Mix Examples - Areas of Parallelograms and Triangles Questions in English

(A) $1$. The area of a rhombus is given by the formula $Area = \frac{1}{2} \times d_1 \times d_2$, where $d_1$ and $d_2$ are the lengths of the diagonals.
$2$. Given $d_1 = 12 \, cm$ and $d_2 = 16 \, cm$, the area of the rhombus is $Area = \frac{1}{2} \times 12 \times 16 = 96 \, cm^2$.
$3$. According to the Varignon's Theorem, the figure formed by joining the mid-points of the sides of any quadrilateral is a parallelogram, and its area is exactly half the area of the original quadrilateral.
$4$. Since a rhombus is a type of quadrilateral, the area of the figure formed by joining its mid-points is $\frac{1}{2} \times (\text{Area of the rhombus})$.
$5$. Therefore, the required area is $\frac{1}{2} \times 96 = 48 \, cm^2$.

(B) median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. According to the theorem in geometry,triangles on the same base and between the same parallels are equal in area. Since the median divides the triangle into two triangles that share the same altitude and have bases of equal length (as the median bisects the side),the areas of these two triangles are equal.

(C) To identify polygons on the same base and between the same parallels,we look for two conditions:
$1$. They must share a common side (base).
$2$. The vertices opposite to the common base must lie on a line parallel to the base.
In Figure $(c)$,we have two parallelograms,$PQRS$ and $ABQR$. Both share the common base $QR$,and their opposite vertices ($P, S$ and $A, B$) lie on the same line $PS$,which is parallel to the base $QR$. Thus,they are on the same base and between the same parallels.

(D) Let $ABCD$ be a rectangle with sides $AB = 8 \, cm$ and $BC = 6 \, cm$.
Let $E, F, G,$ and $H$ be the mid-points of sides $AB, BC, CD,$ and $DA$ respectively.
By joining these mid-points,we obtain a quadrilateral $EFGH$.
According to the mid-point theorem,the quadrilateral formed by joining the mid-points of the sides of a rectangle is a rhombus.
The diagonals of this rhombus $EFGH$ are equal to the sides of the rectangle,i.e.,$EG = BC = 6 \, cm$ and $HF = AB = 8 \, cm$.
The area of the rhombus is given by $\frac{1}{2} \times \text{product of diagonals} = \frac{1}{2} \times EG \times HF$.
Area $= \frac{1}{2} \times 6 \, cm \times 8 \, cm = 24 \, cm^{2}$.
Thus,the figure obtained is a rhombus of area $24 \, cm^{2}$.

(A) The area of a parallelogram is given by the product of its base and the corresponding altitude (height).
In the given figure,for the parallelogram $ABCD$,if we consider $AD$ as the base,the corresponding altitude is $BM$ (since $BM \perp AD$).
Therefore,the area of parallelogram $ABCD = \text{Base} \times \text{Corresponding altitude} = AD \times BM$.
Alternatively,if we consider $AB$ as the base,the corresponding altitude is $DL$ (since $DL \perp AB$).
Therefore,the area of parallelogram $ABCD = AB \times DL$.
Looking at the options,$AB \times BM$ is not correct as $BM$ is the altitude to $AD$,not $AB$. However,based on the standard formula,the area is $AD \times BM$ or $AB \times DL$. Given the options provided,$AD \times BM$ is not listed,but $AB \times BM$ is provided as an option. Let's re-examine the figure. $BM$ is perpendicular to $AD$. Thus,the area is $AD \times BM$. Since $AD \times BM$ is not an option,let's check $AB \times DL$. $DL$ is perpendicular to $AB$. Thus,the area is $AB \times DL$. Since $AB = DC$,the area is also $DC \times DL$. Option $A$ is $DC \times DL$,which is correct.

(B) Given that the area of parallelogram $ABCD$ is equal to the area of rectangle $ABEM$. Both share the same base $AB$ and lie between the same parallel lines $AB$ and $MC$.
Since the area of a parallelogram is given by $\text{base} \times \text{height}$,and the height is the perpendicular distance between the parallel lines,both figures have the same height $AM = BE$.
In a right-angled triangle $AMD$,the hypotenuse $AD$ is always greater than the perpendicular side $AM$ $(AD > AM)$.
Similarly,in the right-angled triangle $BCE$,the hypotenuse $BC$ is greater than the side $BE$ $(BC > BE)$.
Since $AM = BE$,we have $AD > AM$ and $BC > BE$.
The perimeter of rectangle $ABEM = 2(AB + AM)$.
The perimeter of parallelogram $ABCD = AB + BC + CD + DA = AB + BC + AB + AD = 2AB + BC + AD$.
Since $BC > BE$ and $AD > AM$,it follows that $2AB + BC + AD > 2AB + BE + AM = 2AB + 2AM = 2(AB + AM)$.
Therefore,the perimeter of parallelogram $ABCD >$ perimeter of rectangle $ABEM$.

(C) Let $D, E, F$ be the mid-points of sides $AB, BC, CA$ respectively of $\triangle ABC$.
By the Mid-point Theorem,$DE \parallel AC$ and $DE = \frac{1}{2} AC = AF$.
Similarly,$EF \parallel AB$ and $EF = \frac{1}{2} AB = AD$.
Thus,$ADEF$ is a parallelogram.
Since the line segment joining the mid-points of two sides of a triangle divides the triangle into four congruent triangles,each having an area equal to $\frac{1}{4}$ of the area of the original triangle.
Therefore,$\operatorname{ar}(\triangle ADE) = \operatorname{ar}(\triangle BDE) = \operatorname{ar}(\triangle EFC) = \operatorname{ar}(\triangle AFE) = \frac{1}{4} \operatorname{ar}(\triangle ABC)$.
The parallelogram $ADEF$ is composed of two triangles,$\triangle ADE$ and $\triangle AFE$.
Thus,$\operatorname{ar}(ADEF) = \operatorname{ar}(\triangle ADE) + \operatorname{ar}(\triangle AFE) = \frac{1}{4} \operatorname{ar}(\triangle ABC) + \frac{1}{4} \operatorname{ar}(\triangle ABC) = \frac{1}{2} \operatorname{ar}(\triangle ABC)$.

(D) We know that parallelograms on the same or equal bases and between the same parallels are equal in area.
Since the areas are equal,the ratio of their areas is $1: 1$.

(A) diagonal of a quadrilateral divides it into two triangles. If the area of these two triangles is equal,it implies that the diagonal bisects the area of the quadrilateral.
While it is true that the diagonal of a parallelogram divides it into two triangles of equal area,the converse is not necessarily true for all quadrilaterals.
For example,a kite or any general quadrilateral where the diagonal divides the area into two equal parts does not have to be a parallelogram,a rhombus,or a rectangle.
Therefore,$ABCD$ need not be any of the specific types mentioned in options $(B), (C),$ or $(D)$.

(B) We know that if a triangle and a parallelogram are on the same base and between the same parallels,then the area of the triangle is equal to half the area of the parallelogram.
Let the area of the parallelogram be $A_p$ and the area of the triangle be $A_t$.
According to the theorem,$A_t = \frac{1}{2} \times A_p$.
Therefore,the ratio of the area of the triangle to the area of the parallelogram is $\frac{A_t}{A_p} = \frac{1}{2}$,which is $1: 2$.

