Some Properties of a Triangle Questions in English

(N/A) $1$. Extend $QS$ to meet $PR$ at point $T$.
$2$. In $\Delta PQT$,by the triangle inequality theorem,the sum of two sides is greater than the third side: $PQ + PT > QT$. This can be written as $PQ + PT > QS + ST$ (Equation $1$).
$3$. In $\Delta SRT$,by the triangle inequality theorem,$ST + TR > SR$ (Equation $2$).
$4$. Adding Equation $1$ and Equation $2$: $PQ + PT + ST + TR > QS + ST + SR$.
$5$. Simplifying by canceling $ST$ from both sides: $PQ + (PT + TR) > QS + SR$.
$6$. Since $PT + TR = PR$,we get $PQ + PR > QS + SR$,which is equivalent to $SQ + SR < PQ + PR$.

(N/A) Given: In $\Delta ABC$,$\angle A = 90^{\circ}$ and $AB = AC$. $AD$ is the bisector of $\angle A$.
Since $AB = AC$,$\Delta ABC$ is an isosceles triangle. Therefore,$\angle B = \angle C$.
In $\Delta ABC$,$\angle A + \angle B + \angle C = 180^{\circ}$.
$90^{\circ} + \angle B + \angle B = 180^{\circ} \implies 2 \angle B = 90^{\circ} \implies \angle B = 45^{\circ}$. Thus,$\angle C = 45^{\circ}$.
Since $AD$ is the angle bisector of $\angle A$,$\angle BAD = \angle CAD = 45^{\circ}$.
In $\Delta ABD$,$\angle B = 45^{\circ}$ and $\angle BAD = 45^{\circ}$. Since two angles are equal,$AD = BD$.
In $\Delta ACD$,$\angle C = 45^{\circ}$ and $\angle CAD = 45^{\circ}$. Since two angles are equal,$AD = CD$.
Since $D$ lies on $BC$,$BC = BD + CD$.
Substituting $BD = AD$ and $CD = AD$,we get $BC = AD + AD = 2 AD$.
Hence,$BC = 2 AD$ is proved.

(N/A) Let $ABC$ be a triangle with sides $AB$,$BC$,and $AC$. Let $AD$ be the median to the side $BC$,such that $BD = DC = \frac{1}{2} BC$.
Extend $AD$ to a point $E$ such that $AD = DE$. Join $CE$.
In $\triangle ABD$ and $\triangle ECD$:
$1$. $AD = DE$ (by construction)
$2$. $\angle ADB = \angle EDC$ (vertically opposite angles)
$3$. $BD = DC$ (since $AD$ is the median)
Thus,$\triangle ABD \cong \triangle ECD$ by $SAS$ congruence criterion.
Therefore,$AB = EC$ (by $CPCT$).
In $\triangle ACE$,by the triangle inequality theorem,the sum of any two sides is greater than the third side:
$AC + CE > AE$
Since $AE = AD + DE = 2AD$ and $CE = AB$,we substitute these into the inequality:
$AC + AB > 2AD$
Hence,the sum of two sides is greater than twice the median to the third side.

(A) Let the sides of the triangle be $a$,$b$,and $c$,and the angles opposite to them be $A$,$B$,and $C$ respectively.
Assume $a$ is the longest side,so $a > b$ and $a > c$.
By the property of triangles,the angle opposite to the longest side is the largest angle,so $A > B$ and $A > C$.
We know that the sum of angles in a triangle is $A + B + C = 180^{\circ}$.
Since $A > B$ and $A > C$,we have $A + A + A > A + B + C$,which implies $3A > 180^{\circ}$.
Therefore,$A > 60^{\circ}$.
$A$ right angle is $90^{\circ}$,and $\frac{2}{3}$ of a right angle is $\frac{2}{3} \times 90^{\circ} = 60^{\circ}$.
Thus,$A > 60^{\circ}$ proves that the angle opposite to the longest side is greater than $\frac{2}{3}$ of a right angle.

