s$.
Given $4s = 40$, so $s = 10$.
The diagonals of a square are equal in length.
The length of the diagonal $d$ of a square with side $s$ is given by $d = s \sqrt{2}$.
Thus, $AC = BD = 10 \sqrt{2}$.
Therefore, $AC + BD = 10 \sqrt{2} + 10 \sqrt{2} = 20 \sqrt{2}$.

(A) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
First,find the length of the hypotenuse $AC$ using the Pythagorean theorem: $AC^2 = AB^2 + BC^2$.
$AC^2 = (3.6)^2 + (4.8)^2 = 12.96 + 23.04 = 36$.
$AC = \sqrt{36} = 6$.
Since $\overline{BE}$ is the median to the hypotenuse $AC$,its length is $BE = \frac{1}{2} AC$.
$BE = \frac{1}{2} \times 6 = 3$.

(B) Given that in $\Delta ABC$,$m\angle B = 90^{\circ}$.
Let $AB = 3x$ and $BC = 4x$ for some constant $x > 0$.
By the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
$AC^2 = (3x)^2 + (4x)^2 = 9x^2 + 16x^2 = 25x^2$.
Taking the square root,$AC = \sqrt{25x^2} = 5x$.
We need to find the ratio $AB : AC$.
$AB : AC = 3x : 5x = 3 : 5$.
Therefore,the correct option is $B$.

(C) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
Here,$\Delta ABC$ is a right-angled triangle with $m\angle B = 90^{\circ}$.
Since $\overline{BM}$ is the median to the hypotenuse $\overline{AC}$,we have $BM = \frac{1}{2} AC$.
Given $BM = 12.5$,so $12.5 = \frac{1}{2} AC$,which implies $AC = 25$.
Now,using the Pythagorean theorem in $\Delta ABC$:
$AB^2 + BC^2 = AC^2$
$15^2 + BC^2 = 25^2$
$225 + BC^2 = 625$
$BC^2 = 625 - 225 = 400$
$BC = \sqrt{400} = 20$.

(D) According to Apollonius's Theorem for $\Delta ABC$ with median $\overline{AD}$:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
Given $AB^2 + AC^2 = 338$ and $AD = 5$.
Substituting the values into the formula:
$338 = 2(5^2 + BD^2)$
$338 = 2(25 + BD^2)$
$169 = 25 + BD^2$
$BD^2 = 169 - 25 = 144$
$BD = \sqrt{144} = 12$
Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$,so $BC = 2 \times BD$.
$BC = 2 \times 12 = 24$.

(A) The perimeter of an equilateral triangle is given by $P = 3a$, where $a$ is the side length.
Given $P = 24$, we have $3a = 24$, which implies $a = 8$.
The altitude $h$ of an equilateral triangle is given by the formula $h = \frac{\sqrt{3}}{2} a$.
Substituting $a = 8$ into the formula, we get $h = \frac{\sqrt{3}}{2} \times 8 = 4 \sqrt{3}$.
Therefore, the length of the altitude is $4 \sqrt{3}$.

(B) In $\Delta ABC$,we are given $m \angle B = 90^{\circ}$.
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 $m \angle B = 90^{\circ}$,we get $m \angle A + m \angle C = 90^{\circ}$.
Given the ratio $m \angle A : m \angle C = 1 : 2$,let $m \angle A = x$ and $m \angle C = 2x$.
Thus,$x + 2x = 90^{\circ}$,which implies $3x = 90^{\circ}$,so $x = 30^{\circ}$.
Therefore,$m \angle A = 30^{\circ}$ and $m \angle C = 60^{\circ}$.
In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle,the side opposite to $30^{\circ}$ is half the hypotenuse.
Here,$BC$ is opposite to $\angle A = 30^{\circ}$,so $BC = \frac{1}{2} AC$.
Given $BC = 4$,we have $4 = \frac{1}{2} AC$,which implies $AC = 8$.

(D) In $\Delta ABC$,we are given $m \angle B = 90^{\circ}$ and $m \angle C = 30^{\circ}$.
Since the sum of angles in a triangle is $180^{\circ}$,$m \angle A = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ}$.
In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle,the side opposite to $30^{\circ}$ is half the hypotenuse.
Here,$AB$ is opposite to $30^{\circ}$ (at $C$),so $AB = \frac{1}{2} AC$.
Given $AB = 5$,we have $5 = \frac{1}{2} AC$,which implies $AC = 10$.
In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
Therefore,$BD = \frac{1}{2} AC = \frac{1}{2} \times 10 = 5$.

(D) Given that $\Delta XYZ$ is a right-angled triangle with $m \angle Y = 90^{\circ}$.
According to the Pythagorean theorem,$XY^2 + YZ^2 = XZ^2$.
Let $XY = 15k$ and $XZ = 17k$ for some constant $k > 0$.
Substituting the values into the theorem: $(15k)^2 + 4^2 = (17k)^2$.
$225k^2 + 16 = 289k^2$.
$16 = 289k^2 - 225k^2$.
$16 = 64k^2$.
$k^2 = \frac{16}{64} = \frac{1}{4}$.
Therefore,$k = \frac{1}{2} = 0.5$.
Now,calculate the side lengths:
$XY = 15 \times 0.5 = 7.5$.
$XZ = 17 \times 0.5 = 8.5$.
$YZ = 4$.
The perimeter of $\Delta XYZ = XY + YZ + XZ = 7.5 + 4 + 8.5 = 20$.