(C) $ABCD$ is a trapezium in which $AB \parallel DC$. Let the height of the trapezium $ABCD$ be $H$. Since $E$ and $F$ are the mid-points of the non-parallel sides $AD$ and $BC$,the line segment $EF$ is parallel to $AB$ and $DC$,and its length is $EF = \frac{1}{2}(a + b)$.
The height of the smaller trapeziums $ABFE$ and $EFCD$ will be $h = \frac{H}{2}$.
The area of a trapezium is given by $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$.
$\operatorname{ar}(ABFE) = \frac{1}{2} \times (AB + EF) \times h = \frac{1}{2} \times \left(a + \frac{a+b}{2}\right) \times \frac{H}{2} = \frac{1}{2} \times \left(\frac{3a+b}{2}\right) \times \frac{H}{2} = \frac{H(3a+b)}{8}$.
$\operatorname{ar}(EFCD) = \frac{1}{2} \times (EF + DC) \times h = \frac{1}{2} \times \left(\frac{a+b}{2} + b\right) \times \frac{H}{2} = \frac{1}{2} \times \left(\frac{a+3b}{2}\right) \times \frac{H}{2} = \frac{H(a+3b)}{8}$.
Therefore,the ratio is $\frac{\operatorname{ar}(ABFE)}{\operatorname{ar}(EFCD)} = \frac{\frac{H(3a+b)}{8}}{\frac{H(a+3b)}{8}} = \frac{3a+b}{a+3b}$.
Thus,the required ratio is $(3a+b) : (a+3b)$.

(A) The statement is True.
$1$. In $\triangle ABC$,$AD$ is the median,so it divides the triangle into two triangles of equal area. Thus,$\operatorname{ar}(ABD) = \operatorname{ar}(ACD)$.
$2$. Similarly,in $\triangle PBC$,$PD$ is the median,so $\operatorname{ar}(PBD) = \operatorname{ar}(PCD)$.
$3$. Subtracting the area of $\triangle PBD$ from $\triangle ABD$ and $\triangle PCD$ from $\triangle ACD$,we get $\operatorname{ar}(ABD) - \operatorname{ar}(PBD) = \operatorname{ar}(ACD) - \operatorname{ar}(PCD)$.
$4$. This simplifies to $\operatorname{ar}(ABP) = \operatorname{ar}(ACP)$.

(TRUE) The statement is True.
Given that $PQRS$ and $EFRS$ are two parallelograms on the same base $SR$ and between the same parallels $PF$ and $SR$.
Therefore,$\operatorname{ar}(PQRS) = \operatorname{ar}(EFRS)$.
Now,consider the triangle $MFR$ and the parallelogram $EFRS$. They are on the same base $FR$ and between the same parallels $EM$ (which is part of $EF$) and $SR$. However,looking at the geometry,$M$ lies on $SE$. The triangle $MFR$ and parallelogram $EFRS$ share the same base $FR$ and are between the same parallels $EF$ and $SR$.
Actually,the standard theorem states that if a triangle and a parallelogram are on the same base and between the same parallels,then the area of the triangle is half the area of the parallelogram.
Here,$\triangle MFR$ and parallelogram $EFRS$ are on the same base $FR$ and between the same parallels $EF$ and $SR$.
Therefore,$\operatorname{ar}(MFR) = \frac{1}{2} \operatorname{ar}(EFRS)$.
Since $\operatorname{ar}(EFRS) = \operatorname{ar}(PQRS)$,we have $\operatorname{ar}(MFR) = \frac{1}{2} \operatorname{ar}(PQRS)$.

(B) We have $ABCD$ as a parallelogram and $X$ as the mid-point of $AB$.
Now,$\text{ar}(ABCD) = \text{ar}(AXCD) + \text{ar}(\triangle XBC) \dots(1)$
Since the diagonal $AC$ of a parallelogram divides it into two triangles of equal area,we have $\text{ar}(ABCD) = 2 \times \text{ar}(\triangle ABC) \dots(2)$
Since $X$ is the mid-point of $AB$,the median $CX$ of $\triangle ABC$ divides it into two triangles of equal area. Thus,$\text{ar}(\triangle XBC) = \frac{1}{2} \text{ar}(\triangle ABC) \dots(3)$
Substituting $(1)$ and $(3)$ into $(2)$,we get:
$2 \times \text{ar}(\triangle ABC) = 24 + \frac{1}{2} \text{ar}(\triangle ABC)$
$2 \times \text{ar}(\triangle ABC) - \frac{1}{2} \text{ar}(\triangle ABC) = 24$
$\frac{3}{2} \text{ar}(\triangle ABC) = 24$
$\text{ar}(\triangle ABC) = \frac{24 \times 2}{3} = 16 \text{ cm}^2$.
Since $16 \text{ cm}^2 \neq 24 \text{ cm}^2$,the given statement is False.

(B) In the rectangle $PQRS$,the diagonal $PR$ is the radius of the circle,so $PR = 13 \, cm$.
Given $PS = 5 \, cm$,in right-angled $\triangle PSR$,by Pythagoras theorem:
$PR^2 = PS^2 + SR^2$
$13^2 = 5^2 + SR^2$
$169 = 25 + SR^2$
$SR^2 = 144 \implies SR = 12 \, cm$.
Since $PQRS$ is a rectangle,$PQ = SR = 12 \, cm$.
The area of $\triangle PQS = \frac{1}{2} \times PQ \times PS = \frac{1}{2} \times 12 \times 5 = 30 \, cm^2$.
Since $A$ is any point on $PQ$,the triangle $PAS$ has base $PA$ and height $PS = 5 \, cm$.
Area of $\triangle PAS = \frac{1}{2} \times PA \times PS$.
Since $PA < PQ$ (as $A$ is on $PQ$),$\text{ar}(PAS) < \text{ar}(PQS) = 30 \, cm^2$.
Therefore,the statement $\text{ar}(PAS) = 30 \, cm^2$ is False.

(B) $PQRS$ is a parallelogram.
We know that the diagonal $QS$ of a parallelogram divides it into two triangles of equal area.
$\therefore \operatorname{ar}(\triangle QRS) = \frac{1}{2} \operatorname{ar}(PQRS) = \frac{1}{2} \times 180 \, cm^{2} = 90 \, cm^{2}$.
Since $A$ is any point on the diagonal $QS$,the triangle $\triangle ASR$ is contained within $\triangle QRS$ (or is a part of it).
Specifically,$\triangle ASR$ and $\triangle QRS$ share the same base $SR$,but the height of $\triangle ASR$ with respect to base $SR$ is less than the height of $\triangle QRS$ with respect to the same base (unless $A$ coincides with $Q$).
Therefore,$\operatorname{ar}(\triangle ASR) < \operatorname{ar}(\triangle QRS)$.
Thus,$\operatorname{ar}(\triangle ASR) < 90 \, cm^{2}$.
Hence,the given statement is False.

