Mix Examples - Triangles Questions in English

(A) rhombus is a quadrilateral where all sides are equal. Let the side length be $s$.
Given the perimeter is $68$,we have $4s = 68$,which implies $s = 17$.
In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the diagonals $AC$ and $BD$ intersect at point $O$.
Since $AC = 30$,the length $AO = AC / 2 = 15$.
In the right-angled triangle $\triangle AOB$,by the Pythagorean theorem,$AO^2 + BO^2 = AB^2$.
Substituting the values: $15^2 + BO^2 = 17^2$.
$225 + BO^2 = 289$.
$BO^2 = 289 - 225 = 64$.
$BO = \sqrt{64} = 8$.
Since the diagonals bisect each other,$BD = 2 \times BO = 2 \times 8 = 16$.

(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 $\Delta ABC = AB + AC + BC = 18 + 14 + 16 = 48$.

(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 $YZ = 2 \times YM$.
$YZ = 2 \times 14.5 = 29$.

(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: $11^2 + XZ^2 = 2(30.5^2 + 30.5^2)$.
$121 + XZ^2 = 2(930.25 + 930.25)$.
$121 + XZ^2 = 2(1860.5)$.
$121 + XZ^2 = 3721$.
$XZ^2 = 3721 - 121$.
$XZ^2 = 3600$.
$XZ = \sqrt{3600} = 60$.

(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 $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
$\text{Area} = \frac{1}{2} \times PQ \times QR = \frac{1}{2} \times 5 \times 12$.
$\text{Area} = 5 \times 6 = 30$.

(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, $AC + BD = 10 \sqrt{2} + 10 \sqrt{2} = 20 \sqrt{2}$.

(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 $BC = 2 \times BD$.
$BC = 2 \times 12 = 24$.

(D) In a rhombus,the diagonals bisect each other at right angles. Let $AM = x$ and $DM = y$.
Since the diagonals intersect at $M$ at a $90^{\circ}$ angle,$\triangle AMB$ is a right-angled triangle.
By the Pythagorean theorem,$AM^2 + BM^2 = AB^2$.
Given $AM + DM = 17$,we have $x + y = 17$,so $y = 17 - x$.
In $\triangle AMB$,$x^2 + BM^2 = 13^2 = 169$.
In $\triangle AMD$,$x^2 + y^2 = AD^2$. Since all sides of a rhombus are equal,$AD = AB = 13$.
Thus,$x^2 + y^2 = 169$.
Substituting $y = 17 - x$: $x^2 + (17 - x)^2 = 169$.
$x^2 + 289 - 34x + x^2 = 169$.
$2x^2 - 34x + 120 = 0$.
$x^2 - 17x + 60 = 0$.
$(x - 12)(x - 5) = 0$.
So,$x = 12$ or $x = 5$.
If $x = 12$,then $y = 17 - 12 = 5$.
If $x = 5$,then $y = 17 - 5 = 12$.
Since $AC > BD$,$AM > DM$,so $x > y$. Thus,$x = 12$ and $y = 5$.
$DM = y = 5$.
Since the diagonals bisect each other,$BD = 2 \times DM = 2 \times 5 = 10$.

(C) Let $AB = x$.
Given $BC - AB = 7$,so $BC = x + 7$.
Given $AC - BC = 2$,so $AC = BC + 2 = (x + 7) + 2 = x + 9$.
Since $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^{\circ}$,by the Pythagorean theorem:
$AB^2 + BC^2 = AC^2$
$x^2 + (x + 7)^2 = (x + 9)^2$
$x^2 + x^2 + 14x + 49 = x^2 + 18x + 81$
$2x^2 + 14x + 49 = x^2 + 18x + 81$
$x^2 - 4x - 32 = 0$
$(x - 8)(x + 4) = 0$
Since $x$ represents a length,$x = 8$.
Thus,$AB = 8$,$BC = 8 + 7 = 15$,and $AC = 8 + 9 = 17$.
The perimeter of $\Delta ABC = AB + BC + AC = 8 + 15 + 17 = 40$.