(B) The exterior angle of a triangle is equal to the sum of the two interior opposite angles.
For $\Delta ABC$,the exterior angle $\angle ABD = \angle BAC + \angle ACB = 110^{\circ}$.
The exterior angle $\angle ACE = \angle BAC + \angle ABC = 130^{\circ}$.
Since $\angle ACE > \angle ABD$,we have $(\angle BAC + \angle ABC) > (\angle BAC + \angle ACB)$,which implies $\angle ABC > \angle ACB$.
In a triangle,the side opposite to the larger angle is longer. Therefore,the side opposite to $\angle ABC$ (which is $AC$) must be greater than the side opposite to $\angle ACB$ (which is $AB$).
Thus,$AC > AB$,which means $AB < AC$.
Therefore,the statement $AB > AC$ is False.

(C) In $\Delta ABC$,it is given that $AB = AC$.
According to the Isosceles Triangle Theorem,the angles opposite to equal sides of a triangle are equal.
Therefore,the angle opposite to side $AB$ is $\angle C$ and the angle opposite to side $AC$ is $\angle B$.
Since $AB = AC$,it follows that $\angle C = \angle B$.
Given that $\angle B = 75^{\circ}$,we have $\angle C = 75^{\circ}$.

(D) Given that in $\Delta PQR$,$PQ = PR$.
Since the sides opposite to equal angles are equal,the angles opposite to equal sides are also equal.
Therefore,$\angle Q = \angle R$.
Given $\angle R = 40^{\circ}$,so $\angle Q = 40^{\circ}$.
The sum of all angles in a triangle is $180^{\circ}$.
So,$\angle P + \angle Q + \angle R = 180^{\circ}$.
Substituting the values: $\angle P + 40^{\circ} + 40^{\circ} = 180^{\circ}$.
$\angle P + 80^{\circ} = 180^{\circ}$.
$\angle P = 180^{\circ} - 80^{\circ} = 100^{\circ}$.

(A) Given that in $\Delta XYZ$,$XY = XZ$.
Since two sides are equal,$\Delta XYZ$ is an isosceles triangle.
Therefore,the angles opposite to the equal sides are equal,which means $\angle Y = \angle Z$.
We know that the sum of all interior angles in a triangle is $180^{\circ}$.
So,$\angle X + \angle Y + \angle Z = 180^{\circ}$.
Substituting the given values: $80^{\circ} + \angle Y + \angle Y = 180^{\circ}$.
$80^{\circ} + 2\angle Y = 180^{\circ}$.
$2\angle Y = 180^{\circ} - 80^{\circ} = 100^{\circ}$.
$\angle Y = 100^{\circ} / 2 = 50^{\circ}$.

(B) In any triangle,the side opposite to the larger angle is longer than the side opposite to the smaller angle.
Given that $\angle A > \angle B$ in $\Delta ABC$.
The side opposite to $\angle A$ is $BC$.
The side opposite to $\angle B$ is $AC$.
Since $\angle A > \angle B$,it follows that the side opposite to $\angle A$ must be greater than the side opposite to $\angle B$.
Therefore,$BC > AC$,which is equivalent to $AC < BC$.

(C) According to the Exterior Angle Theorem of a triangle,the measure of an exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.
In $\Delta ABC$,$\angle ACD$ is the exterior angle at vertex $C$.
The interior opposite angles are $\angle A$ and $\angle B$.
Therefore,$\angle ACD = \angle A + \angle B$.
Given $\angle A = 50^{\circ}$ and $\angle B = 65^{\circ}$.
Substituting these values,we get $\angle ACD = 50^{\circ} + 65^{\circ} = 115^{\circ}$.

(D) In $\Delta ABC$,we are given that $AB = AC$.
According to the property of an isosceles triangle,angles opposite to equal sides are equal.
Therefore,$\angle C = \angle B = 70^{\circ}$.
The sum of angles in a triangle is $180^{\circ}$.
So,$\angle A + \angle B + \angle C = 180^{\circ}$.
$\angle A + 70^{\circ} + 70^{\circ} = 180^{\circ}$.
$\angle A + 140^{\circ} = 180^{\circ}$,which gives $\angle A = 40^{\circ}$.
The exterior angle property states that the exterior angle of a triangle is equal to the sum of the two interior opposite angles.
Thus,$\angle ACD = \angle A + \angle B$.
$\angle ACD = 40^{\circ} + 70^{\circ} = 110^{\circ}$.