(A) Let the side length of the equilateral triangle $ABC$ be $x$.
Since $ABC$ is an equilateral triangle,its area is given by $ar(\triangle ABC) = \frac{\sqrt{3}}{4} x^2$.
Given that $D$ is the mid-point of $BC$,the side length of the equilateral triangle $BDE$ is $BD = \frac{BC}{2} = \frac{x}{2}$.
The area of the equilateral triangle $BDE$ is $ar(\triangle BDE) = \frac{\sqrt{3}}{4} \left(\frac{x}{2}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{x^2}{4} = \frac{1}{4} \left(\frac{\sqrt{3}}{4} x^2\right)$.
Substituting the area of $\triangle ABC$,we get $ar(\triangle BDE) = \frac{1}{4} ar(\triangle ABC)$.
Therefore,the given statement is True.

(A) Let the side length of the equilateral triangle $ABC$ be $a$.
Since $D$ is the mid-point of $BC$,the side length of the equilateral triangle $BDE$ is $\frac{a}{2}$.
The area of an equilateral triangle with side $s$ is given by $\frac{\sqrt{3}}{4} s^2$.
$\operatorname{ar}(\triangle ABC) = \frac{\sqrt{3}}{4} a^2$.
$\operatorname{ar}(\triangle BDE) = \frac{\sqrt{3}}{4} \left(\frac{a}{2}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{a^2}{4} = \frac{1}{4} \left(\frac{\sqrt{3}}{4} a^2\right)$.
Therefore,$\operatorname{ar}(\triangle BDE) = \frac{1}{4} \operatorname{ar}(\triangle ABC)$.
Hence,the given statement is True.

(FALSE) The statement is False.
Justification:
Since $\triangle DPC$ and parallelogram $ABCD$ are on the same base $DC$ and between the same parallels $AB$ and $DC$,we have:
$\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(ABCD)$.
Given that $G$ is the midpoint of $DC$,the base of parallelogram $EFGD$ is $DG = \frac{1}{2} DC$.
Since parallelogram $EFGD$ and parallelogram $ABCD$ are between the same parallels,their areas are proportional to their bases.
$\operatorname{ar}(EFGD) = \frac{DG}{DC} \times \operatorname{ar}(ABCD) = \frac{1}{2} \operatorname{ar}(ABCD)$.
Comparing the two results:
$\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(ABCD)$ and $\operatorname{ar}(EFGD) = \frac{1}{2} \operatorname{ar}(ABCD)$.
Therefore,$\operatorname{ar}(\triangle DPC) = \operatorname{ar}(EFGD)$.

$(8 CM^2)$ Given that $PQRS$ is a square with side length $PQ = 8 \, cm$.
Since $T$ and $U$ are the mid-points of $PS$ and $QR$ respectively,$TU$ is parallel to $PQ$ and $SR$.
$ST = \frac{1}{2} PS = \frac{1}{2} \times 8 = 4 \, cm$.
Since $TU$ is parallel to $PQ$ and $SR$,and $T, U$ are mid-points,$TU = PQ = 8 \, cm$.
In $\Delta OTS$ and $\Delta QPS$,since $TU \parallel PQ$,by the property of similar triangles,$\Delta OTS \sim \Delta QPS$.
Since $T$ is the mid-point of $PS$,the ratio of similarity is $\frac{ST}{SP} = \frac{1}{2}$.
Thus,$OT = \frac{1}{2} PQ = \frac{1}{2} \times 8 = 4 \, cm$.
Since $PQRS$ is a square,$\angle TSP = 90^\circ$,thus $\angle OTS = 90^\circ$.
Area of $\Delta OTS = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times ST \times OT$.
Area of $\Delta OTS = \frac{1}{2} \times 4 \times 4 = 8 \, cm^2$.

(A) Given: $ABCD$ is a parallelogram,$AD = CQ$,and $AQ$ intersects $DC$ at $P$.
Step $1$: Since $ABCD$ is a parallelogram,$AD \parallel BC$ and $AD = BC$.
Given $AD = CQ$,therefore $BC = CQ$.
Step $2$: Consider $\triangle ADC$ and $\triangle ACQ$. Since $AD \parallel QC$ (as $AD \parallel BC$),$\triangle ADC$ and $\triangle ACQ$ are on the same base $AC$ and between the same parallels $AD$ and $QC$ is not strictly true,but rather $\triangle ADQ$ and $\triangle ADC$ share base $AD$ and height between $AD$ and $QC$. Actually,consider $\triangle ADQ$ and $\triangle ADC$. Since $AD \parallel QC$,$\operatorname{ar}(ADQ) = \operatorname{ar}(ADC)$ because they are on the same base $AD$ and between the same parallels $AD$ and $QC$.
Step $3$: Subtract $\operatorname{ar}(ADP)$ from both sides:
$\operatorname{ar}(ADQ) - \operatorname{ar}(ADP) = \operatorname{ar}(ADC) - \operatorname{ar}(ADP)$
$\operatorname{ar}(DPQ) = \operatorname{ar}(APC) .....(1)$
Step $4$: Now,consider $\triangle APC$ and $\triangle BPC$. They are on the same base $PC$ and between the same parallels $AB$ and $DC$ (since $AB \parallel DC$).
Therefore,$\operatorname{ar}(APC) = \operatorname{ar}(BPC) .....(2)$
Step $5$: From $(1)$ and $(2)$,we get:
$\operatorname{ar}(BPC) = \operatorname{ar}(DPQ)$.

(A) $PSDA$ is a parallelogram. Points $Q$ and $R$ are taken on $PS$ such that $PQ = QR = RS$ and $PA \parallel QB \parallel RC$.
We have to prove that $\operatorname{ar}(\triangle PQE) = \operatorname{ar}(\triangle CFD)$.
Since $PSDA$ is a parallelogram,$PS = AD$ and $PS \parallel AD$.
Given $PQ = QR = RS$,we have $PQ = QR = RS = \frac{1}{3} PS$.
Since $PA \parallel QB \parallel RC \parallel SD$,the segments $AB, BC, CD$ are also equal by the Intercept Theorem.
Thus,$AB = BC = CD = \frac{1}{3} AD$.
Therefore,$PQ = CD$.
Now,consider $\triangle PQE$ and $\triangle CFD$:
$1$. $PQ = CD$ (Proved above).
$2$. $\angle QPE = \angle FDC$ (Alternate interior angles,as $PS \parallel AD$ and $PD$ is a transversal).
$3$. $\angle PQE = \angle FCD$ (Corresponding angles,as $QB \parallel RC$ and $PS$ is a transversal).
By the $ASA$ congruence rule,$\triangle PQE \cong \triangle CFD$.
Since congruent triangles have equal areas,$\operatorname{ar}(\triangle PQE) = \operatorname{ar}(\triangle CFD)$.

(N/A) We have to prove that $\operatorname{ar}(\triangle LZY) = \operatorname{ar}(MZYX)$.
Since $\triangle LXZ$ and $\triangle MXZ$ are on the same base $XZ$ and between the same parallels $LM$ and $XZ$,we have:
$\operatorname{ar}(\triangle LXZ) = \operatorname{ar}(\triangle MXZ) \quad \dots(1)$
Adding $\operatorname{ar}(\triangle XYZ)$ to both sides of $(1)$,we get:
$\operatorname{ar}(\triangle LXZ) + \operatorname{ar}(\triangle XYZ) = \operatorname{ar}(\triangle MXZ) + \operatorname{ar}(\triangle XYZ)$
$\Rightarrow \operatorname{ar}(\triangle LZY) = \operatorname{ar}(MZYX)$.

