Properties of a Parallelogram Questions in English

(N/A) $ABCD$ is a parallelogram and diagonal $AC$ bisects $\angle A$. We have to show that $ABCD$ is a rhombus.
$\angle 1 = \angle 2$ $\dots(1)$ [$\because AC$ bisects $\angle A$]
$\angle 2 = \angle 4$ $\dots(2)$ [Alternate interior angles,since $AB \parallel DC$]
From $(1)$ and $(2)$,we get:
$\angle 1 = \angle 4$
Now,in $\triangle ABC$,we have:
$\angle 1 = \angle 4$ [Proved above]
$BC = AB$ [$\because$ Sides opposite to equal angles are equal]
Also,$AB = DC$ and $AD = BC$ [$\because$ Opposite sides of a parallelogram are equal]
Therefore,$AB = BC = CD = AD$.
Since all sides of the parallelogram $ABCD$ are equal,it is a rhombus.

(C) parallelogram with equal diagonals is a rectangle.
In a rectangle,all interior angles are equal to $90^{\circ}$.
Therefore,$\angle PQR = 90^{\circ}$.

(N/A) Given: $A$ parallelogram $ABCD$.
To Prove: $A$ diagonal,say $AC$,of parallelogram $ABCD$ divides it into congruent triangles $ABC$ and $CDA$. i.e.,$\Delta ABC \cong \Delta CDA$.
Construction: Join $AC$.
Proof:
Since $ABCD$ is a parallelogram,
$AB \parallel CD$ and $AD \parallel BC$.
Now,$AD \parallel BC$ and transversal $AC$ intersects them at $A$ and $C$ respectively.
Therefore,$\angle DAC = \angle BCA$ (Alternate interior angles)........$(1)$
Again,$AB \parallel DC$ and transversal $AC$ intersects them at $A$ and $C$ respectively.
Therefore,$\angle BAC = \angle DCA$ (Alternate interior angles)........$(2)$
Now,in $\Delta ABC$ and $\Delta CDA$,we have:
$\angle BCA = \angle DAC$ [From $(1)$]
$AC = CA$ (Common side)
$\angle BAC = \angle DCA$ [From $(2)$]
So,by $ASA$ congruence criterion,we obtain:
$\Delta ABC \cong \Delta CDA$.

(A) In quadrilateral $ABCD$,the sum of the interior angles is $360^{\circ}$.
Given the ratio $\angle A : \angle B : \angle C : \angle D = 4 : 5 : 4 : 5$.
Let the angles be $4x, 5x, 4x$,and $5x$.
Sum of angles: $4x + 5x + 4x + 5x = 360^{\circ}$.
$18x = 360^{\circ}$,which gives $x = 20^{\circ}$.
Therefore,the angles are:
$\angle A = 4 \times 20^{\circ} = 80^{\circ}$
$\angle B = 5 \times 20^{\circ} = 100^{\circ}$
$\angle C = 4 \times 20^{\circ} = 80^{\circ}$
$\angle D = 5 \times 20^{\circ} = 100^{\circ}$
Since $\angle A = \angle C$ and $\angle B = \angle D$,both pairs of opposite angles are equal.
Therefore,$ABCD$ is a parallelogram.

(N/A) Let $ABCD$ be a square. We need to prove that $AC = BD$,$OA = OC$,$OB = OD$,and $\angle AOB = 90^{\circ}$.
$1$. To prove diagonals are equal: Consider $\triangle ABC$ and $\triangle BAD$. In these triangles,$AB = BA$ (common side),$BC = AD$ (sides of a square),and $\angle ABC = \angle BAD = 90^{\circ}$. By $SAS$ congruence,$\triangle ABC \cong \triangle BAD$. Thus,$AC = BD$ $(CPCT)$.
$2$. To prove diagonals bisect each other: Consider $\triangle OAB$ and $\triangle OCD$. Here,$\angle OAB = \angle OCD$ (alternate interior angles),$\angle OBA = \angle ODC$ (alternate interior angles),and $AB = CD$ (sides of a square). By $ASA$ congruence,$\triangle OAB \cong \triangle OCD$. Thus,$OA = OC$ and $OB = OD$ $(CPCT)$.
$3$. To prove diagonals intersect at right angles: Consider $\triangle OAB$ and $\triangle OCB$. Here,$OA = OC$ (proved above),$AB = CB$ (sides of a square),and $OB = OB$ (common side). By $SSS$ congruence,$\triangle OAB \cong \triangle OCB$. Thus,$\angle AOB = \angle COB$ $(CPCT)$. Since $\angle AOB + \angle COB = 180^{\circ}$ (linear pair),$2 \angle AOB = 180^{\circ}$,which implies $\angle AOB = 90^{\circ}$.