(D) In a rectangle $\square ABCD$,opposite sides are equal,so $AB = CD$ and $BC = DA$.
Given the equation: $AB^{2} + BC^{2} + CD^{2} + DA^{2} = 128$.
Substituting the equal sides,we get: $AB^{2} + BC^{2} + AB^{2} + BC^{2} = 128$.
This simplifies to: $2(AB^{2} + BC^{2}) = 128$.
Dividing by $2$,we get: $AB^{2} + BC^{2} = 64$.
In the right-angled triangle $\triangle ABC$,by the Pythagorean theorem: $AC^{2} = AB^{2} + BC^{2}$.
Therefore,$AC^{2} = 64$.
Taking the square root of both sides,$AC = \sqrt{64} = 8$.

(B) The area of an equilateral triangle with side length $a$ is given by the formula: $\text{Area} = \frac{\sqrt{3}}{4} a^2$.
Given that the area is $36 \sqrt{3}$,we can set up the equation:
$\frac{\sqrt{3}}{4} a^2 = 36 \sqrt{3}$.
Divide both sides by $\sqrt{3}$:
$\frac{1}{4} a^2 = 36$.
Multiply both sides by $4$:
$a^2 = 36 \times 4 = 144$.
Taking the square root of both sides:
$a = \sqrt{144} = 12$.
Therefore,the measure of its side is $12$.

(D) Given that $\Delta ABC$ is an isosceles triangle with $AB = AC = 25$ and base $BC = 14$.
Since $\overline{AM}$ is an altitude to the base $BC$,in an isosceles triangle,the altitude from the vertex to the base also acts as the median.
Therefore,$M$ is the midpoint of $BC$.
Thus,$BM = MC = \frac{BC}{2} = \frac{14}{2} = 7$.
In the right-angled triangle $\Delta ABM$,by the Pythagorean theorem:
$AB^2 = AM^2 + BM^2$
$25^2 = AM^2 + 7^2$
$625 = AM^2 + 49$
$AM^2 = 625 - 49 = 576$
$AM = \sqrt{576} = 24$.
Hence,the length of the altitude $AM$ is $24$.

(B) From the diagram,there are $4$ steps in total.
Total horizontal distance (base) $= 4 \times 20 \, cm = 80 \, cm = 0.8 \, m$.
Total vertical distance (height) $= 4 \times 15 \, cm = 60 \, cm = 0.6 \, m$.
The distance between points $A$ and $B$ forms the hypotenuse of a right-angled triangle with base $0.8 \, m$ and height $0.6 \, m$.
Using the Pythagorean theorem: $AB = \sqrt{(0.8)^2 + (0.6)^2} = \sqrt{0.64 + 0.36} = \sqrt{1.00} = 1 \, m$.

(N/A) $1$. In $\Delta ABC$,by the Pythagorean theorem,$AC^{2} = AB^{2} + BC^{2}$.
$2$. In right-angled $\Delta ABD$,$AD^{2} = AB^{2} + BD^{2}$. Since $D$ is the midpoint of $BC$,$BD = \frac{1}{2} BC$. Thus,$AD^{2} = AB^{2} + (\frac{1}{2} BC)^{2} = AB^{2} + \frac{1}{4} BC^{2}$.
$3$. In right-angled $\Delta CBF$,$CF^{2} = BC^{2} + BF^{2}$. Since $F$ is the midpoint of $AB$,$BF = \frac{1}{2} AB$. Thus,$CF^{2} = BC^{2} + (\frac{1}{2} AB)^{2} = BC^{2} + \frac{1}{4} AB^{2}$.
$4$. Adding the two equations: $AD^{2} + CF^{2} = (AB^{2} + \frac{1}{4} BC^{2}) + (BC^{2} + \frac{1}{4} AB^{2})$.
$5$. Simplifying: $AD^{2} + CF^{2} = (1 + \frac{1}{4}) AB^{2} + (1 + \frac{1}{4}) BC^{2} = \frac{5}{4} AB^{2} + \frac{5}{4} BC^{2}$.
$6$. Factoring out $\frac{5}{4}$: $AD^{2} + CF^{2} = \frac{5}{4} (AB^{2} + BC^{2})$.
$7$. Since $AB^{2} + BC^{2} = AC^{2}$,we get $AD^{2} + CF^{2} = \frac{5}{4} AC^{2}$.