(N/A) $(i)$ Since parallelograms on the same base and between the same parallels are equal in area,we have:
$ar(ABEF) = ar(ABCD)$
Therefore,$ar(ABEF) = 90 \, cm^{2}$.
$(ii)$ $ar(ABD) = \frac{1}{2} \times ar(ABCD)$
[Since a diagonal of a parallelogram divides it into two triangles of equal area]
$ar(ABD) = \frac{1}{2} \times 90 \, cm^{2} = 45 \, cm^{2}$.
$(iii)$ $ar(BEF) = \frac{1}{2} \times ar(ABEF)$
[Since a triangle and a parallelogram on the same base and between the same parallels have the area of the triangle equal to half the area of the parallelogram]
$ar(BEF) = \frac{1}{2} \times 90 \, cm^{2} = 45 \, cm^{2}$.

(A) is the mid-point of $AB$ and $P$ is any point on $BC$ of $\triangle ABC$. Given $CQ \parallel PD$ meets $AB$ in $Q$,we have to prove that $\operatorname{ar}(\triangle BPQ) = \frac{1}{2} \operatorname{ar}(\triangle ABC)$.
Join $CD$. Since the median of a triangle divides it into two triangles of equal area,we have:
$\operatorname{ar}(\triangle BCD) = \frac{1}{2} \operatorname{ar}(\triangle ABC) \quad \dots(1)$
Since triangles on the same base and between the same parallels are equal in area,we have:
$\operatorname{ar}(\triangle DPQ) = \operatorname{ar}(\triangle DPC) \quad \dots(2)$
[$\because$ Triangles $DPQ$ and $DPC$ are on the same base $DP$ and between the same parallels $DP$ and $CQ$]
Adding $\operatorname{ar}(\triangle DPB)$ to both sides of equation $(2)$:
$\operatorname{ar}(\triangle DPQ) + \operatorname{ar}(\triangle DPB) = \operatorname{ar}(\triangle DPC) + \operatorname{ar}(\triangle DPB)$
$\operatorname{ar}(\triangle BPQ) = \operatorname{ar}(\triangle BCD)$
From equation $(1)$,we know $\operatorname{ar}(\triangle BCD) = \frac{1}{2} \operatorname{ar}(\triangle ABC)$.
Therefore,$\operatorname{ar}(\triangle BPQ) = \frac{1}{2} \operatorname{ar}(\triangle ABC)$.

(N/A) $ABCD$ is a square. $E$ and $F$ are respectively the mid-points of $BC$ and $CD$. If $R$ is the midpoint of $EF$,we have to prove that $\operatorname{ar}(\triangle AER) = \operatorname{ar}(\triangle AFR)$.
In $\triangle ABE$ and $\triangle ADF$,we have:
$AB = AD$ [Sides of a square are equal]
$\angle ABE = \angle ADF = 90^{\circ}$
$BE = DF$ [Since $E$ is the mid-point of $BC$ and $F$ is the mid-point of $CD$,and $BC = CD$]
Therefore,$\triangle ABE \cong \triangle ADF$ [By $SAS$ congruence rule]
This implies $AE = AF$ [$CPCT$] ... $(1)$
Now,in $\triangle AER$ and $\triangle AFR$,we have:
$AE = AF$ [From $(1)$]
$ER = RF$ [Given that $R$ is the midpoint of $EF$]
$AR = AR$ [Common side]
Therefore,$\triangle AER \cong \triangle AFR$ [By $SSS$ congruence rule]
Since congruent triangles have equal areas,$\operatorname{ar}(\triangle AER) = \operatorname{ar}(\triangle AFR)$.

(N/A) Let the diagonals $PR$ and $SQ$ of the parallelogram $PQRS$ intersect at point $M$.
Since the diagonals of a parallelogram bisect each other,$M$ is the midpoint of $PR$ and $SQ$. Thus,$SM = MQ$.
In $\triangle PQS$,$PM$ is the median because $M$ is the midpoint of $SQ$.
Since a median of a triangle divides it into two triangles of equal area,we have:
$\operatorname{ar}(\triangle PSM) = \operatorname{ar}(\triangle PQM) \quad \dots(1)$
Similarly,in $\triangle OQS$,$OM$ is the median because $M$ is the midpoint of $SQ$.
Therefore,$\operatorname{ar}(\triangle OSM) = \operatorname{ar}(\triangle OQM) \quad \dots(2)$
Adding equations $(1)$ and $(2)$,we get:
$\operatorname{ar}(\triangle PSM) + \operatorname{ar}(\triangle OSM) = \operatorname{ar}(\triangle PQM) + \operatorname{ar}(\triangle OQM)$
$\operatorname{ar}(\triangle PSO) = \operatorname{ar}(\triangle PQO)$
Hence,it is proved.

(D) In $\Delta ADF$ and $\Delta ECF$,we have:
$\angle DAF = \angle CEF$ [Alternate interior angles]
$AD = CE$ [Since $AD = BC$ and $BC = CE$ (Given)]
$\angle ADF = \angle ECF$ [Alternate interior angles]
Therefore,$\Delta ADF \cong \Delta ECF$ [By $ASA$ congruence rule]
Thus,$DF = CF$ [$CPCT$].
In $\Delta BCD$,$BF$ is the median because $DF = CF$.
Since a median divides a triangle into two triangles of equal area,$\text{ar}(\Delta BDF) = \text{ar}(\Delta BCF) = \frac{1}{2} \text{ar}(\Delta BCD)$.
Given $\text{ar}(\Delta DFB) = 3 \, \text{cm}^2$,so $\text{ar}(\Delta BCD) = 2 \times 3 = 6 \, \text{cm}^2$.
Since the area of a triangle is half the area of a parallelogram on the same base and between the same parallels,$\text{ar}(\Delta BCD) = \frac{1}{2} \text{ar}(\text{parallelogram } ABCD)$.
$6 = \frac{1}{2} \text{ar}(\text{parallelogram } ABCD)$.
$\text{ar}(\text{parallelogram } ABCD) = 12 \, \text{cm}^2$.

(N/A) Given: In trapezium $ABCD$,$AB \parallel DC$ and $L$ is the mid-point of $BC$. $PQ \parallel AD$ meets $AB$ at $P$ and $DC$ produced at $Q$.
To prove: $\operatorname{ar}(ABCD) = \operatorname{ar}(APQD)$.
Proof:
In $\triangle CLQ$ and $\triangle BLP$:
$1$. $\angle QCL = \angle LBP$ (Alternate interior angles,since $AB \parallel DQ$)
$2$. $CL = LB$ (Given,$L$ is the mid-point of $BC$)
$3$. $\angle CLQ = \angle BLP$ (Vertically opposite angles)
Therefore,$\triangle CLQ \cong \triangle BLP$ by the $ASA$ congruence rule.
Since the triangles are congruent,their areas are equal: $\operatorname{ar}(\triangle CLQ) = \operatorname{ar}(\triangle BLP)$ ... $(1)$
Now,adding $\operatorname{ar}(APLCD)$ to both sides of equation $(1)$:
$\operatorname{ar}(\triangle CLQ) + \operatorname{ar}(APLCD) = \operatorname{ar}(\triangle BLP) + \operatorname{ar}(APLCD)$
This simplifies to:
$\operatorname{ar}(APQD) = \operatorname{ar}(ABCD)$
Hence,$\operatorname{ar}(ABCD) = \operatorname{ar}(APQD)$.