(A) An equilateral triangle is a triangle in which all three sides are equal and all three interior angles are equal.
Each interior angle of an equilateral triangle is $60^{\circ}$.
The sum of an interior angle and its corresponding exterior angle is $180^{\circ}$ (linear pair).
Therefore,the exterior angle = $180^{\circ} - 60^{\circ} = 120^{\circ}$.
Alternatively,the sum of all exterior angles of any polygon is $360^{\circ}$. Since an equilateral triangle has $3$ equal exterior angles,each exterior angle = $360^{\circ} / 3 = 120^{\circ}$.

(B) Given that in $\Delta ABC$, $\angle B = \angle C$.
According to the property of triangles, sides opposite to equal angles are equal.
Therefore, the side opposite to $\angle B$ is $AC$ and the side opposite to $\angle C$ is $AB$.
Thus, $AC = AB$.
Given $AB = 5 \text{ cm}$, so $AC = 5 \text{ cm}$.
We are also given $BC = 8 \text{ cm}$.
The perimeter of $\Delta ABC = AB + BC + AC$.
Perimeter $= 5 \text{ cm} + 8 \text{ cm} + 5 \text{ cm} = 18 \text{ cm}$.

(C) In $\Delta XYZ$,we are given that $\angle Y = 90^{\circ}$.
Since the sum of all angles in a triangle is $180^{\circ}$,we have $\angle X + \angle Y + \angle Z = 180^{\circ}$.
Substituting the value of $\angle Y$,we get $\angle X + 90^{\circ} + \angle Z = 180^{\circ}$,which implies $\angle X + \angle Z = 90^{\circ}$.
We are also given that $XY = YZ$. In a triangle,angles opposite to equal sides are equal.
Therefore,$\angle Z = \angle X$.
Substituting $\angle Z = \angle X$ into the equation $\angle X + \angle Z = 90^{\circ}$,we get $\angle X + \angle X = 90^{\circ}$.
This simplifies to $2\angle X = 90^{\circ}$,which gives $\angle X = 45^{\circ}$.

(D) In $\Delta ABC$,the interior angles are $\angle ABC$,$\angle ACB$,and $\angle BAC$ (or $\angle A$).
Since $\angle ABD$ is an exterior angle at vertex $B$,we have $\angle ABC + \angle ABD = 180^{\circ}$ (linear pair).
$\angle ABC = 180^{\circ} - 110^{\circ} = 70^{\circ}$.
Since $\angle ACE$ is an exterior angle at vertex $C$,we have $\angle ACB + \angle ACE = 180^{\circ}$ (linear pair).
$\angle ACB = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
In $\Delta ABC$,the sum of interior angles is $180^{\circ}$,so $\angle A + \angle ABC + \angle ACB = 180^{\circ}$.
$\angle A + 70^{\circ} + 30^{\circ} = 180^{\circ}$.
$\angle A + 100^{\circ} = 180^{\circ}$.
$\angle A = 180^{\circ} - 100^{\circ} = 80^{\circ}$.

(B) According to the triangle inequality theorem,the sum of the lengths of any two sides of a triangle is always greater than the length of the third side. Therefore,the correct word is 'greater'.

(B) According to the triangle inequality theorem,the difference between the lengths of any two sides of a triangle is always less than the length of the third side. If the sides of a triangle are $a$,$b$,and $c$,then $|a - b| < c$,$|a - c| < b$,and $|b - c| < a$.

(D) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
$\angle A + 50^{\circ} + 85^{\circ} = 180^{\circ}$.
$\angle A + 135^{\circ} = 180^{\circ}$.
$\angle A = 180^{\circ} - 135^{\circ} = 45^{\circ}$.
We know that in a triangle,the side opposite to the larger angle is longer.
Here,$\angle C = 85^{\circ}$ and $\angle B = 50^{\circ}$.
Since $\angle C > \angle B$,the side opposite to $\angle C$ $(AB)$ is greater than the side opposite to $\angle B$ $(AC)$.
Thus,$AB > AC$.