(N/A) $1$. In parallelogram $PQRS$,we have $\angle P = \angle R$ (opposite angles are equal) and $PQ \parallel RS$.
$2$. Let the bisector of $\angle P$ be $PM$ and the bisector of $\angle R$ be $RN$.
$3$. Since $PM$ bisects $\angle P$,$\angle SPM = \angle RPM = \frac{1}{2} \angle P$.
$4$. Since $RN$ bisects $\angle R$,$\angle QRN = \angle PRN = \frac{1}{2} \angle R$.
$5$. Since $\angle P = \angle R$,it follows that $\frac{1}{2} \angle P = \frac{1}{2} \angle R$,so $\angle RPM = \angle PRN$.
$6$. These are alternate interior angles for lines $PM$ and $RN$ with transversal $PR$. Therefore,$PM \parallel RN$.
$7$. Also,$PQ \parallel RS$ implies $PN \parallel RM$ (since $N$ lies on $PQ$ and $M$ lies on $RS$).
$8$. Since both pairs of opposite sides are parallel ($PM \parallel RN$ and $PN \parallel RM$),$PNRM$ is a parallelogram.

(N/A) Given: $PQRS$ is a parallelogram where diagonal $PR$ bisects $\angle P$,i.e.,$\angle SPR = \angle QPR$.
Step $1$: Since $PQRS$ is a parallelogram,$PS \parallel QR$ and $PQ \parallel SR$.
Step $2$: Because $PS \parallel QR$ and $PR$ is a transversal,$\angle SPR = \angle PRQ$ (alternate interior angles).
Step $3$: Because $PQ \parallel SR$ and $PR$ is a transversal,$\angle QPR = \angle PRS$ (alternate interior angles).
Step $4$: Since $\angle SPR = \angle QPR$ (given),it follows that $\angle PRQ = \angle PRS$. Thus,$PR$ bisects $\angle R$.
Step $5$: In $\triangle PQR$ and $\triangle PSR$,we have $\angle QPR = \angle SPR$ and $\angle PRQ = \angle PRS$. Since $PR = PR$ (common side),the triangles are congruent by $ASA$ congruence.
Step $6$: By $CPCT$,$PQ = PS$. Since adjacent sides of a parallelogram are equal,$PQRS$ is a rhombus.

(A) In a parallelogram,adjacent angles are supplementary,so $\angle A + \angle B = 180^{\circ}$.
Given that $\angle A = \angle B + 50^{\circ}$.
Substituting the value of $\angle A$ in the first equation: $(\angle B + 50^{\circ}) + \angle B = 180^{\circ}$.
$2 \angle B + 50^{\circ} = 180^{\circ}$.
$2 \angle B = 130^{\circ}$,which gives $\angle B = 65^{\circ}$.
Then,$\angle A = 65^{\circ} + 50^{\circ} = 115^{\circ}$.
Since opposite angles of a parallelogram are equal,$\angle C = \angle A = 115^{\circ}$ and $\angle D = \angle B = 65^{\circ}$.
Thus,the angles are $\angle A = 115^{\circ}, \angle B = 65^{\circ}, \angle C = 115^{\circ}, \angle D = 65^{\circ}$.

(A) In a trapezium $ABCD$ with $AB || CD$,the sum of consecutive interior angles between parallel lines is $180^{\circ}$.
Therefore,$\angle A + \angle D = 180^{\circ}$ and $\angle B + \angle C = 180^{\circ}$.
Substituting the given values:
$(y + 60^{\circ}) + (3y - 80^{\circ}) = 180^{\circ} \implies 4y - 20^{\circ} = 180^{\circ} \implies 4y = 200^{\circ} \implies y = 50^{\circ}$.
$(x + 60^{\circ}) + (3x - 40^{\circ}) = 180^{\circ} \implies 4x + 20^{\circ} = 180^{\circ} \implies 4x = 160^{\circ} \implies x = 40^{\circ}$.
Now,calculating the angles:
$\angle A = 50^{\circ} + 60^{\circ} = 110^{\circ}$.
$\angle B = 40^{\circ} + 60^{\circ} = 100^{\circ}$.
$\angle C = 3(40^{\circ}) - 40^{\circ} = 120^{\circ} - 40^{\circ} = 80^{\circ}$.
$\angle D = 3(50^{\circ}) - 80^{\circ} = 150^{\circ} - 80^{\circ} = 70^{\circ}$.