(N/A) In $\Delta ABC$,$\overline{AM}$ is the altitude to side $\overline{BC}$,so $\angle AMB = 90^{\circ}$ and $\angle AMC = 90^{\circ}$.
In $\Delta AMC$,by Pythagoras theorem: $AC^{2} = AM^{2} + MC^{2}$.
Since $M$ lies on $BC$,we have $MC = BC - BM$.
Substituting this into the equation: $AC^{2} = AM^{2} + (BC - BM)^{2}$.
Expanding the square: $AC^{2} = AM^{2} + BC^{2} + BM^{2} - 2 \cdot BC \cdot BM$.
In $\Delta AMB$,by Pythagoras theorem: $AB^{2} = AM^{2} + BM^{2}$.
Substituting $AM^{2} + BM^{2} = AB^{2}$ into the equation: $AC^{2} = AB^{2} + BC^{2} - 2 \cdot BC \cdot BM$.
Hence,the result is proved.

(N/A) In $\Delta AMC$,$\angle AMC = 90^{\circ}$. By Pythagoras theorem,$AC^{2} = AM^{2} + MC^{2}$.
Since $MC = MB + BC$,we have $AC^{2} = AM^{2} + (MB + BC)^{2}$.
Expanding the square,$AC^{2} = AM^{2} + MB^{2} + BC^{2} + 2 \cdot MB \cdot BC$.
In $\Delta AMB$,$\angle AMB = 90^{\circ}$. By Pythagoras theorem,$AB^{2} = AM^{2} + MB^{2}$.
Substituting $AB^{2}$ for $AM^{2} + MB^{2}$ in the equation,we get $AC^{2} = AB^{2} + BC^{2} + 2 \cdot BC \cdot BM$.

(D) triangle is a right-angled triangle if the square of the longest side is equal to the sum of the squares of the other two sides (Pythagoras theorem: $a^2 + b^2 = c^2$).
For option $A$: $3^2 + 4^2 = 9 + 16 = 25 = 5^2$. This is a right-angled triangle.
For option $B$: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. This is a right-angled triangle.
For option $C$: $7^2 + 24^2 = 49 + 576 = 625 = 25^2$. This is a right-angled triangle.
For option $D$: $8^2 + 24^2 = 64 + 576 = 640$,while $26^2 = 676$. Since $640 \neq 676$,this is not a right-angled triangle.

(D) Given that $\Delta XYZ$ is a right-angled triangle with $m\angle X = 90^{\circ}$.
Here,$YZ$ is the hypotenuse and $XY$ is one of the legs.
Using the Pythagorean theorem: $XY^2 + XZ^2 = YZ^2$.
Substituting the given values: $8^2 + XZ^2 = 17^2$.
$64 + XZ^2 = 289$.
$XZ^2 = 289 - 64 = 225$.
$XZ = \sqrt{225} = 15$.
The area of a right-angled triangle is given by $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
$\text{Area} = \frac{1}{2} \times XY \times XZ$.
$\text{Area} = \frac{1}{2} \times 8 \times 15 = 4 \times 15 = 60$.

(A) According to Apollonius's theorem for $\Delta ABC$ with median $\overline{AD}$:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
Given that $\overline{AD}$ is a median to side $\overline{BC}$,$D$ is the midpoint of $\overline{BC}$.
Therefore,$BD = DC = \frac{BC}{2} = \frac{18}{2} = 9$.
Substituting the given values into the theorem:
$8^2 + AC^2 = 2(7^2 + 9^2)$
$64 + AC^2 = 2(49 + 81)$
$64 + AC^2 = 2(130)$
$64 + AC^2 = 260$
$AC^2 = 260 - 64$
$AC^2 = 196$
$AC = \sqrt{196} = 14$.