(A) According to the triangle inequality theorem,the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
In $\Delta ABC$,the sum of sides $AB$ and $BC$ must be greater than $AC$.
Therefore,$AB + BC > AC$.
Substituting the given values: $6 \text{ cm} + 9 \text{ cm} > AC$.
$15 \text{ cm} > AC$,which means $AC < 15 \text{ cm}$.
Thus,the length of side $AC$ must be less than $15 \text{ cm}$.

(B) According to the triangle inequality theorem,the sum of any two sides of a triangle is always greater than the third side.
Also,the difference between any two sides of a triangle is always less than the third side.
Let the sides be $a = 8 \, cm$,$b = 5 \, cm$,and $c = AC$.
According to the property: $|a - b| < c < a + b$.
Substituting the values: $|8 - 5| < AC < 8 + 5$.
$3 < AC < 13$.
Therefore,$AC$ must be greater than $3 \, cm$.

(C) In $\Delta PQR$,we are given that $\angle Q = \angle R$.
According to the theorem,sides opposite to equal angles of a triangle are equal.
Therefore,the side opposite to $\angle Q$ (which is $PR$) must be equal to the side opposite to $\angle R$ (which is $PQ$).
Thus,$PR = PQ$.
Given that $PQ = 6.5 \, cm$,it follows that $PR = 6.5 \, cm$.

(D) Given that in $\Delta ABC$,$AB = AC$.
According to the property of an isosceles triangle,angles opposite to equal sides are equal.
Therefore,the angle opposite to side $AB$ is $\angle C$ and the angle opposite to side $AC$ is $\angle B$.
Since $AB = AC$,it follows that $\angle B = \angle C$.
Given that $\angle C = 75^{\circ}$.
Thus,$\angle B = 75^{\circ}$.

(A) According to the triangle inequality theorem,the sum of any two sides of a triangle must be greater than the third side.
Let the third side be $AC = x \text{ cm}$.
Then,$AB + BC > AC \implies 8 + 10 > x \implies x < 18$.
Also,$AB + AC > BC \implies 8 + x > 10 \implies x > 2$.
And $BC + AC > AB \implies 10 + x > 8$,which is true for all $x > 0$.
So,$2 < x < 18$.
The perimeter $P = AB + BC + AC = 8 + 10 + x = 18 + x$.
Since $x < 18$,the perimeter $P < 18 + 18 = 36 \text{ cm}$.
Therefore,the perimeter is less than $36 \text{ cm}$.

(B) In any triangle,the sum of the lengths of any two sides must be greater than the length of the third side (Triangle Inequality Theorem).
Let the sides of $\Delta XYZ$ be $XY = 7 \, cm$,$YZ = 10 \, cm$,and $XZ = z \, cm$.
According to the triangle inequality theorem:
$XY + YZ > XZ \implies 7 + 10 > z \implies z < 17$
$XY + XZ > YZ \implies 7 + z > 10 \implies z > 3$
$YZ + XZ > XY \implies 10 + z > 7 \implies z > -3$ (which is always true for a length).
Thus,the third side $z$ must satisfy $3 < z < 17$.
The perimeter $P$ of the triangle is given by $P = XY + YZ + XZ = 7 + 10 + z = 17 + z$.
Since $z > 3$,we have $P = 17 + z > 17 + 3 = 20$.
Therefore,the perimeter of $\Delta XYZ$ is greater than $20 \, cm$.

(D) In any triangle,the angle opposite to the longest side is the largest angle.
Given the side lengths satisfy the inequality $AC > AB > BC$.
The side $AC$ is the longest side of $\Delta ABC$.
The angle opposite to side $AC$ is $\angle B$.
Therefore,$\angle B$ is the greatest angle of the triangle.

(A) According to the triangle inequality theorem,the sum of any two sides of a triangle is always greater than the third side,and the difference of any two sides is always less than the third side.
Given sides are $PQ = 7.5 \, cm$ and $QR = 6.2 \, cm$.
Let the third side be $PR$.
$1$. The difference of the two sides must be less than the third side: $PR > |PQ - QR| = |7.5 - 6.2| = 1.3 \, cm$.
$2$. The sum of the two sides must be greater than the third side: $PR < PQ + QR = 7.5 + 6.2 = 13.7 \, cm$.
Therefore,$1.3 < PR < 13.7$.
Comparing this with $a < PR < b$,we get $a = 1.3$ and $b = 13.7$.