(N/A) $1$. In $\triangle ABP$ and $\triangle CDQ$:
$AB = CD$ (Opposite sides of parallelogram $ABCD$ are equal).
$\angle BAP = \angle DCQ$ (Alternate interior angles as $AB \parallel CD$ and $AC$ is a transversal).
$AP = CQ$ (Given).
By $SAS$ congruence criterion, $\triangle ABP \cong \triangle CDQ$.
Thus, $BP = DQ$ and $\angle ABP = \angle CDQ$.
$2$. In $\triangle ADP$ and $\triangle CBQ$:
$AD = CB$ (Opposite sides of parallelogram $ABCD$ are equal).
$\angle DAP = \angle BCQ$ (Alternate interior angles as $AD \parallel BC$ and $AC$ is a transversal).
$AP = CQ$ (Given).
By $SAS$ congruence criterion, $\triangle ADP \cong \triangle CBQ$.
Thus, $DP = BQ$ and $\angle ADP = \angle CBQ$.
$3$. Since $BP = DQ$ and $DP = BQ$, quadrilateral $PBQD$ is a parallelogram (opposite sides are equal).
Therefore, $BQ \parallel DP$.
$4$. In parallelogram $PBQD$, diagonals $BD$ and $PQ$ bisect each other. Hence, $BD$ bisects $PQ$.

(A) In quadrilateral $AMCN$,the sum of angles is $360^{\circ}$.
Given $AM \perp BC$ and $AN \perp CD$,we have $\angle AMC = 90^{\circ}$ and $\angle ANC = 90^{\circ}$.
In quadrilateral $AMCN$,$\angle MCN + \angle AMC + \angle ANC + \angle MAN = 360^{\circ}$.
$\angle C + 90^{\circ} + 90^{\circ} + 60^{\circ} = 360^{\circ}$.
$\angle C + 240^{\circ} = 360^{\circ}$,so $\angle C = 120^{\circ}$.
In a parallelogram,opposite angles are equal,so $\angle A = \angle C = 120^{\circ}$.
Adjacent angles are supplementary,so $\angle B = 180^{\circ} - 120^{\circ} = 60^{\circ}$ and $\angle D = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
Thus,the angles are $120^{\circ}, 60^{\circ}, 120^{\circ}, 60^{\circ}$.

(A) The statement is True.
By definition,a rhombus is a quadrilateral where all four sides are equal in length.
$A$ key property of a rhombus is that its diagonals are perpendicular bisectors of each other.
This means that the diagonals intersect at a $90^{\circ}$ angle and divide each other into two equal segments.

(D) In a rhombus $ABCD$,all sides are equal,so $AB = AD$.
Since $AB = AD$,$\triangle ABD$ is an isosceles triangle.
In a rhombus,adjacent angles are supplementary,so $\angle A + \angle B = 180^{\circ}$.
Given $\angle B = 80^{\circ}$,then $\angle A = 180^{\circ} - 80^{\circ} = 100^{\circ}$.
In $\triangle ABD$,the sum of angles is $180^{\circ}$,so $\angle A + \angle ADB + \angle ABD = 180^{\circ}$.
Since $AB = AD$,the base angles are equal: $\angle ADB = \angle ABD$.
Therefore,$100^{\circ} + 2(\angle ADB) = 180^{\circ}$.
$2(\angle ADB) = 80^{\circ}$.
$\angle ADB = 40^{\circ}$.

(C) In a parallelogram,opposite angles are equal.
Therefore,$\angle A = \angle C$.
Given that $\angle A = 5x - 40^{\circ}$ and $\angle C = 3x + 10^{\circ}$.
Equating the two expressions: $5x - 40 = 3x + 10$.
Subtract $3x$ from both sides: $2x - 40 = 10$.
Add $40$ to both sides: $2x = 50$.
Divide by $2$: $x = 25$.