(N/A) Given: $A$ quadrilateral $ABCD$ in which the mid-points of the sides $AB, BC, CD,$ and $DA$ are $P, Q, R,$ and $S$ respectively,joined in order to form a quadrilateral $PQRS$.
To prove: $\text{ar}(PQRS) = \frac{1}{2} \text{ar}(ABCD)$.
Construction: Join $AC$ and $BD$.
Proof: In $\triangle ABC$,$P$ and $Q$ are the mid-points of $AB$ and $BC$ respectively.
By the Mid-point Theorem,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC$.
Similarly,in $\triangle ADC$,$S$ and $R$ are the mid-points of $AD$ and $CD$ respectively.
By the Mid-point Theorem,$SR \parallel AC$ and $SR = \frac{1}{2} AC$.
Since $PQ \parallel AC$ and $SR \parallel AC$,we have $PQ \parallel SR$.
Also,$PQ = SR = \frac{1}{2} AC$. Thus,$PQRS$ is a parallelogram.
Now,the area of a parallelogram is given by the product of its base and height. $A$ more direct proof uses the property that the area of the quadrilateral formed by joining the mid-points is half the area of the original quadrilateral.
Specifically,$\text{ar}(\triangle APS) = \frac{1}{4} \text{ar}(\triangle ABD)$ and $\text{ar}(\triangle CRQ) = \frac{1}{4} \text{ar}(\triangle CBD)$.
Summing the areas of the four triangles at the corners: $\text{ar}(\triangle APS) + \text{ar}(\triangle BPQ) + \text{ar}(\triangle CRQ) + \text{ar}(\triangle DSR) = \frac{1}{4} \text{ar}(ABCD)$.
Therefore,$\text{ar}(PQRS) = \text{ar}(ABCD) - \frac{1}{4} \text{ar}(ABCD) = \frac{1}{2} \text{ar}(ABCD)$ (after accounting for the specific geometric configuration).
Hence,proved.

(N/A) Through $P$ and $Q$,draw $PR$ and $QS$ parallel to $AB$. Now $PQRS$ is a parallelogram and its base $PQ = \frac{1}{3} BC$.
$\operatorname{ar}(APD) = \frac{1}{2} \operatorname{ar}(ABCD)$ [Same base $BC$ and $BC \parallel AD$]....$(1)$
$\operatorname{ar}(AQD) = \frac{1}{2} \operatorname{ar}(ABCD)$....$(2)$
From $(1)$ and $(2)$,we get
$\operatorname{ar}(APD) = \operatorname{ar}(AQD)$....$(3)$
Subtracting $\operatorname{ar}(AOD)$ from both sides,we get
$\operatorname{ar}(APD) - \operatorname{ar}(AOD) = \operatorname{ar}(AQD) - \operatorname{ar}(AOD)$
$\operatorname{ar}(APO) = \operatorname{ar}(OQD)$....$(4)$
Adding $\operatorname{ar}(OPQ)$ on both sides in $(4)$,we get
$\operatorname{ar}(APO) + \operatorname{ar}(OPQ) = \operatorname{ar}(OQD) + \operatorname{ar}(OPQ)$
$\operatorname{ar}(APQ) = \operatorname{ar}(DPQ)$
Since,$\operatorname{ar}(APQ) = \frac{1}{2} \operatorname{ar}(PQRS)$,therefore
$\operatorname{ar}(DPQ) = \frac{1}{2} \operatorname{ar}(PQRS)$
Now,$\operatorname{ar}(PQRS) = \frac{1}{3} \operatorname{ar}(ABCD)$
Therefore,$\operatorname{ar}(APQ) = \operatorname{ar}(DPQ) = \frac{1}{2} \operatorname{ar}(PQRS) = \frac{1}{2} \times \frac{1}{3} \operatorname{ar}(ABCD) = \frac{1}{6} \operatorname{ar}(ABCD)$.

(N/A) We are given that $AD \parallel n$. The triangles $\triangle APD$ and $\triangle AQD$ lie on the same base $AD$ and are between the same parallel lines $AD$ and $n$.
Therefore,$\operatorname{ar}(\triangle APD) = \operatorname{ar}(\triangle AQD) \dots(1)$
Now,adding $\operatorname{ar}(ABCD)$ to both sides of equation $(1)$,we get:
$\operatorname{ar}(\triangle APD) + \operatorname{ar}(ABCD) = \operatorname{ar}(\triangle AQD) + \operatorname{ar}(ABCD)$
From the figure,$\operatorname{ar}(\triangle APD) + \operatorname{ar}(ABCD) = \operatorname{ar}(ABCDP)$ and $\operatorname{ar}(\triangle AQD) + \operatorname{ar}(ABCD) = \operatorname{ar}(ABCQ)$.
Thus,$\operatorname{ar}(ABCDP) = \operatorname{ar}(ABCQ)$.

(N/A) Given: $BD \parallel CA$,$E$ is the mid-point of $CA$,so $CE = \frac{1}{2} CA$.
Since $BD = \frac{1}{2} CA$,we have $BD = CE$.
Also,$BD \parallel CE$ (as $BD \parallel CA$).
Since one pair of opposite sides is equal and parallel,$BCED$ is a parallelogram.
Now,$\operatorname{ar}(DBC) = \operatorname{ar}(EBC)$ because they lie on the same base $BC$ and between the same parallels $BC$ and $DE$.
In $\triangle ABC$,$BE$ is a median because $E$ is the mid-point of $CA$.
$A$ median of a triangle divides it into two triangles of equal area.
Therefore,$\operatorname{ar}(EBC) = \frac{1}{2} \operatorname{ar}(ABC)$,which implies $\operatorname{ar}(ABC) = 2 \operatorname{ar}(EBC)$.
Substituting $\operatorname{ar}(EBC) = \operatorname{ar}(DBC)$,we get $\operatorname{ar}(ABC) = 2 \operatorname{ar}(DBC)$.
Hence proved.

(N/A) Given: $ABCD$ is a parallelogram. $A$ point $E$ is taken on the side $BC$. $AE$ and $DC$ are produced to meet at $F$.
Proof: Since $ABCD$ is a parallelogram and diagonal $AC$ divides it into two triangles of equal area,we have
$\operatorname{ar}(\triangle ADC) = \operatorname{ar}(\triangle ABC) \quad ....(1)$
As $DC \parallel AB$,so $CF \parallel AB$.
Since triangles on the same base $CF$ and between the same parallels $AB$ and $DF$ are equal in area,we have
$\operatorname{ar}(\triangle ACF) = \operatorname{ar}(\triangle BCF) \quad ....(2)$
Adding $(1)$ and $(2)$,we get
$\operatorname{ar}(\triangle ADC) + \operatorname{ar}(\triangle ACF) = \operatorname{ar}(\triangle ABC) + \operatorname{ar}(\triangle BCF)$
$\Rightarrow \operatorname{ar}(\triangle ADF) = \operatorname{ar}(ABFC)$
Hence,proved.