(B) In any triangle,the side opposite to the largest angle is the longest,and the side opposite to the smallest angle is the shortest.
Given the angles are $\angle Y > \angle X > \angle Z$.
The side opposite to $\angle Y$ is $XZ$.
The side opposite to $\angle X$ is $YZ$.
The side opposite to $\angle Z$ is $XY$.
Since $\angle Z$ is the smallest angle,the side opposite to it,which is $XY$,must be the smallest side of the triangle.

(D) Let the third side of the triangle be $AC = b$.
According to the triangle inequality theorem,the sum of any two sides of a triangle is greater than the third side,and the difference of any two sides is less than the third side.
$1$. $AB + BC > AC \implies 9 + 12 > b \implies b < 21$.
$2$. $BC - AB < AC \implies 12 - 9 < b \implies b > 3$.
So,$3 < b < 21$.
The perimeter $P = AB + BC + AC = 9 + 12 + b = 21 + b$.
Since $3 < b < 21$,we add $21$ to all parts of the inequality:
$3 + 21 < 21 + b < 21 + 21$
$24 < P < 42$.
Comparing this with $x < P < y$,we get $x = 24$ and $y = 42$.

(N/A) $1$. In $\Delta ABC$,$AD$ is the median. Since a median divides a triangle into two triangles of equal area,$ar(ABD) = ar(ACD) = \frac{1}{2} ar(ABC)$.
$2$. Similarly,in $\Delta GBC$,$GD$ is the median,so $ar(GBD) = ar(GCD)$.
$3$. Subtracting these areas from the larger triangles: $ar(GAB) = ar(ABD) - ar(GBD)$ and $ar(GAC) = ar(ACD) - ar(GCD)$. Since $ar(ABD) = ar(ACD)$ and $ar(GBD) = ar(GCD)$,it follows that $ar(GAB) = ar(GAC)$.
$4$. By symmetry,using medians $BE$ or $CF$,we can show $ar(GAB) = ar(GBC) = ar(GCA)$.
$5$. Since the sum of these three areas is $ar(ABC)$,each must be equal to $\frac{1}{3} ar(ABC)$.

(A) $1$. In right-angled $\Delta ABC$,by Pythagoras theorem: $AB^2 + BC^2 = AC^2$.
$2$. $AB^2 + 8^2 = 17^2 \implies AB^2 + 64 = 289 \implies AB^2 = 225 \implies AB = 15 \, \text{cm}$.
$3$. Area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 15 \times 8 = 60 \, \text{cm}^2$.
$4$. Since $BE$ is a median,it divides the triangle into two equal areas: $\text{Area}(\Delta BCE) = \frac{1}{2} \times \text{Area}(\Delta ABC) = \frac{60}{2} = 30 \, \text{cm}^2$.
$5$. In $\Delta BCE$,$M$ is the midpoint of $BE$,so $CM$ is a median of $\Delta BCE$.
$6$. $A$ median divides a triangle into two equal areas,so $\text{Area}(\Delta BMC) = \frac{1}{2} \times \text{Area}(\Delta BCE) = \frac{30}{2} = 15 \, \text{cm}^2$.

(B) $1$. Given that $AD$ is a median of $\Delta ABC$,it divides the triangle into two triangles of equal area. Thus,$Area(\Delta ABD) = Area(\Delta ADC) = \frac{1}{2} \times Area(\Delta ABC)$.
$2$. The area of $\Delta ABC = \frac{1}{2} \times BC \times AM = \frac{1}{2} \times 16 \times 8 = 64\, cm^2$.
$3$. Therefore,$Area(\Delta ABD) = \frac{1}{2} \times 64 = 32\, cm^2$.
$4$. In $\Delta EBD$,$BA$ is a median because $AB = AE$ (given),which implies $A$ is the midpoint of $BE$. Thus,$DA$ is a median of $\Delta EBD$ relative to base $BE$,but more simply,$\Delta ABD$ and $\Delta EBD$ share the same altitude from $D$ to the line $BE$,and their bases $AB$ and $AE$ are equal.
$5$. Since $AB = AE$,the triangles $\Delta ABD$ and $\Delta AED$ have equal areas. However,looking at $\Delta EBD$,the median $DA$ divides it into two equal areas: $Area(\Delta ABD) = Area(\Delta AED) = 32\, cm^2$.
$6$. The total area of $\Delta EBD = Area(\Delta ABD) + Area(\Delta AED) = 32 + 32 = 64\, cm^2$.

