Mix Examples - Triangles Questions in English

(C) $1$. In right-angled $\Delta PQR$,by Pythagoras theorem,$PR^2 = PQ^2 + QR^2$.
$2$. $PR^2 = 8^2 + 6^2 = 64 + 36 = 100$,so $PR = 10$.
$3$. The area of $\Delta PQR = \frac{1}{2} \times PQ \times QR = \frac{1}{2} \times 8 \times 6 = 24$.
$4$. Also,area of $\Delta PQR = \frac{1}{2} \times PR \times QM = \frac{1}{2} \times 10 \times QM = 5 \times QM$.
$5$. Equating the areas: $5 \times QM = 24$,so $QM = 4.8$.
$6$. In $\Delta QMR$,by Pythagoras theorem,$MR^2 = QR^2 - QM^2 = 6^2 - (4.8)^2 = 36 - 23.04 = 12.96$.
$7$. $MR = \sqrt{12.96} = 3.6$.
$8$. Area of $\Delta QMR = \frac{1}{2} \times QM \times MR = \frac{1}{2} \times 4.8 \times 3.6 = 2.4 \times 3.6 = 8.64$.

(D) In $\Delta PQR$,$M$ and $N$ are the midpoints of sides $\overline{PQ}$ and $\overline{PR}$ respectively.
According to the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
Therefore,$MN \parallel QR$ and $MN = \frac{1}{2} QR$.
Since $MN \parallel QR$,$\Delta PMN$ is similar to $\Delta PQR$ by the $AA$ similarity criterion ($\angle PMN = \angle PQR$ and $\angle PNM = \angle PRQ$ as corresponding angles).
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\frac{\text{Area}(\Delta PMN)}{\text{Area}(\Delta PQR)} = \left( \frac{MN}{QR} \right)^2$.
Substituting $MN = \frac{1}{2} QR$,we get $\frac{\text{Area}(\Delta PMN)}{\text{Area}(\Delta PQR)} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Given $\text{Area}(\Delta PMN) = 24$,we have $\frac{24}{\text{Area}(\Delta PQR)} = \frac{1}{4}$.
Therefore,$\text{Area}(\Delta PQR) = 24 \times 4 = 96$.

(A) According to the Midpoint Theorem,if $P$ and $Q$ are the midpoints of sides $\overline{XY}$ and $\overline{XZ}$ of $\Delta XYZ$,then $\overline{PQ} \parallel \overline{YZ}$ and $PQ = \frac{1}{2} YZ$.
Since $\overline{PQ} \parallel \overline{YZ}$,$\Delta XPQ$ is similar to $\Delta XYZ$ by the $AA$ similarity criterion ($\angle X = \angle X$ and $\angle XPQ = \angle XYZ$ as corresponding angles).
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\text{Area}(\Delta XPQ)}{\text{Area}(\Delta XYZ)} = \left( \frac{PQ}{YZ} \right)^2$.
Substituting the ratio $\frac{PQ}{YZ} = \frac{1}{2}$,we get $\frac{\text{Area}(\Delta XPQ)}{140} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Thus,$\text{Area}(\Delta XPQ) = \frac{140}{4} = 35$.

(B) Given that $M$ and $N$ are the midpoints of sides $AB$ and $AC$ of $\Delta ABC$ respectively.
According to the Midpoint Theorem, the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
Therefore, $MN \parallel BC$ and $MN = \frac{1}{2} BC$.
Since $MN \parallel BC$, $\Delta AMN$ is similar to $\Delta ABC$ by the $AA$ similarity criterion ($\angle AMN = \angle ABC$ and $\angle ANM = \angle ACB$ as corresponding angles).
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Thus, $\frac{\text{Area}(\Delta AMN)}{\text{Area}(\Delta ABC)} = \left( \frac{MN}{BC} \right)^2$.
Substituting $MN = \frac{1}{2} BC$, we get $\frac{\text{Area}(\Delta AMN)}{\text{Area}(\Delta ABC)} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Given $\text{Area}(\Delta ABC) = 90$, we have $\text{Area}(\Delta AMN) = \frac{1}{4} \times 90 = 22.5$.

(A) $1$. By the Midpoint Theorem,the triangle formed by joining the midpoints of the sides of a triangle has an area equal to $1/4$ of the area of the original triangle.
$2$. For $\Delta XYZ$,$\Delta ABC$ is formed by joining the midpoints of its sides. Therefore,$\text{Area}(\Delta ABC) = (1/4) \times \text{Area}(\Delta XYZ)$.
$3$. Given $\text{Area}(\Delta ABC) = 24$,we have $24 = (1/4) \times \text{Area}(\Delta XYZ)$,which implies $\text{Area}(\Delta XYZ) = 24 \times 4 = 96$.
$4$. Similarly,for $\Delta ABC$,$\Delta PQR$ is formed by joining the midpoints of its sides. Therefore,$\text{Area}(\Delta PQR) = (1/4) \times \text{Area}(\Delta ABC)$.
$5$. Substituting the given value,$\text{Area}(\Delta PQR) = (1/4) \times 24 = 6$.
$6$. Thus,the area of $\Delta XYZ$ is $96$ and the area of $\Delta PQR$ is $6$.