(A) In a parallelogram $ABCD$,the sum of adjacent angles is $180^o$. Therefore,$\angle A + \angle B = 180^o$.
Given that $\angle A = 5 \angle B$,we substitute this into the equation:
$5 \angle B + \angle B = 180^o$
$6 \angle B = 180^o$
$\angle B = 30^o$.
Since $\angle A = 5 \angle B$,then $\angle A = 5 \times 30^o = 150^o$.
In a parallelogram,opposite angles are equal,so $\angle C = \angle A$.
Therefore,$\angle C = 150^o$.

(C) rhombus is a quadrilateral where all sides are equal and diagonals bisect each other at right angles.
By the properties of a rhombus,the diagonals $XZ$ and $YW$ are perpendicular to each other.
Since the diagonals intersect at point $P$,the angle formed at the intersection,$\angle XPY$,must be $90^o$.
Therefore,$\angle XPY = 90^o$.

(D) In a parallelogram,the diagonals bisect each other.
This means that the point of intersection $P$ is the midpoint of both diagonals $AC$ and $BD$.
Therefore,$BD = 2 \times PB$.
Given that $PB = 5.2 \, cm$,we have:
$BD = 2 \times 5.2 \, cm = 10.4 \, cm$.
Thus,the length of diagonal $BD$ is $10.4 \, cm$.

(A) In a rhombus $ABCD$,adjacent angles are supplementary,meaning $\angle A + \angle B = 180^{\circ}$.
Given that $\angle A = \angle B - 30^{\circ}$,we can substitute this into the equation:
$(\angle B - 30^{\circ}) + \angle B = 180^{\circ}$
$2\angle B = 210^{\circ}$
$\angle B = 105^{\circ}$.
Since $\angle A = \angle B - 30^{\circ}$,then $\angle A = 105^{\circ} - 30^{\circ} = 75^{\circ}$.
In a rhombus,opposite angles are equal,so $\angle C = \angle A$.
Therefore,$\angle C = 75^{\circ}$.

(C) In a parallelogram $PQRS$,the opposite sides are equal in length.
Therefore,$PQ = RS = 8 \, \text{cm}$ and $QR = PS = 5 \, \text{cm}$.
The perimeter of a parallelogram is the sum of all its sides.
Perimeter $= PQ + QR + RS + PS$
Perimeter $= 8 \, \text{cm} + 5 \, \text{cm} + 8 \, \text{cm} + 5 \, \text{cm} = 26 \, \text{cm}$.
Alternatively,the perimeter of a parallelogram is given by $2 \times (\text{sum of adjacent sides}) = 2 \times (PQ + QR) = 2 \times (8 + 5) = 2 \times 13 = 26 \, \text{cm}$.

(N/A) $(1)$ The diagonals of a rhombus are perpendicular to each other.
$(2)$ The diagonals of a parallelogram bisect each other.

(A) In a parallelogram $ABCD$,the adjacent angles are supplementary,meaning their sum is $180^{\circ}$.
Therefore,$\angle A + \angle B = 180^{\circ}$.
Given that $\angle A = \angle B - 30^{\circ}$,we can express $\angle B$ as $\angle B = \angle A + 30^{\circ}$.
Substituting this into the sum equation: $\angle A + (\angle A + 30^{\circ}) = 180^{\circ}$.
$2\angle A + 30^{\circ} = 180^{\circ}$.
$2\angle A = 180^{\circ} - 30^{\circ} = 150^{\circ}$.
$\angle A = 150^{\circ} / 2 = 75^{\circ}$.

(C) The perimeter of a parallelogram is given by the formula $P = 2(a + b)$,where $a$ and $b$ are the lengths of adjacent sides.
Given,$AB = 12.5 \, cm$ and $BC = 7 \, cm$.
Here,$a = 12.5 \, cm$ and $b = 7 \, cm$.
Perimeter $P = 2(12.5 + 7) \, cm$.
$P = 2(19.5) \, cm$.
$P = 39 \, cm$.
Therefore,the perimeter of parallelogram $ABCD$ is $39 \, cm$.

(A) In a trapezium $ABCD$,the sides $AB$ and $CD$ are parallel $(AB \parallel CD)$.
According to the properties of parallel lines,the consecutive interior angles (also known as co-interior angles) between parallel lines are supplementary.
Therefore,$\angle A + \angle D = 180^{\circ}$ and $\angle B + \angle C = 180^{\circ}$.
Given $\angle A = 80^{\circ}$,we can find $\angle D$ as follows:
$\angle A + \angle D = 180^{\circ}$
$80^{\circ} + \angle D = 180^{\circ}$
$\angle D = 180^{\circ} - 80^{\circ}$
$\angle D = 100^{\circ}$.