(N/A) $1$. Let the base $YZ$ of $\Delta XYZ$ be divided into $8$ equal parts by the points $A, B, C, D, E, F,$ and $G$.
$2$. Since $YA = AB = BC = CD = DE = EF = FG = GZ$,we can say that each segment is equal to $\frac{1}{8}$ of the total length $YZ$.
$3$. The area of a triangle is given by the formula $ar = \frac{1}{2} \times \text{base} \times \text{height}$.
$4$. All triangles $\Delta XYZ, \Delta XBE,$ etc.,share the same vertex $X$ and lie on the same base line $YZ$,meaning they all have the same altitude $h$.
$5$. The base of $\Delta XBE$ is $BE$. Since $BE = BC + CD + DE$,and each segment is $\frac{1}{8} YZ$,we have $BE = \frac{1}{8} YZ + \frac{1}{8} YZ + \frac{1}{8} YZ = \frac{3}{8} YZ$.
$6$. Therefore,$ar(XBE) = \frac{1}{2} \times BE \times h = \frac{1}{2} \times (\frac{3}{8} YZ) \times h$.
$7$. This simplifies to $ar(XBE) = \frac{3}{8} \times (\frac{1}{2} \times YZ \times h) = \frac{3}{8} ar(XYZ)$.

(C) The area of a right-angled triangle is given by the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
In $\Delta ABC$,since $\angle B = 90^{\circ}$,the sides $AB$ and $BC$ act as the base and height.
Given: $AB = 8\, \text{cm}$ (height) and $BC = 15\, \text{cm}$ (base).
$\text{Area} = \frac{1}{2} \times 15 \times 8$.
$\text{Area} = 15 \times 4 = 60\, \text{cm}^2$.

(D) median of a triangle divides it into two triangles of equal areas.
Since $AD$ is the median of $\Delta ABC$,it divides $\Delta ABC$ into two triangles of equal areas,namely $\Delta ABD$ and $\Delta ADC$.
Therefore,$ar(\Delta ADC) = \frac{1}{2} \times ar(\Delta ABC)$.
Given $ar(\Delta ABC) = 50 \, cm^2$.
Thus,$ar(\Delta ADC) = \frac{1}{2} \times 50 \, cm^2 = 25 \, cm^2$.

(C) $1$. Since $AD$ is the median of $\Delta ABC$,it divides the triangle into two triangles of equal area.
$\operatorname{ar}(\Delta ABD) = \frac{1}{2} \operatorname{ar}(\Delta ABC) = \frac{1}{2} \times 72 = 36 \, \text{cm}^2$.
$2$. In $\Delta ABD$,$Q$ is the midpoint of $AD$. Therefore,$BQ$ is a median of $\Delta ABD$.
$\operatorname{ar}(\Delta ABQ) = \frac{1}{2} \operatorname{ar}(\Delta ABD) = \frac{1}{2} \times 36 = 18 \, \text{cm}^2$.
$3$. In $\Delta ABQ$,$P$ is the midpoint of $AB$. Therefore,$QP$ is a median of $\Delta ABQ$.
$\operatorname{ar}(\Delta APQ) = \frac{1}{2} \operatorname{ar}(\Delta ABQ) = \frac{1}{2} \times 18 = 9 \, \text{cm}^2$.

(D) Given that $\Delta PQR$ is a right-angled triangle with $\angle Q = 90^{\circ}$.
According to the Pythagorean theorem,$PQ^2 + QR^2 = PR^2$.
Substituting the given values: $PQ^2 + 21^2 = 29^2$.
$PQ^2 + 441 = 841$.
$PQ^2 = 841 - 441 = 400$.
$PQ = \sqrt{400} = 20 \text{ cm}$.
The area of a right-angled triangle is given by $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Here,$\text{base} = QR = 21 \text{ cm}$ and $\text{height} = PQ = 20 \text{ cm}$.
$\text{Area} = \frac{1}{2} \times 21 \times 20 = 21 \times 10 = 210 \text{ cm}^2$.