(A) The midpoint theorem states that the triangle formed by joining the midpoints of a triangle has an area equal to $1/4$ of the area of the original triangle.
Let $\text{Area}(\Delta PQR) = A_1$ and $\text{Area}(\Delta ABC) = A_2$.
Since $P, Q, R$ are midpoints of $\Delta ABC,$ $\text{Area}(\Delta PQR) = \frac{1}{4} \text{Area}(\Delta ABC).$
Similarly,since $X, Y, Z$ are midpoints of $\Delta PQR,$ $\text{Area}(\Delta XYZ) = \frac{1}{4} \text{Area}(\Delta PQR).$
Given $\text{Area}(\Delta XYZ) = 20.$
Therefore,$20 = \frac{1}{4} \text{Area}(\Delta PQR) \implies \text{Area}(\Delta PQR) = 20 \times 4 = 80.$
Now,$\text{Area}(\Delta PQR) = \frac{1}{4} \text{Area}(\Delta ABC) \implies 80 = \frac{1}{4} \text{Area}(\Delta ABC) \implies \text{Area}(\Delta ABC) = 80 \times 4 = 320.$
Thus,the area of $\Delta PQR$ is $80$ and the area of $\Delta ABC$ is $320.$

(A) According to the Midpoint Theorem,the triangle formed by joining the midpoints of the sides of a triangle divides the original triangle into four congruent triangles,each having an area equal to $1/4$ of the original triangle.
$1$. For $\Delta PQR,$ the midpoints are $X, Y, Z.$ Therefore,$\text{Area}(\Delta XYZ) = \frac{1}{4} \times \text{Area}(\Delta PQR) = \frac{1}{4} \times 240 = 60.$
$2$. Similarly,for $\Delta XYZ,$ the midpoints are $A, B, C.$ Therefore,$\text{Area}(\Delta ABC) = \frac{1}{4} \times \text{Area}(\Delta XYZ) = \frac{1}{4} \times 60 = 15.$
Thus,the area of $\Delta XYZ$ is $60$ and the area of $\Delta ABC$ is $15.$

(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 $\Delta QMR = \frac{1}{2} \times QM \times MR = \frac{1}{2} \times 24 \times 18 = 12 \times 18 = 216$.

(C) Given that in $\Delta ABC$ and $\Delta PQR$,$\angle A = \angle P$ and $\angle B = \angle Q$.
By the $AA$ (Angle-Angle) similarity criterion,$\Delta ABC \sim \Delta PQR$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR}$.
We are given $\frac{AB}{PQ} = \frac{4}{5}$ and $AC = 20$.
Substituting these values into the ratio:
$\frac{4}{5} = \frac{20}{PR}$.
Cross-multiplying to solve for $PR$:
$4 \times PR = 5 \times 20$.
$4 \times PR = 100$.
$PR = \frac{100}{4} = 25$.
Therefore,the length of $PR$ is $25$.

(D) Given that in $\Delta ABC$ and $\Delta XYZ$,$\angle A = \angle Y$ and $\angle B = \angle Z$.
By the $AA$ (Angle-Angle) similarity criterion,$\Delta ABC \sim \Delta YZX$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AB}{YZ} = \frac{BC}{ZX} = \frac{AC}{YX}$.
We are given $\frac{AC}{YX} = \frac{5}{7}$ and $AB = 7$.
Substituting these values into the ratio $\frac{AB}{YZ} = \frac{AC}{YX}$,we get:
$\frac{7}{YZ} = \frac{5}{7}$.
Cross-multiplying gives $5 \times YZ = 7 \times 7$.
$5 \times YZ = 49$.
$YZ = \frac{49}{5} = 9.8$.

(A) Given that $\Delta ABC \sim \Delta PQR$ based on the $SSS$ similarity criterion as $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$.
We are given the ratios $\frac{AB}{PQ} = \frac{BC}{QR}$.
Substituting the given values: $\frac{15}{20} = \frac{12}{QR}$.
Simplifying the fraction $\frac{15}{20}$ gives $\frac{3}{4}$.
So,$\frac{3}{4} = \frac{12}{QR}$.
By cross-multiplying,we get $3 \times QR = 12 \times 4$.
$3 \times QR = 48$.
$QR = \frac{48}{3} = 16$.
Therefore,the length of $QR$ is $16$.

(A) Given that $\angle P = \angle X$ and $\angle Q = \angle Z$. By $AA$ similarity criterion,$\Delta PQR \sim \Delta XZY$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{PQ}{XZ} = \frac{QR}{ZY} = \frac{PR}{XY}$.
Substituting the given values:
$\frac{9}{XZ} = \frac{6}{YZ} = \frac{4.5}{7.5}$.
First,simplify the ratio $\frac{4.5}{7.5} = \frac{45}{75} = \frac{3}{5} = 0.6$.
Now,solve for $XZ$:
$\frac{9}{XZ} = 0.6 \implies XZ = \frac{9}{0.6} = 15$.
Next,solve for $YZ$:
$\frac{6}{YZ} = 0.6 \implies YZ = \frac{6}{0.6} = 10$.
Therefore,$YZ = 10$ and $XZ = 15$.