(A) Given that $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^{\circ}$ and $AB = BC$.
Let $AB = BC = x$.
According to the Pythagorean theorem,$AB^2 + BC^2 = AC^2$.
Substituting the values,we get $x^2 + x^2 = (14\sqrt{2})^2$.
$2x^2 = 196 \times 2$.
$2x^2 = 392$.
$x^2 = 196$.
$x = 14$.
Thus,the base $BC = 14$ and the height $AB = 14$.
The area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14 \times 14 = 98$ square units.

(B) In a right-angled triangle,if an altitude is drawn from the right-angle vertex to the hypotenuse,the altitude is the geometric mean of the segments of the hypotenuse.
According to the geometric mean theorem: $BM^2 = AM \cdot CM$.
Given $AM = 2x$,$BM = 3x + 5$,and $CM = 8x$.
Substituting these values into the equation: $(3x + 5)^2 = (2x)(8x)$.
Expanding the left side: $9x^2 + 30x + 25 = 16x^2$.
Rearranging the terms to form a quadratic equation: $7x^2 - 30x - 25 = 0$.
Factoring the quadratic equation: $7x^2 - 35x + 5x - 25 = 0$.
$7x(x - 5) + 5(x - 5) = 0$.
$(7x + 5)(x - 5) = 0$.
This gives $x = 5$ or $x = -5/7$.
Since $x$ represents a length,it must be positive,so $x = 5$.

(C) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,if $\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$,then by the geometric mean theorem (or property of similar triangles $\Delta AMB \sim \Delta ABC$),we have the relation: $AB^2 = AM \cdot AC$.
Given values are $AB = x - 2$,$AM = x - 6$,and $AC = 2x - 4$.
Substituting these into the equation: $(x - 2)^2 = (x - 6)(2x - 4)$.
Expanding both sides: $x^2 - 4x + 4 = 2x^2 - 4x - 12x + 24$.
Simplifying: $x^2 - 4x + 4 = 2x^2 - 16x + 24$.
Rearranging the terms to form a quadratic equation: $x^2 - 12x + 20 = 0$.
Factoring the quadratic equation: $(x - 10)(x - 2) = 0$.
This gives $x = 10$ or $x = 2$.
If $x = 2$,then $AB = 2 - 2 = 0$,which is impossible for a triangle side length. Thus,$x = 10$ is the only valid solution.

(D) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,if $\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$,then by the geometric mean theorem,$BM^2 = AM \cdot CM$.
Substituting the given values: $(x + 1)^2 = (x - 1)(x + 4)$.
Expanding both sides: $x^2 + 2x + 1 = x^2 + 4x - x - 4$.
Simplifying the equation: $x^2 + 2x + 1 = x^2 + 3x - 4$.
Subtracting $x^2$ from both sides: $2x + 1 = 3x - 4$.
Rearranging the terms: $3x - 2x = 1 + 4$.
Thus,$x = 5$.

(A) In a right-angled triangle $\Delta ABC$ with $\angle B = 90^\circ$,if $\overline{BM}$ is the altitude to the hypotenuse $AC$,then by the Geometric Mean Theorem (or properties of similar triangles $\Delta BMC \sim \Delta BCA$),we have the relation: $BC^2 = CM \cdot AC$.
Given $BC = 2x + 2$,$CM = x + 1$,and $AC = 5x$.
Substituting these values into the equation: $(2x + 2)^2 = (x + 1)(5x)$.
Expanding the left side: $(2(x + 1))^2 = 4(x + 1)^2 = 4(x^2 + 2x + 1)$.
So,$4(x^2 + 2x + 1) = 5x^2 + 5x$.
$4x^2 + 8x + 4 = 5x^2 + 5x$.
Rearranging the terms to form a quadratic equation: $x^2 - 3x - 4 = 0$.
Factoring the quadratic: $(x - 4)(x + 1) = 0$.
This gives $x = 4$ or $x = -1$.
Since $x$ represents a length component and $BC, CM, AC$ must be positive,we discard $x = -1$.
Therefore,$x = 4$.

(B) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the diagonals $AC$ and $BD$ intersect at point $O$.
Since $AC = 24$,the length of $AO = AC / 2 = 24 / 2 = 12$.
In the right-angled triangle $\triangle AOB$,by the Pythagorean theorem,$AB^2 = AO^2 + BO^2$.
Given $AB = 13$ and $AO = 12$,we have $13^2 = 12^2 + BO^2$.
$169 = 144 + BO^2$.
$BO^2 = 169 - 144 = 25$.
$BO = \sqrt{25} = 5$.
Since the diagonals bisect each other,$BD = 2 \times BO = 2 \times 5 = 10$.