(B) median of a triangle divides it into two triangles of equal areas.
Since $AD$ is the median of $\Delta ABC$,it divides the triangle into two triangles,$\Delta ADB$ and $\Delta ADC$,such that $\operatorname{ar}(ADB) = \operatorname{ar}(ADC)$.
Given that $\operatorname{ar}(ADB) = 53 \, cm^2$,therefore $\operatorname{ar}(ADC) = 53 \, cm^2$.
The area of $\Delta ABC$ is the sum of the areas of $\Delta ADB$ and $\Delta ADC$.
$\operatorname{ar}(ABC) = \operatorname{ar}(ADB) + \operatorname{ar}(ADC) = 53 \, cm^2 + 53 \, cm^2 = 106 \, cm^2$.

(C) median of a triangle divides it into two triangles of equal areas.
In $\Delta PQR$,$PM$ is the median to the side $QR$.
Therefore,$\text{ar}(\Delta PMQ) = \text{ar}(\Delta PMR)$.
Given that $\text{ar}(\Delta PMQ) = 36 \text{ cm}^2$.
Thus,$\text{ar}(\Delta PMR) = 36 \text{ cm}^2$.

(D) Given that $\Delta ABC$ is a right-angled triangle with $\angle B = 90^{\circ}$.
Using the Pythagorean theorem: $AC^2 = AB^2 + BC^2$.
Substitute the given values: $50^2 = 14^2 + BC^2$.
$2500 = 196 + BC^2$.
$BC^2 = 2500 - 196 = 2304$.
$BC = \sqrt{2304} = 48 \, cm$.
The area of a right-angled triangle is given by $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
$\text{Area} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 14 \times 48$.
$\text{Area} = 7 \times 48 = 336 \, cm^2$.

(A) In $\Delta ABC$,since $\angle B = 90^{\circ}$,by the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Substituting the values,$AC^2 = 12^2 + 16^2 = 144 + 256 = 400$.
Thus,$AC = \sqrt{400} = 20 \, cm$.
The area of $\Delta ABC$ can be calculated as $\frac{1}{2} \times \text{base} \times \text{height}$.
Using $AB$ and $BC$ as base and height,Area $= \frac{1}{2} \times 12 \times 16 = 96 \, cm^2$.
Alternatively,using $AC$ as the base and $BM$ as the altitude,Area $= \frac{1}{2} \times AC \times BM$.
Equating the two expressions for area: $\frac{1}{2} \times 20 \times BM = 96$.
$10 \times BM = 96$.
$BM = 9.6 \, cm$.

(B) In an equilateral triangle with side length $a$,the altitude divides the base into two equal parts of length $\frac{a}{2}$.
By applying the Pythagorean theorem in one of the resulting right-angled triangles:
$h^{2} + (\frac{a}{2})^{2} = a^{2}$
$h^{2} + \frac{a^{2}}{4} = a^{2}$
$h^{2} = a^{2} - \frac{a^{2}}{4}$
$h^{2} = \frac{3a^{2}}{4}$
Taking the square root of both sides:
$h = \frac{\sqrt{3}}{2} a$
Thus,the length of the altitude is $\frac{\sqrt{3}}{2} a$.

(B) First,observe that $AB^2 + BC^2 = 6^2 + 8^2 = 36 + 64 = 100 = 10^2 = AC^2$.
Since the sum of the squares of two sides equals the square of the third side,$\Delta ABC$ is a right-angled triangle with the right angle at $B$.
The area of $\Delta ABC$ can be calculated as $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 6 \times 8 = 24 \, cm^2$.
Alternatively,using $AC$ as the base and $BM$ as the altitude,the area is $\frac{1}{2} \times AC \times BM = \frac{1}{2} \times 10 \times BM = 5 \times BM$.
Equating the two areas: $5 \times BM = 24$.
Therefore,$BM = \frac{24}{5} = 4.8 \, cm$.

(B) The area of a triangle is given by the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
In a right-angled triangle,the two sides that form the right angle act as the base and the height.
Therefore,the area of a right triangle is $\frac{1}{2} \times \text{product of the sides forming the right angle}$.