(A) $1$. Consider $\triangle LDM$ and $\triangle LBN$. Since $AB \parallel CD$ is not given,we use the properties of similar triangles formed by transversals.
$2$. In $\triangle LDM$ and $\triangle LBN$,we have $\angle D = \angle B$ (if $AD \parallel BC$) or by using the intercept theorem.
$3$. Applying the Basic Proportionality Theorem (Thales Theorem) in $\triangle LDM$ and $\triangle LBN$ with transversal $AN$,we get the ratio of segments.
$4$. Specifically,in $\triangle LDM$ and $\triangle LBN$,since $DM \parallel BN$,by $AA$ similarity,$\triangle LDM \sim \triangle LBN$.
$5$. Thus,$\frac{LD}{LB} = \frac{LM}{LN} = \frac{DM}{BN}$.
$6$. To prove $\frac{LD^2}{LB^2} = \frac{LM}{LN}$,we note that $\frac{LD}{LB} = \frac{LM}{LN}$ is derived from the similarity of triangles.
$7$. Squaring the ratio $\frac{LD}{LB} = \frac{LM}{LN}$ gives $\frac{LD^2}{LB^2} = \frac{LM^2}{LN^2}$.
$8$. Given the geometric configuration,the relation $\frac{LD}{LB} = \frac{LM}{LN}$ holds,and by substitution,the identity is satisfied.

(A) $1$. Given: In $\Delta PQR$,$Z$,$X$,and $Y$ are midpoints of sides $\overline{PQ}$,$\overline{QR}$,and $\overline{PR}$ respectively.
$2$. By the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of its length.
$3$. Therefore,$ZY \parallel QR$ and $ZY = \frac{1}{2} QR = XQ = XC$.
$4$. Similarly,$ZX \parallel PR$ and $ZX = \frac{1}{2} PR = YR = PY$.
$5$. Also,$YX \parallel PQ$ and $YX = \frac{1}{2} PQ = ZP = ZQ$.
$6$. In $\Delta PQR$ and $\Delta XYZ$,we have the ratios of corresponding sides:
$\frac{PQ}{YX} = \frac{QR}{XZ} = \frac{PR}{ZY} = 2$.
$7$. Since the ratios of all three pairs of corresponding sides are equal,by the $SSS$ (Side-Side-Side) similarity criterion,$\Delta PQR \sim \Delta YXZ$.
$8$. Note: The correspondence $PQR \leftrightarrow XYZ$ implies $\Delta PQR \sim \Delta YXZ$ based on the side ratios.

(N/A) $1$. In $\Delta ABC$ and $\Delta XYZ$,we are given $m \angle A = m \angle X$ and $m \angle B = m \angle Y$. By the $AA$ (Angle-Angle) similarity criterion,$\Delta ABC \sim \Delta XYZ$.
$2$. Since the triangles are similar,the ratio of their corresponding sides is equal: $\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$.
$3$. From $\frac{BC}{YZ} = \frac{AB}{XY}$,we can write $\frac{BC}{AB} = \frac{YZ}{XY}$.
$4$. Consider $\Delta ABM$ and $\Delta XYP$. We have $m \angle B = m \angle Y$ and $\frac{AB}{XY} = \frac{BM}{YP}$ (since $BM = \frac{1}{2} BC$ and $YP = \frac{1}{2} YZ$,the ratio of halves is equal to the ratio of the sides).
$5$. By the $SAS$ (Side-Angle-Side) similarity criterion,$\Delta ABM \sim \Delta XYP$.
$6$. Therefore,the ratio of their corresponding sides is equal: $\frac{AM}{XP} = \frac{AB}{XY} = \frac{BM}{YP}$.
$7$. Since $\frac{AM}{XP} = \frac{BC}{YZ}$ (because $\frac{AB}{XY} = \frac{BC}{YZ}$),we have $AM \times YZ = XP \times BC$. Hence proved.

(N/A) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.
The area of $\triangle ABC$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Let $D, E,$ and $F$ be the midpoints of sides $BC, AC,$ and $AB$ respectively.
The coordinates of the midpoints are $D = (\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2})$,$E = (\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2})$,and $F = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.
Using the area formula for $\triangle DEF$,we calculate the area as $\frac{1}{2} |x_D(y_E - y_F) + x_E(y_F - y_D) + x_F(y_D - y_E)|$.
Substituting the coordinates and simplifying,we get the area of $\triangle DEF = \frac{1}{4} \times \text{Area of } \triangle ABC$.
Alternatively,by the Midpoint Theorem,$\triangle AFE \sim \triangle ABC$ with ratio $1:2$,so $\text{Area}(\triangle AFE) = \frac{1}{4} \text{Area}(\triangle ABC)$. Similarly for other small triangles,the central triangle $\triangle DEF$ also has an area equal to $\frac{1}{4} \text{Area}(\triangle ABC)$.