(C) First,identify the longest side of $\Delta ABC$. We have $AB = 4$,$BC = 2\sqrt{3} \approx 3.46$,and $AC = 2\sqrt{7} \approx 5.29$. Since $AC$ is the longest side,we need to find the length of the median $BD$ drawn to side $AC$.
Using Apollonius Theorem for $\Delta ABC$ with median $BD$ on side $AC$:
$AB^2 + BC^2 = 2(BD^2 + AD^2)$
Since $D$ is the midpoint of $AC$,$AD = AC/2 = (2\sqrt{7})/2 = \sqrt{7}$.
Substitute the values: $4^2 + (2\sqrt{3})^2 = 2(BD^2 + (\sqrt{7})^2)$
$16 + 12 = 2(BD^2 + 7)$
$28 = 2(BD^2 + 7)$
$14 = BD^2 + 7$
$BD^2 = 7$
$BD = \sqrt{7}$.

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

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

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

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

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

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

(D) In $\Delta ABC$,$m \angle A = 90^{\circ}$ and $\overline{AD}$ is an altitude to the hypotenuse $\overline{BC}$.
According to the geometric mean theorem for right-angled triangles,$AB^{2} = BD \cdot BC$.
Substituting the given values: $(\sqrt{5})^{2} = 2 \cdot BC$.
$5 = 2 \cdot BC$.
$BC = \frac{5}{2} = 2.5$.
Therefore,the length of the hypotenuse $\overline{BC}$ is $2.5$.

(A) In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$.
According to the geometric mean theorem for right-angled triangles,$AB^2 = AM \cdot AC$.
Given $AB = \sqrt{10}$ and $AM = 2.5$.
Substituting the values: $(\sqrt{10})^2 = 2.5 \cdot AC$.
$10 = 2.5 \cdot AC$.
$AC = \frac{10}{2.5} = 4$.
Since $AC = AM + MC$,we have $4 = 2.5 + MC$.
$MC = 4 - 2.5 = 1.5$.

(C) In $\Delta PQR$,since $m\angle Q = 90^{\circ}$,the triangle is a right-angled triangle.
By the Pythagoras theorem,$PR^{2} = PQ^{2} + QR^{2}$.
Therefore,$PQ^{2} = PR^{2} - QR^{2}$.
Substituting the given values:
$PQ^{2} = (a^{2} + b^{2})^{2} - (a^{2} - b^{2})^{2}$.
Using the algebraic identity $(x + y)^{2} - (x - y)^{2} = 4xy$,where $x = a^{2}$ and $y = b^{2}$:
$PQ^{2} = 4(a^{2})(b^{2}) = 4a^{2}b^{2}$.
Taking the square root on both sides:
$PQ = \sqrt{4a^{2}b^{2}} = 2ab$.

(B) Let $AB = AC = x$.
In $\Delta ABC$,$m \angle A = 90^\circ$.
By the Pythagorean theorem,$AB^2 + AC^2 = BC^2$.
Substituting the given values: $x^2 + x^2 = (\sqrt{2} a)^2$.
$2x^2 = 2a^2$.
$x^2 = a^2$,so $x = a$ (since $a \in R^+$).
Thus,$AB = AC = a$.
The area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times AC$.
Area $= \frac{1}{2} \times a \times a = \frac{1}{2} a^2$.