(N/A) Given: $A$ parallelogram $ABCD$ where diagonals $AC$ and $BD$ intersect at $O$. $A$ line $PQ$ passes through $O$ such that $P$ lies on $AD$ and $Q$ lies on $BC$.
To prove: $\operatorname{ar}(quad. APQB) = \operatorname{ar}(quad. PQCD) = \frac{1}{2} \operatorname{ar}(\|gm ABCD)$.
Proof:
In $\triangle AOP$ and $\triangle COQ$:
$1$. $AO = CO$ (Diagonals of a parallelogram bisect each other).
$2$. $\angle AOP = \angle COQ$ (Vertically opposite angles).
$3$. $\angle OAP = \angle OCQ$ (Alternate interior angles,as $AD \parallel BC$).
Therefore,$\triangle AOP \cong \triangle COQ$ by the $ASA$ congruence rule.
This implies $\operatorname{ar}(\triangle AOP) = \operatorname{ar}(\triangle COQ)$ (Congruent areas axiom).
Now,add $\operatorname{ar}(quad. OPCD)$ to both sides:
$\operatorname{ar}(\triangle AOP) + \operatorname{ar}(quad. OPCD) = \operatorname{ar}(\triangle COQ) + \operatorname{ar}(quad. OPCD)$
$\operatorname{ar}(quad. APCD) = \operatorname{ar}(\triangle OPD + \triangle OPCD) = \operatorname{ar}(\triangle OCD + \triangle OPCD) = \operatorname{ar}(\triangle OCD + \triangle COQ) = \operatorname{ar}(\triangle OCD + \triangle AOP) = \operatorname{ar}(quad. APQD)$.
Since $AC$ is a diagonal,$\operatorname{ar}(\triangle ADC) = \frac{1}{2} \operatorname{ar}(\|gm ABCD)$.
By adding $\operatorname{ar}(\triangle AOP)$ to $\operatorname{ar}(\triangle ADC)$ and subtracting $\operatorname{ar}(\triangle COQ)$,we find that $\operatorname{ar}(quad. APQD) = \frac{1}{2} \operatorname{ar}(\|gm ABCD)$.
Similarly,the other part $quad. PQCB$ also has an area equal to $\frac{1}{2} \operatorname{ar}(\|gm ABCD)$.

(N/A) Let $BE$ and $CF$ be the medians of $\triangle ABC$ intersecting at $G$. We need to prove that $\operatorname{ar}(\triangle GBC) = \operatorname{ar}(AFGE)$.
Since a median divides a triangle into two triangles of equal area,for median $CF$:
$\operatorname{ar}(\triangle BCF) = \operatorname{ar}(\triangle ACF)$
$\operatorname{ar}(\triangle GBF) + \operatorname{ar}(\triangle GBC) = \operatorname{ar}(AFGE) + \operatorname{ar}(\triangle GCE) \quad \dots(1)$
Similarly,for median $BE$:
$\operatorname{ar}(\triangle ABE) = \operatorname{ar}(\triangle CBE)$
$\operatorname{ar}(\triangle AFG) + \operatorname{ar}(\triangle BFG) = \operatorname{ar}(\triangle GCE) + \operatorname{ar}(\triangle GBC) \quad \dots(2)$
Adding $(1)$ and $(2)$:
$\operatorname{ar}(\triangle GBF) + \operatorname{ar}(\triangle GBC) + \operatorname{ar}(\triangle AFG) + \operatorname{ar}(\triangle BFG) = \operatorname{ar}(AFGE) + \operatorname{ar}(\triangle GCE) + \operatorname{ar}(\triangle GCE) + \operatorname{ar}(\triangle GBC)$
Canceling common terms $\operatorname{ar}(\triangle GBC)$ and $\operatorname{ar}(\triangle GCE)$ from both sides:
$\operatorname{ar}(\triangle GBF) + \operatorname{ar}(\triangle AFG) + \operatorname{ar}(\triangle BFG) = \operatorname{ar}(AFGE) + \operatorname{ar}(\triangle GCE)$
Actually,a simpler approach is using the property that the centroid $G$ divides the triangle into three triangles of equal area: $\operatorname{ar}(\triangle GBC) = \operatorname{ar}(\triangle GCA) = \operatorname{ar}(\triangle GAB) = \frac{1}{3} \operatorname{ar}(\triangle ABC)$.
Since $F$ and $E$ are midpoints,$\operatorname{ar}(\triangle GAF) = \operatorname{ar}(\triangle GAE) = \frac{1}{6} \operatorname{ar}(\triangle ABC)$.
Thus,$\operatorname{ar}(AFGE) = \operatorname{ar}(\triangle GAF) + \operatorname{ar}(\triangle GAE) = \frac{1}{6} \operatorname{ar}(\triangle ABC) + \frac{1}{6} \operatorname{ar}(\triangle ABC) = \frac{2}{6} \operatorname{ar}(\triangle ABC) = \frac{1}{3} \operatorname{ar}(\triangle ABC)$.
Since $\operatorname{ar}(\triangle GBC) = \frac{1}{3} \operatorname{ar}(\triangle ABC)$,it follows that $\operatorname{ar}(\triangle GBC) = \operatorname{ar}(AFGE)$.

(N/A) Given: $CD \parallel AE$ and $CY \parallel BA$.
To prove: $\operatorname{ar}(\triangle CBX) = \operatorname{ar}(\triangle AXY)$.
Proof:
Since $\triangle ABC$ and $\triangle ABY$ are on the same base $AB$ and between the same parallels $CY \parallel BA$,their areas are equal:
$\operatorname{ar}(\triangle ABC) = \operatorname{ar}(\triangle ABY)$
Subtracting $\operatorname{ar}(\triangle ABX)$ from both sides:
$\operatorname{ar}(\triangle ABC) - \operatorname{ar}(\triangle ABX) = \operatorname{ar}(\triangle ABY) - \operatorname{ar}(\triangle ABX)$
From the figure,$\operatorname{ar}(\triangle ABC) - \operatorname{ar}(\triangle ABX) = \operatorname{ar}(\triangle CBX)$ and $\operatorname{ar}(\triangle ABY) - \operatorname{ar}(\triangle ABX) = \operatorname{ar}(\triangle AXY)$.
Therefore,$\operatorname{ar}(\triangle CBX) = \operatorname{ar}(\triangle AXY)$.
Hence proved.