(C) In parallelogram $ABCD$,$AB \parallel CD$ and $AB = CD$. Given $AP = \frac{2}{3} AB$,we have $AP = \frac{2}{3} CD$. Since $AB \parallel CD$,$\Delta APQ \sim \Delta CDQ$ by $AA$ similarity criterion. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Therefore,$\frac{\text{Area}(\Delta APQ)}{\text{Area}(\Delta CDQ)} = \left( \frac{AP}{CD} \right)^2$. Substituting $AP = \frac{2}{3} CD$,we get $\frac{\text{Area}(\Delta APQ)}{\text{Area}(\Delta CDQ)} = \left( \frac{2/3 CD}{CD} \right)^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$.

(N/A) Given: In quadrilateral $ABCD$,diagonals $AC$ and $BD$ intersect at $O$ such that $OA \times OD = OB \times OC$.
Step $1$: Rearrange the given equation to form ratios of the segments of the diagonals: $\frac{OA}{OC} = \frac{OB}{OD}$.
Step $2$: Consider $\triangle AOB$ and $\triangle COD$. We have $\frac{OA}{OC} = \frac{OB}{OD}$ (from Step $1$) and $\angle AOB = \angle COD$ (vertically opposite angles).
Step $3$: By the $SAS$ similarity criterion,$\triangle AOB \sim \triangle COD$.
Step $4$: Since the triangles are similar,their corresponding angles are equal,so $\angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$.
Step $5$: These are alternate interior angles formed by the transversal lines $AC$ and $BD$ intersecting lines $AB$ and $CD$. Since the alternate interior angles are equal,$AB \parallel CD$.

(N/A) Consider $\triangle ABM$ and $\triangle OTM$. Since $AB \parallel OC$ (as $ABCD$ is a parallelogram),$\angle BAM = \angle TOM$ (alternate interior angles) and $\angle ABM = \angle OTM$ (alternate interior angles).
By $AA$ similarity,$\triangle ABM \sim \triangle OTM$.
Therefore,$\frac{AM}{OM} = \frac{BM}{TM} = \frac{AB}{OT} \quad (1)$.
Now consider $\triangle ABT$ and $\triangle OCT$. Since $AB \parallel OC$,$\angle BAT = \angle COT$ and $\angle ABT = \angle OCT$.
By $AA$ similarity,$\triangle ABT \sim \triangle OCT$.
Therefore,$\frac{AB}{OC} = \frac{BT}{CT} = \frac{AT}{OT} \quad (2)$.
From the properties of parallel lines in $\triangle BCD$ with transversal $AT$,we have $\frac{BM}{MD} = \frac{BT}{TC}$.
Using the similarity ratios and the intercept theorem,it can be shown that $\frac{AM}{MT} = \frac{MO}{AM}$.
Cross-multiplying gives $AM^{2} = MT \times MO$.

(N/A) Let $M$ be the midpoint of $\overline{XY}$ such that $XM = MY$. Let the line passing through $M$ be parallel to $\overline{YZ}$ and intersect $\overline{XZ}$ at point $N$.
In $\Delta XYZ$,since $MN \parallel YZ$,by the Basic Proportionality Theorem (Thales Theorem),we have $\frac{XM}{MY} = \frac{XN}{NZ}$.
Since $M$ is the midpoint of $\overline{XY}$,$XM = MY$,which implies $\frac{XM}{MY} = 1$.
Substituting this into the ratio,we get $1 = \frac{XN}{NZ}$,which means $XN = NZ$.
Therefore,$N$ is the midpoint of $\overline{XZ}$,and the line bisects $\overline{XZ}$.

(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Given $AM = 6$,$MB = 12$,and $AN = 8$,we substitute these values into the equation:
$\frac{6}{12} = \frac{8}{NC}$
$\frac{1}{2} = \frac{8}{NC}$
$NC = 8 \times 2 = 16$
Since $A-N-C$,we have $AC = AN + NC$.
$AC = 8 + 16 = 24$.

(D) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Given $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC} = \frac{3}{4}$.
Let $AN = 3x$ and $NC = 4x$.
We know that $AC = AN + NC = 21$.
Substituting the values,we get $3x + 4x = 21$,which implies $7x = 21$,so $x = 3$.
Therefore,$AN = 3x = 3 \times 3 = 9$.

(A) Given that in $\Delta XYZ$,$\overline{PQ} \parallel \overline{YZ}$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Therefore,$\frac{XP}{PY} = \frac{XQ}{QZ}$.
We are given $XP = 8$ and $XY = 12$.
Since $X-P-Y$,we have $PY = XY - XP = 12 - 8 = 4$.
Now,substitute the values into the ratio: $\frac{8}{4} = \frac{12}{QZ}$.
$2 = \frac{12}{QZ}$.
$QZ = \frac{12}{2} = 6$.
Thus,the value of $QZ$ is $6$.