(C) In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$.
By the properties of right-angled triangles with an altitude to the hypotenuse,we have:
$AB^{2} = AM \cdot AC$ and $BC^{2} = MC \cdot AC$
Subtracting these two equations:
$AB^{2} - BC^{2} = AM \cdot AC - MC \cdot AC$
$AB^{2} - BC^{2} = AC(AM - MC)$
Given that $AB^{2} - BC^{2} = 175$ and $AM - MC = 7$,we substitute these values:
$175 = AC(7)$
$AC = \frac{175}{7}$
$AC = 25$

(A) Given that $\square PQRS$ is a rectangle.
Area of rectangle $PQRS = PQ \cdot QR = 12$.
Sum of adjacent sides $PQ + QR = 7$.
In $\triangle PQR$,$\angle Q = 90^\circ$.
By Pythagoras theorem,$PR^2 = PQ^2 + QR^2$.
We know that $(PQ + QR)^2 = PQ^2 + QR^2 + 2(PQ \cdot QR)$.
Therefore,$PQ^2 + QR^2 = (PQ + QR)^2 - 2(PQ \cdot QR)$.
Substituting the given values: $PR^2 = (7)^2 - 2(12)$.
$PR^2 = 49 - 24 = 25$.
Taking the square root on both sides,$PR = 5$.

(B) Given that in $\Delta ABC$,$m \angle A = m \angle B + m \angle C$.
We know that the sum of angles in a triangle is $180^{\circ}$,so $m \angle A + m \angle B + m \angle C = 180^{\circ}$.
Substituting $m \angle B + m \angle C$ with $m \angle A$,we get $m \angle A + m \angle A = 180^{\circ}$,which implies $2 m \angle A = 180^{\circ}$,so $m \angle A = 90^{\circ}$.
Since $\Delta ABC$ is a right-angled triangle with the right angle at $A$,$BC$ is the hypotenuse.
By the Pythagorean theorem,$BC^2 = AB^2 + AC^2$.
Given $AB = 7$ and $BC = 25$,we have $25^2 = 7^2 + AC^2$.
$625 = 49 + AC^2$,so $AC^2 = 625 - 49 = 576$.
Thus,$AC = \sqrt{576} = 24$.
The perimeter of $\Delta ABC = AB + BC + AC = 7 + 25 + 24 = 56$.

(D) Let $\overline{AC}$ represent the ladder and $\overline{AB}$ represent the height of the wall.
In $\Delta ABC$,$\angle B = 90^{\circ}$.
By the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Given $AC = 6.5 \, m$ and $AB = 6 \, m$.
$6.5^2 = 6^2 + BC^2$
$42.25 = 36 + BC^2$
$BC^2 = 42.25 - 36 = 6.25$
$BC = \sqrt{6.25} = 2.5 \, m$.
Thus,the lower end of the ladder is $2.5 \, m$ away from the wall.

(C) In $\Delta ABC$,$D$ is the midpoint of $\overline{BC}$. Therefore,$\overline{AD}$ is a median.
By Apollonius' theorem:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
$7^2 + 5^2 = 2(5^2 + BD^2)$
$49 + 25 = 2(25 + BD^2)$
$74 = 2(25 + BD^2)$
$37 = 25 + BD^2$
$BD^2 = 12$
$BD = \sqrt{12} = 2\sqrt{3}$
Since $D$ is the midpoint of $\overline{BC}$,$BC = 2 \times BD = 2 \times 2\sqrt{3} = 4\sqrt{3}$.

(C) In $\Delta ABC$,$\overline{AD}$ is a median to side $\overline{BC}$.
According to Apollonius's Theorem,$AB^2 + AC^2 = 2(AD^2 + BD^2)$.
Given $AB^2 + AC^2 = 148$ and $AD = 7$.
Substituting the values: $148 = 2(7^2 + BD^2)$.
$148 = 2(49 + BD^2)$.
$74 = 49 + BD^2$.
$BD^2 = 74 - 49 = 25$.
$BD = \sqrt{25} = 5$.
Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$,so $BC = 2 \times BD$.
$BC = 2 \times 5 = 10$.