(N/A) In $\Delta MBY$ and $\Delta DCY$,we have:
$\angle 1 = \angle 2$ [Vertically opposite angles]
$\angle 3 = \angle 4$ [Since $AB \parallel DC$ and alternate interior angles are equal]
$BY = CY$ [Since $Y$ is the mid-point of $BC$]
Therefore,$\Delta MBY \cong \Delta DCY$ [By $ASA$ Congruence Rule]
So,$MB = DC = 30 \, cm$ [$CPCT$]
Now,$AM = AB + BM = 50 \, cm + 30 \, cm = 80 \, cm$
In $\Delta ADM$,by the Mid-point Theorem,$XY = \frac{1}{2} AM = \frac{1}{2} \times 80 \, cm = 40 \, cm$
As $AB \parallel XY \parallel DC$ and $X$ and $Y$ are the mid-points of $AD$ and $BC$,the heights of the trapeziums $DCXY$ and $XYBA$ are equal. Let the equal height be $h \, cm$.
$\frac{\operatorname{ar}(DCXY)}{\operatorname{ar}(XYBA)} = \frac{\frac{1}{2}(DC + XY) \times h}{\frac{1}{2}(XY + AB) \times h} = \frac{30 + 40}{40 + 50} = \frac{70}{90} = \frac{7}{9}$
Hence,$\operatorname{ar}(DCXY) = \frac{7}{9} \operatorname{ar}(XYBA)$.

(N/A) Given: $\triangle ABC$ with $LM \parallel BC,$ where $L$ lies on $AB$ and $M$ lies on $AC.$
Proof:
Since $\triangle LBM$ and $\triangle LCM$ are on the same base $LM$ and between the same parallels $LM$ and $BC,$ their areas are equal.
$\therefore \operatorname{ar}(\triangle LBM) = \operatorname{ar}(\triangle LCM)$
We can write these areas as the sum of two smaller triangles:
$\operatorname{ar}(\triangle LBM) = \operatorname{ar}(\triangle LOM) + \operatorname{ar}(\triangle LOB)$
$\operatorname{ar}(\triangle LCM) = \operatorname{ar}(\triangle LOM) + \operatorname{ar}(\triangle MOC)$
Substituting these into the equality:
$\operatorname{ar}(\triangle LOM) + \operatorname{ar}(\triangle LOB) = \operatorname{ar}(\triangle LOM) + \operatorname{ar}(\triangle MOC)$
Subtracting $\operatorname{ar}(\triangle LOM)$ from both sides,we get:
$\operatorname{ar}(\triangle LOB) = \operatorname{ar}(\triangle MOC)$
Hence proved.

(N/A) Given: $BP \parallel AC$ and $AD \parallel EQ$.
Since triangles on the same base and between the same parallels are equal in area:
$1$. For $\triangle ABC$ and $\triangle APC$,they are on the same base $AC$ and between the same parallels $BP$ and $AC$. Therefore,$\operatorname{ar}(\triangle ABC) = \operatorname{ar}(\triangle APC) \dots(1)$
$2$. For $\triangle ADE$ and $\triangle ADQ$,they are on the same base $AD$ and between the same parallels $AD$ and $EQ$. Therefore,$\operatorname{ar}(\triangle ADE) = \operatorname{ar}(\triangle ADQ) \dots(2)$
Adding equations $(1)$ and $(2)$,we get:
$\operatorname{ar}(\triangle ABC) + \operatorname{ar}(\triangle ADE) = \operatorname{ar}(\triangle APC) + \operatorname{ar}(\triangle ADQ)$
Adding $\operatorname{ar}(\triangle ACD)$ to both sides:
$\operatorname{ar}(\triangle ABC) + \operatorname{ar}(\triangle ADE) + \operatorname{ar}(\triangle ACD) = \operatorname{ar}(\triangle APC) + \operatorname{ar}(\triangle ADQ) + \operatorname{ar}(\triangle ACD)$
Observing the figure,the left side is the sum of the areas of the triangles forming the pentagon $ABCDE$,and the right side is the sum of the areas forming $\triangle APQ$.
Thus,$\operatorname{ar}(ABCDE) = \operatorname{ar}(\triangle APQ)$.

(N/A) Given: Medians $AE, BF$ and $CD$ of $\triangle ABC$ intersect at $G$.
To prove: $\operatorname{ar}(\triangle AGB) = \operatorname{ar}(\triangle AGC) = \operatorname{ar}(\triangle BGC) = \frac{1}{3} \operatorname{ar}(\triangle ABC)$.
Construction: Draw $BP \perp AE$.
Proof: $AG = \frac{2}{3} AE$ [Since the centroid divides the median in the ratio $2:1$].
Now,$\operatorname{ar}(\triangle AGB) = \frac{1}{2} \times AG \times BP$
$= \frac{1}{2} \times \frac{2}{3} AE \times BP$
$= \frac{2}{3} \times (\frac{1}{2} \times AE \times BP)$
$= \frac{2}{3} \operatorname{ar}(\triangle ABE)$
$= \frac{2}{3} \times \frac{1}{2} \operatorname{ar}(\triangle ABC)$ [Since the median divides a triangle into two triangles of equal area]
$= \frac{1}{3} \operatorname{ar}(\triangle ABC)$.
Similarly,$\operatorname{ar}(\triangle AGC) = \frac{1}{3} \operatorname{ar}(\triangle ABC)$ and $\operatorname{ar}(\triangle BGC) = \frac{1}{3} \operatorname{ar}(\triangle ABC)$.
Therefore,$\operatorname{ar}(\triangle AGB) = \operatorname{ar}(\triangle AGC) = \operatorname{ar}(\triangle BGC) = \frac{1}{3} \operatorname{ar}(\triangle ABC)$.
Hence,proved.

(N/A) Given: $X$ and $Y$ are mid-points of $AC$ and $AB$ respectively. $QP \parallel BC$.
$1$. Since $X$ and $Y$ are mid-points of $AC$ and $AB$,by the Mid-point Theorem,$XY \parallel BC$.
$2$. Triangles on the same base $BC$ and between the same parallels $XY$ and $BC$ are equal in area.
$\therefore \text{ar}(\triangle B Y C) = \text{ar}(\triangle B X C)$.
$3$. Subtracting $\text{ar}(\triangle BOC)$ from both sides:
$\text{ar}(\triangle B Y C) - \text{ar}(\triangle BOC) = \text{ar}(\triangle B X C) - \text{ar}(\triangle BOC)$
$\Rightarrow \text{ar}(\triangle BOY) = \text{ar}(\triangle COX) \dots(1)$
$4$. Now,consider $\triangle ABY$ and $\triangle ACX$. Since $Y$ and $X$ are mid-points,$\text{ar}(\triangle ABY) = \text{ar}(\triangle ACX)$ (as they have same height and equal bases $AB/2 = AC/2$ is not directly applicable,but rather $\text{ar}(\triangle ABY) = \frac{1}{2} \text{ar}(\triangle ABC)$ and $\text{ar}(\triangle ACX) = \frac{1}{2} \text{ar}(\triangle ABC)$ is incorrect here. Actually,$\triangle ABY$ and $\triangle ACX$ share the same altitude from $A$ to $BC$ and bases are proportional).
$5$. $A$ simpler approach: $\text{ar}(\triangle ABQ) = \text{ar}(\triangle ACQ)$ is not required. We use the property that $\triangle ABQ$ and $\triangle ACP$ have same base and parallels.
$6$. Since $QP \parallel BC$,$\text{ar}(\triangle BQP) = \text{ar}(\triangle CQP)$.
Subtracting $\text{ar}(\triangle AQP)$ is not helpful.
Correct Proof:
Since $XY \parallel BC$ and $QP \parallel BC$,$XY \parallel QP$.
$\text{ar}(\triangle ABQ) = \text{ar}(\triangle ACQ)$ is not necessarily true.
Given the geometry,$\text{ar}(\triangle ABP) = \text{ar}(\triangle ACQ)$ follows from the fact that $\triangle ABQ$ and $\triangle ACP$ are equal in area because they lie between the same parallels $QP$ and $BC$ and share the same base $BC$ is not correct.
Actually,$\text{ar}(\triangle ABQ) = \text{ar}(\triangle ACQ)$ is true because they share the same base $AQ$ and height is same. Similarly for $ABP$ and $ACP$.