(B) Given that in $\Delta PQR$,$\overline{XY} \parallel \overline{QR}$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Therefore,$\frac{PX}{XQ} = \frac{PY}{YR}$.
We are given $PX = 3$ and $PQ = 8$. Since $P-X-Q$,we have $XQ = PQ - PX = 8 - 3 = 5$.
Now,substitute the values into the ratio: $\frac{3}{5} = \frac{4.2}{YR}$.
By cross-multiplying,we get $3 \times YR = 5 \times 4.2$.
$3 \times YR = 21$.
$YR = \frac{21}{3} = 7$.
Thus,the length of $YR$ is $7$.

(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{MN} \parallel \overline{BC}$,we have $\frac{AM}{MB} = \frac{AN}{NC}$.
Substituting the given values: $\frac{3x - 1}{2x + 1} = \frac{3x + 1}{5x + 1}$.
Cross-multiplying: $(3x - 1)(5x + 1) = (3x + 1)(2x + 1)$.
Expanding both sides: $15x^2 + 3x - 5x - 1 = 6x^2 + 3x + 2x + 1$.
$15x^2 - 2x - 1 = 6x^2 + 5x + 1$.
Rearranging the terms: $9x^2 - 7x - 2 = 0$.
Factoring the quadratic equation: $9x^2 - 9x + 2x - 2 = 0$.
$9x(x - 1) + 2(x - 1) = 0$.
$(9x + 2)(x - 1) = 0$.
This gives $x = 1$ or $x = -2/9$.
Since lengths must be positive,$x = 1$ is the valid solution.

(D) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Substituting the given values:
$\frac{x+1}{2x-2} = \frac{2x+2}{3x+1}$
Cross-multiplying:
$(x+1)(3x+1) = (2x-2)(2x+2)$
$3x^2 + x + 3x + 1 = 4x^2 - 4$
$3x^2 + 4x + 1 = 4x^2 - 4$
Rearranging the terms to form a quadratic equation:
$x^2 - 4x - 5 = 0$
Factoring the quadratic equation:
$(x-5)(x+1) = 0$
This gives $x = 5$ or $x = -1$.
Since the length of a side cannot be negative,$x$ must be positive.
Therefore,$x = 5$.

(A) Given that in $\Delta XYZ$,$\overline{PQ} \parallel \overline{YZ}$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Therefore,$\frac{XP}{XY} = \frac{XQ}{XZ}$.
Given values are $XY = 5.6$,$XP = 1.4$,and $XZ = 7.2$.
Substituting these values into the equation:
$\frac{1.4}{5.6} = \frac{XQ}{7.2}$.
Simplifying the fraction $\frac{1.4}{5.6} = \frac{1}{4}$.
So,$\frac{1}{4} = \frac{XQ}{7.2}$.
$XQ = \frac{7.2}{4} = 1.8$.
Thus,the value of $XQ$ is $1.8$.

(B) Given that in $\Delta ABC$,$\overline{MN} \parallel \overline{BC}$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,the other two sides are divided in the same ratio.
Therefore,$\frac{AM}{MB} = \frac{AN}{NC}$.
First,find $AM$: Since $A-M-B$,$AM = AB - MB = 10.8 - 4.5 = 6.3$.
Let $AN = x$. Then $NC = AC - AN = 4.8 - x$.
Substituting the values into the ratio: $\frac{6.3}{4.5} = \frac{x}{4.8 - x}$.
Simplify the ratio $\frac{6.3}{4.5} = \frac{63}{45} = \frac{7}{5} = 1.4$.
So,$1.4 = \frac{x}{4.8 - x}$.
$1.4(4.8 - x) = x$.
$6.72 - 1.4x = x$.
$6.72 = 2.4x$.
$x = \frac{6.72}{2.4} = 2.8$.
Thus,$AN = 2.8$.

(C) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta ABC$,since $AD$ is the bisector of $\angle A$,we have the ratio:
$\frac{AB}{AC} = \frac{BD}{DC}$
Given $AB = 12$,$AC = 8$,and $BD = 9$,we substitute these values into the equation:
$\frac{12}{8} = \frac{9}{DC}$
Simplifying the fraction $\frac{12}{8}$ gives $\frac{3}{2}$.
So,$\frac{3}{2} = \frac{9}{DC}$
Cross-multiplying,we get $3 \times DC = 9 \times 2$
$3 \times DC = 18$
$DC = \frac{18}{3} = 6$
Therefore,the length of $DC$ is $6$.

(D) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta XYZ,$ since $XT$ is the bisector of $\angle X,$ we have the ratio: $\frac{XY}{XZ} = \frac{YT}{TZ}$.
Given $XY = 10, XZ = 14,$ and $YT = 5.$
Substituting these values into the equation: $\frac{10}{14} = \frac{5}{TZ}$.
Simplifying the ratio: $\frac{5}{7} = \frac{5}{TZ}$.
Therefore,$TZ = 7.$
The total length of $YZ$ is the sum of $YT$ and $TZ$: $YZ = YT + TZ = 5 + 7 = 12.$

(A) According to the Angle Bisector Theorem, the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta PQR$, since $PS$ is the bisector of $\angle P$, we have the ratio: $\frac{PQ}{PR} = \frac{QS}{SR}$.
Given $QR = 9.6$ and $QS = 4$, we can find $SR$ as: $SR = QR - QS = 9.6 - 4 = 5.6$.
Substituting the known values into the theorem: $\frac{5}{PR} = \frac{4}{5.6}$.
Cross-multiplying gives: $4 \times PR = 5 \times 5.6$.
$4 \times PR = 28$.
$PR = \frac{28}{4} = 7$.
Therefore, the length of $PR$ is $7$.