(B) In $\Delta ADC$,$\angle D = 90^{\circ}$.
By Pythagoras theorem,$AC^{2} = AD^{2} + CD^{2}$.
$25^{2} = AD^{2} + 7^{2}$.
$625 = AD^{2} + 49$.
$AD^{2} = 625 - 49 = 576$.
$AD = \sqrt{576} = 24$.
The perimeter of rectangle $ABCD = 2(AD + CD) = 2(24 + 7) = 2(31) = 62$.

(B) In rhombus $XYZW$,let the diagonals $XZ$ and $YW$ intersect at point $O$.
The diagonals of a rhombus bisect each other at right angles.
Therefore,$OX = \frac{1}{2} XZ = \frac{1}{2} \times 14 = 7$ and $OY = \frac{1}{2} YW = \frac{1}{2} \times 48 = 24$.
In the right-angled triangle $\Delta XOY$,by the Pythagorean theorem:
$XY^2 = OX^2 + OY^2$
$XY^2 = 7^2 + 24^2$
$XY^2 = 49 + 576 = 625$
$XY = \sqrt{625} = 25$.

(C) Given that in $\Delta PQR$,$m \angle Q : m \angle R : m \angle P = 1 : 2 : 1$.
Let the angles be $x, 2x, x$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $x + 2x + x = 180^{\circ}$,which gives $4x = 180^{\circ}$,so $x = 45^{\circ}$.
Thus,$m \angle Q = 45^{\circ}$,$m \angle R = 90^{\circ}$,and $m \angle P = 45^{\circ}$.
Since $m \angle P = m \angle Q$,the sides opposite to these angles are equal,so $QR = PR$.
In the right-angled $\Delta PQR$ (where $m \angle R = 90^{\circ}$),by the Pythagorean theorem:
$PQ^2 = PR^2 + QR^2$
Since $QR = PR$,we have $PQ^2 = PR^2 + PR^2 = 2PR^2$.
Given $PQ = 2 \sqrt{6}$,so $PQ^2 = (2 \sqrt{6})^2 = 4 \times 6 = 24$.
Therefore,$24 = 2PR^2$,which implies $PR^2 = 12$.
Taking the square root,$PR = \sqrt{12} = 2 \sqrt{3}$.

(D) In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$,which means $\Delta ABC$ is an isosceles triangle.
In an isosceles triangle,the median to the base is also the altitude to the base.
Therefore,$\overline{AD} \perp \overline{BC}$,so $\overline{AD}$ is the height of the triangle with respect to base $\overline{BC}$.
Area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height}$
Area $= \frac{1}{2} \times BC \times AD$
Area $= \frac{1}{2} \times 18 \times 12$
Area $= 9 \times 12 = 108$.

(A) In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^\circ$.
Given $m \angle A + m \angle C = m \angle B$,substituting this into the sum gives $m \angle B + m \angle B = 180^\circ$,so $2 m \angle B = 180^\circ$,which means $m \angle B = 90^\circ$.
Since $\Delta ABC$ is a right-angled triangle at $B$,by the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Let $AC = 17x$ and $AB = 15x$. Given $BC = 8$,we have $(17x)^2 = (15x)^2 + 8^2$.
$289x^2 = 225x^2 + 64$.
$64x^2 = 64$,so $x^2 = 1$,which gives $x = 1$.
Thus,$AB = 15(1) = 15$ and $AC = 17(1) = 17$.
The area of the right-angled triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AB$.
Area $= \frac{1}{2} \times 8 \times 15 = 4 \times 15 = 60$.

(B) The perimeter of an equilateral triangle is given by the formula: $P = 3 \times \text{side}$.
Given $P = 12$,we have $12 = 3 \times \text{side}$.
Therefore,the side length is $\text{side} = \frac{12}{3} = 4$.
The area of an equilateral triangle is given by the formula: $\text{Area} = \frac{\sqrt{3}}{4} \times \text{side}^2$.
Substituting the side length: $\text{Area} = \frac{\sqrt{3}}{4} \times 4^2 = \frac{\sqrt{3}}{4} \times 16 = 4\sqrt{3}$.