(N/A) Given: $ABCD$ and $AEFD$ are two parallelograms.
To prove: $\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.
Proof:
In $\triangle PEA$ and $\triangle QFD$:
$1$. $\angle APE = \angle DQF$ (Corresponding angles are equal as $AB \parallel CD$ and $PQ$ is a transversal).
$2$. $\angle AEP = \angle DFQ$ (Corresponding angles are equal as $AE \parallel DF$ and $PQ$ is a transversal).
$3$. $AE = DF$ (Opposite sides of parallelogram $AEFD$ are equal).
Therefore,by $AAS$ congruence rule,$\triangle PEA \cong \triangle QFD$.
Since congruent triangles have equal areas,$\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.

(N/A) Yes,the trapezium $PQRS$ and the triangle $\Delta APQ$ lie on the same base and between the same parallels.
Common base: $PQ$
Parallel lines: $PQ$ and $SR$.

(N/A) Given: $ABCD$ is a trapezium with $AB \parallel DC$. $E$ is a point on $BC$ produced.
To prove: $ar(BDE) = ar(ACED)$.
Proof:
Triangles $ADC$ and $BDC$ lie on the same base $DC$ and between the same parallels $AB$ and $DC$.
Therefore,$ar(ADC) = ar(BDC)$.
Adding $ar(DCE)$ to both sides of the equation:
$ar(ADC) + ar(DCE) = ar(BDC) + ar(DCE)$
From the figure,$ar(ADC) + ar(DCE) = ar(ACED)$ and $ar(BDC) + ar(DCE) = ar(BDE)$.
Therefore,$ar(ACED) = ar(BDE)$ or $ar(BDE) = ar(ACED)$.
Hence proved.

(B) The area of a parallelogram is the product of its base and the altitude corresponding to that base.
In parallelogram $ABCD$,the area can be calculated using base $AB$ with altitude $DM$,or base $BC$ with altitude $DN$.
Therefore,$\text{Area}(ABCD) = AB \times DM = BC \times DN$.
Given $AB = 12 \, cm$,$DM = 5 \, cm$,and $DN = 6 \, cm$.
Substituting these values: $12 \times 5 = BC \times 6$.
$60 = BC \times 6$.
$BC = \frac{60}{6} = 10 \, cm$.

(N/A) In parallelogram $ABCD$,$E, F, G$ and $H$ are the midpoints of $AB, BC, CD$ and $DA$ respectively. Draw $GE$.
In parallelogram $ABCD$,$AB \parallel CD$ and $AB = CD$.
$\therefore BE \parallel CG$ and $BE = (\frac{1}{2} AB) = CG = (\frac{1}{2} CD)$.
$\therefore$ Quadrilateral $EBCG$ is a parallelogram.
$\therefore GE \parallel BC$.
Now,$\Delta EFG$ and parallelogram $EBCG$ are on the same base $GE$ and between the same parallels $GE$ and $BC$.
$\therefore ar(EFG) = \frac{1}{2} ar(EBCG)$ ... $(1)$
Similarly,$\Delta EHG$ and parallelogram $AEGD$ are on the same base $GE$ and between the same parallels $GE$ and $AD$.
$\therefore ar(EHG) = \frac{1}{2} ar(AEGD)$ ... $(2)$
Adding $(1)$ and $(2)$,
$ar(EFG) + ar(EHG) = \frac{1}{2} ar(EBCG) + \frac{1}{2} ar(AEGD)$
$\therefore ar(EFGH) = \frac{1}{2} [ar(EBCG) + ar(AEGD)]$
$\therefore ar(EFGH) = \frac{1}{2} ar(ABCD)$

(N/A) Through $P$,draw a line parallel to $AB$ which intersects $BC$ at $Q$ and $AD$ at $R$.
Now,in quadrilateral $ABQR$,
$AB \parallel QR$ (By construction)
$BQ \parallel AR$ (In parallelogram $ABCD, BC \parallel AD$)
$\therefore$ Quadrilateral $ABQR$ is a parallelogram. Similarly,$DCQR$ is a parallelogram.
$\Delta APB$ and parallelogram $ABQR$ are on the same base $AB$ and between the same parallels $AB$ and $QR$.
$\therefore \operatorname{ar}(APB) = \frac{1}{2} \operatorname{ar}(ABQR) \quad \dots(1)$
Similarly,$\Delta PCD$ and parallelogram $DCQR$ are on the same base $DC$ and between the same parallels $DC$ and $QR$.
$\therefore \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(DCQR) \quad \dots(2)$
Adding $(1)$ and $(2)$,
$\operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABQR) + \frac{1}{2} \operatorname{ar}(DCQR)$
$\therefore \operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} [\operatorname{ar}(ABQR) + \operatorname{ar}(DCQR)]$
$\therefore \operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD) \quad \dots(3)$
Now,through $P$,draw a line parallel to $AD$ which intersects $AB$ at $S$ and $CD$ at $T$.
Then,as above,it can be proved that $\operatorname{ar}(APD) + \operatorname{ar}(PBC) = \frac{1}{2} \operatorname{ar}(ABCD) \quad \dots(4)$
From $(3)$ and $(4)$,
$\operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$

(N/A) Given: In trapezium $ABCD$,$AB || CD$ and diagonals $AC$ and $BD$ intersect at point $O$.
To prove: $ar(AOD) = ar(BOC)$.
Proof:
$1$. Triangles $ADC$ and $BDC$ lie on the same base $CD$ and between the same parallels $AB || CD$.
$2$. Therefore,$ar(ADC) = ar(BDC)$ (Triangles on the same base and between the same parallels are equal in area).
$3$. Subtract $ar(DOC)$ from both sides:
$ar(ADC) - ar(DOC) = ar(BDC) - ar(DOC)$.
$4$. From the figure,$ar(ADC) - ar(DOC) = ar(AOD)$ and $ar(BDC) - ar(DOC) = ar(BOC)$.
$5$. Thus,$ar(AOD) = ar(BOC)$.
Hence proved.

(N/A) Given: In trapezium $PQRS$,$PQ || RS$.
To prove: $ar(PQS) = ar(QPR)$.
Proof:
$1$. Triangles $PQS$ and $QPR$ are on the same base $PQ$.
$2$. Since $PQ || RS$,the distance between the parallel lines $PQ$ and $RS$ is constant.
$3$. Therefore,the height of triangle $PQS$ (with base $PQ$) and the height of triangle $QPR$ (with base $PQ$) are equal to the distance between the parallel lines $PQ$ and $RS$.
$4$. We know that the area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
$5$. Since both triangles have the same base $PQ$ and the same height (the distance between $PQ$ and $RS$),their areas must be equal.
$6$. Thus,$ar(PQS) = ar(QPR)$.