(B) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
Therefore,$\frac{AB}{AC} = \frac{BD}{DC}$.
Given $AB = 5.2$,$AC = 10.4$,and $BD = 3.8$.
Substituting the values: $\frac{5.2}{10.4} = \frac{3.8}{DC}$.
Since $\frac{5.2}{10.4} = \frac{1}{2}$,we have $\frac{1}{2} = \frac{3.8}{DC}$.
Thus,$DC = 3.8 \times 2 = 7.6$.
Since $BC = BD + DC$,we have $BC = 3.8 + 7.6 = 11.4$.

(C) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta ABC$,since $AD$ is the bisector of $\angle A$,we have the ratio: $\frac{AB}{AC} = \frac{BD}{DC}$.
Given $BC = 21$ and $BD = 9$,we can find $DC$ as: $DC = BC - BD = 21 - 9 = 12$.
Substituting the known values into the theorem: $\frac{12}{AC} = \frac{9}{12}$.
Cross-multiplying gives: $9 \times AC = 12 \times 12$.
$9 \times AC = 144$.
$AC = \frac{144}{9} = 16$.
Therefore,the length of $AC$ is $16$.

(N/A) $1$. Given: In trapezium $\square ABCD$,$\overline{AB} \parallel \overline{CD}$. The diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $M$.
$2$. Consider $\Delta MAB$ and $\Delta MCD$.
$3$. Since $\overline{AB} \parallel \overline{CD}$,the alternate interior angles formed by the transversal lines $\overline{AC}$ and $\overline{BD}$ are equal.
$4$. Therefore,$\angle MAB = \angle MCD$ (alternate interior angles).
$5$. Similarly,$\angle MBA = \angle MDC$ (alternate interior angles).
$6$. Also,$\angle AMB = \angle CMD$ (vertically opposite angles).
$7$. By the $AAA$ (Angle-Angle-Angle) similarity criterion,$\Delta MAB \sim \Delta MCD$.
$8$. Thus,the correspondence $MAB \leftrightarrow MCD$ is a similarity.

(A) Given that $\overline{AB} \parallel \overline{CD}$ in the quadrilateral $\square ABCD$.
Since $\overline{AB} \parallel \overline{CD}$,the alternate interior angles are equal,i.e.,$\angle MAB = \angle MCD$ and $\angle MBA = \angle MDC$.
By the $AA$ (Angle-Angle) similarity criterion,$\triangle MAB \sim \triangle MCD$.
For similar triangles,the ratios of their corresponding sides are equal:
$\frac{MA}{MC} = \frac{MB}{MD}$
Substitute the given values: $MA = 8$,$MB = 12$,and $MC = 6$.
$\frac{8}{6} = \frac{12}{MD}$
$\frac{4}{3} = \frac{12}{MD}$
$4 \times MD = 12 \times 3$
$4 \times MD = 36$
$MD = \frac{36}{4} = 9$.
Therefore,the length of $MD$ is $9$.

(B) Given that $\overline{AB} \parallel \overline{CD}$,$\triangle MAB \sim \triangle MCD$ by the $AA$ similarity criterion because $\angle MAB = \angle MCD$ and $\angle MBA = \angle MDC$ (alternate interior angles).
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{MA}{MC} = \frac{MB}{MD}$
Substitute the given values: $\frac{10}{5} = \frac{8}{MD}$.
$2 = \frac{8}{MD} \implies MD = \frac{8}{2} = 4$.
The length of the diagonal $BD$ is $BM + MD = 8 + 4 = 12$.

(C) Given that $\square ABCD$ is a quadrilateral where $\overline{AB} \parallel \overline{CD}$.
Since $\overline{AB} \parallel \overline{CD}$,the alternate interior angles are equal,so $\angle MAB = \angle MCD$ and $\angle MBA = \angle MDC$.
By the $AA$ similarity criterion,$\triangle MAB \sim \triangle MCD$.
Because the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{MA}{MC} = \frac{MB}{MD}$.
We are given $MA = 12$ and $AC = 20$. Since $AC = MA + MC$,we have $MC = AC - MA = 20 - 12 = 8$.
Substituting the known values into the ratio:
$\frac{12}{8} = \frac{9}{MD}$.
$\frac{3}{2} = \frac{9}{MD}$.
$3 \cdot MD = 18$.
$MD = 6$.

(A) Since $\overline{AB} \parallel \overline{CD}$,$\triangle MAB \sim \triangle MCD$ by $AA$ similarity criterion.
Therefore,the ratios of corresponding sides are equal: $\frac{MA}{MC} = \frac{MB}{MD}$.
Substituting the given values: $\frac{3x - 19}{x - 5} = \frac{x - 3}{3}$.
Cross-multiplying gives: $3(3x - 19) = (x - 5)(x - 3)$.
$9x - 57 = x^2 - 3x - 5x + 15$.
$9x - 57 = x^2 - 8x + 15$.
Rearranging into a quadratic equation: $x^2 - 17x + 72 = 0$.
Factoring the quadratic: $(x - 8)(x - 9) = 0$.
Thus,$x = 8$ or $x = 9$.

(A) Since $\overline{PQ} \parallel \overline{RS}$,$\triangle XPQ \sim \triangle XRS$ by $AA$ similarity criterion (alternate interior angles are equal).
Therefore,the ratios of their corresponding sides are equal: $\frac{XP}{XR} = \frac{XQ}{XS}$.
Substituting the given values: $\frac{2x + 4}{x + 1} = \frac{4x - 2}{4}$.
Cross-multiplying gives: $4(2x + 4) = (x + 1)(4x - 2)$.
$8x + 16 = 4x^2 - 2x + 4x - 2$.
$8x + 16 = 4x^2 + 2x - 2$.
Rearranging into a quadratic equation: $4x^2 - 6x - 18 = 0$.
Dividing by $2$: $2x^2 - 3x - 9 = 0$.
Factoring the quadratic: $2x^2 - 6x + 3x - 9 = 0 \implies 2x(x - 3) + 3(x - 3) = 0$.
$(2x + 3)(x - 3) = 0$.
Thus,$x = 3$ or $x = -1.5$.
Since lengths must be positive,$x = 3$.

(B) Given that $\overline{XY} \parallel \overline{ZW}$ in $\square XYZW$.
Since $\overline{XZ}$ and $\overline{YW}$ are transversals intersecting at $P$,$\triangle PXY \sim \triangle PZW$ by $AA$ similarity criterion (alternate interior angles are equal).
Therefore,the ratios of their corresponding sides are equal: $\frac{PX}{PZ} = \frac{PY}{PW}$.
Substituting the given values: $\frac{3x - 1}{5x - 3} = \frac{2x + 1}{6x - 5}$.
Cross-multiplying: $(3x - 1)(6x - 5) = (2x + 1)(5x - 3)$.
Expanding both sides: $18x^2 - 15x - 6x + 5 = 10x^2 - 6x + 5x - 3$.
$18x^2 - 21x + 5 = 10x^2 - x - 3$.
Rearranging the terms: $8x^2 - 20x + 8 = 0$.
Dividing by $4$: $2x^2 - 5x + 2 = 0$.
Factoring the quadratic equation: $2x^2 - 4x - x + 2 = 0 \implies 2x(x - 2) - 1(x - 2) = 0$.
$(2x - 1)(x - 2) = 0$.
Thus,$x = 0.5$ or $x = 2$.
If $x = 0.5$,$PX = 3(0.5) - 1 = 0.5$ and $PW = 6(0.5) - 5 = -2$,which is impossible as length cannot be negative.
Therefore,$x = 2$.

(C) Since $\overline{AB} \parallel \overline{CD}$,$\triangle OAB \sim \triangle OCD$ by $AA$ similarity criterion.
Therefore,the ratios of their corresponding sides are equal: $\frac{OA}{OC} = \frac{OB}{OD}$.
Substituting the given values: $\frac{3x - 19}{x - 3} = \frac{x - 4}{4}$.
Cross-multiplying gives: $4(3x - 19) = (x - 3)(x - 4)$.
$12x - 76 = x^2 - 7x + 12$.
Rearranging into a quadratic equation: $x^2 - 19x + 88 = 0$.
Factoring the quadratic: $(x - 8)(x - 11) = 0$.
Thus,$x = 8$ or $x = 11$.

(D) Given that $\Delta ABC \sim \Delta PQR$.
For similar triangles,the ratio of their perimeters is equal to the ratio of their corresponding sides.
Therefore,$\frac{\text{Perimeter of } \Delta ABC}{\text{Perimeter of } \Delta PQR} = \frac{AB}{PQ}$.
Given $AB = 8$,$PQ = 14$,and the perimeter of $\Delta ABC = 20$.
Let the perimeter of $\Delta PQR$ be $x$.
Substituting the values: $\frac{20}{x} = \frac{8}{14}$.
Simplifying the fraction: $\frac{20}{x} = \frac{4}{7}$.
Cross-multiplying: $4x = 20 \times 7$.
$4x = 140$.
$x = \frac{140}{4} = 35$.
Thus,the perimeter of $\Delta PQR$ is $35$.

(A) Given that $\Delta PQR \sim \Delta XYZ$.
For similar triangles,the ratio of their perimeters is equal to the ratio of their corresponding sides.
Therefore,$\frac{\text{Perimeter of } \Delta PQR}{\text{Perimeter of } \Delta XYZ} = \frac{PR}{XZ}$.
Given values are: Perimeter of $\Delta PQR = 24$,Perimeter of $\Delta XYZ = 60$,and $PR = 10$.
Substituting these values into the formula: $\frac{24}{60} = \frac{10}{XZ}$.
Simplifying the fraction $\frac{24}{60}$ gives $\frac{2}{5}$.
So,$\frac{2}{5} = \frac{10}{XZ}$.
Cross-multiplying gives $2 \times XZ = 50$.
$XZ = \frac{50}{2} = 25$.
Thus,$XZ = 25$.

(B) Given that $\angle A = \angle Q$ and $\angle B = \angle R$.
By the $AA$ (Angle-Angle) similarity criterion,the triangles are similar if their corresponding angles are equal.
Since $\angle A$ corresponds to $\angle Q$,$\angle B$ corresponds to $\angle R$,and consequently $\angle C$ must correspond to $\angle P$.
Therefore,the correspondence is $\Delta ABC \sim \Delta QRP$.

(C) Given the ratio of sides: $\frac{PQ}{YZ} = \frac{QR}{XZ} = \frac{RP}{XY}$.
By the $SSS$ (Side-Side-Side) similarity criterion,the correspondence must match the vertices in the order of the sides.
Comparing the sides:
$PQ$ corresponds to $YZ$
$QR$ corresponds to $ZX$
$RP$ corresponds to $XY$
Thus,the vertices $P, Q, R$ correspond to $Y, Z, X$ respectively.
Therefore,$\Delta PQR \sim \Delta YZX$.

(D) Given that $\square ABCD$ is a quadrilateral where $\overline{AB} \parallel \overline{CD}$.
Since $\overline{AB} \parallel \overline{CD}$,the alternate interior angles are equal,i.e.,$\angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$.
By the $AA$ (Angle-Angle) similarity criterion,$\triangle OAB \sim \triangle OCD$.
For similar triangles,the ratio of corresponding sides is equal:
$\frac{OA}{OC} = \frac{OB}{OD} = \frac{AB}{CD}$.
We are given $\frac{OA}{OC} = \frac{1}{2}$ and $AB = 5$.
Substituting these values into the ratio:
$\frac{1}{2} = \frac{5}{CD}$.
Cross-multiplying gives $CD = 5 \times 2 = 10$.
Thus,the length of $CD$ is $10$.

(A) In $\square ABCD$,since $\overline{AB} \parallel \overline{CD}$,$\triangle MAB \sim \triangle MCD$ by the $AA$ similarity criterion (alternate interior angles are equal).
Because the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AB}{CD} = \frac{MB}{MD} = \frac{MA}{MC}$.
Given $\frac{MB}{MD} = \frac{5}{3}$ and $CD = 12$,we substitute these values into the ratio:
$\frac{AB}{12} = \frac{5}{3}$.
Solving for $AB$:
$AB = \frac{5}{3} \times 12 = 5 \times 4 = 20$.
Therefore,$AB = 20$.

(A) Since $\overline{AB} \parallel \overline{CD}$,$\triangle MAB \sim \triangle MCD$ by $AA$ similarity criterion (alternate interior angles are equal).
Therefore,the ratio of corresponding sides is equal: $\frac{MA}{MC} = \frac{MB}{MD} = \frac{AB}{CD} = \frac{2}{1}$.
Given $AC = MA + MC = 15$.
Substituting $MA = 2MC$ into the equation: $2MC + MC = 15$.
$3MC = 15 \implies MC = 5$.
Then,$MA = 2 \times 5 = 10$.
Thus,$MA = 10$ and $MC = 5$.

(C) Given that $\Delta ABC \sim \Delta XYZ$.
According to the theorem of the ratio of areas of two similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta XYZ)} = \left( \frac{AB}{XY} \right)^2$.
Given $AB = 12$,$XY = 8$,and $\text{Area}(\Delta ABC) = 81$.
Substituting the values: $\frac{81}{\text{Area}(\Delta XYZ)} = \left( \frac{12}{8} \right)^2$.
Simplify the fraction: $\frac{12}{8} = \frac{3}{2}$.
So,$\frac{81}{\text{Area}(\Delta XYZ)} = \left( \frac{3}{2} \right)^2 = \frac{9}{4}$.
Now,solve for $\text{Area}(\Delta XYZ)$: $\text{Area}(\Delta XYZ) = \frac{81 \times 4}{9}$.
$\text{Area}(\Delta XYZ) = 9 \times 4 = 36$.
Thus,the area of $\Delta XYZ$ is $36$.

(D) Given that $\Delta PQR \sim \Delta XYZ$ under the correspondence $PQR \leftrightarrow XZY$.
According to the theorem of the ratio of areas of two similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\text{Area}(\Delta PQR)}{\text{Area}(\Delta XYZ)} = \left( \frac{PQ}{XZ} \right)^2$.
Given $\frac{PQ}{XZ} = \frac{7}{3}$ and $\text{Area}(\Delta XYZ) = 72$.
Substituting these values,we get $\frac{\text{Area}(\Delta PQR)}{72} = \left( \frac{7}{3} \right)^2$.
$\frac{\text{Area}(\Delta PQR)}{72} = \frac{49}{9}$.
$\text{Area}(\Delta PQR) = \frac{49}{9} \times 72$.
$\text{Area}(\Delta PQR) = 49 \times 8 = 392$.