(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) The area of a parallelogram is given by the formula: $\text{Area} = \text{base} \times \text{corresponding altitude}$.
For base $PQ = 15 \, cm$ and altitude $SM = 6 \, cm$,the area is: $\text{Area} = 15 \times 6 = 90 \, cm^2$.
Since the area of the parallelogram is constant,we can also write: $\text{Area} = QR \times SN$.
Substituting the known values: $90 = QR \times 10$.
Therefore,$QR = \frac{90}{10} = 9 \, cm$.
The perimeter of a parallelogram is given by $2 \times (\text{sum of adjacent sides}) = 2 \times (PQ + QR)$.
Perimeter $= 2 \times (15 + 9) = 2 \times 24 = 48 \, cm$.

(A) The area of a parallelogram is given by the formula $\text{Area} = \text{base} \times \text{corresponding altitude}$.
For base $AB = 20 \, cm$ and corresponding altitude $DX = 12 \, cm$,the area is:
$\text{Area} = AB \times DX = 20 \, cm \times 12 \, cm = 240 \, cm^2$.
Since the area is constant,for base $BC$ and corresponding altitude $AY = 15 \, cm$,we have:
$\text{Area} = BC \times AY = 240 \, cm^2$.
$BC \times 15 \, cm = 240 \, cm^2 \implies BC = \frac{240}{15} \, cm = 16 \, cm$.
The perimeter of a parallelogram is $2 \times (\text{sum of adjacent sides})$.
$\text{Perimeter} = 2 \times (AB + BC) = 2 \times (20 \, cm + 16 \, cm) = 2 \times 36 \, cm = 72 \, cm$.

(N/A) The area of a parallelogram is given by the formula: $\text{Area} = \text{base} \times \text{corresponding altitude}$.
For base $XY = 24 \, cm$ and altitude $WP = 6 \, cm$,the area is: $\text{Area} = 24 \times 6 = 144 \, cm^2$.
Since the area of the parallelogram is constant,we can also write: $\text{Area} = YZ \times WQ$.
Given $WQ = 8 \, cm$,we have: $144 = YZ \times 8$.
Therefore,$YZ = 144 / 8 = 18 \, cm$.
The perimeter of a parallelogram is given by $2 \times (\text{sum of adjacent sides})$.
Perimeter $= 2 \times (XY + YZ) = 2 \times (24 + 18) = 2 \times 42 = 84 \, cm$.

(N/A) Let $ABCD$ be a parallelogram where $E$ and $F$ are the midpoints of sides $AB$ and $CD$ respectively.
Since $AB \parallel CD$ and $AB = CD$,it follows that $AE = EB = \frac{1}{2} AB$ and $CF = FD = \frac{1}{2} CD$.
Thus,$AE = FC$ and $AE \parallel FC$,which implies that $AEFC$ is a parallelogram.
Similarly,$EBFD$ is a parallelogram because $EB = FD$ and $EB \parallel FD$.
Since $ABCD$ is a parallelogram,the height $h$ between sides $AB$ and $CD$ is constant.
The area of parallelogram $AEFC = AE \times h$ and the area of parallelogram $EBFD = EB \times h$.
Since $AE = EB$,it follows that $\text{Area}(AEFC) = \text{Area}(EBFD)$.
Thus,the line segment $EF$ divides the parallelogram $ABCD$ into two parallelograms of equal area.

(B) The area of a parallelogram is given by the formula: $\text{Area} = \text{base} \times \text{height}$.
Here,the base is $AB = 15 \text{ cm}$ and the corresponding altitude is $DM$.
Given,$\text{ar}(ABCD) = 360 \text{ cm}^2$.
Substituting the values into the formula: $360 = 15 \times DM$.
To find $DM$,divide $360$ by $15$: $DM = \frac{360}{15} = 24 \text{ cm}$.
Therefore,the length of the altitude $DM$ is $24 \text{ cm}$.

(A) diagonal of a parallelogram divides it into two triangles of equal area.
In parallelogram $ABCD$,the diagonal $AC$ divides the parallelogram into two triangles,$\triangle ABC$ and $\triangle ADC$.
Therefore,$\operatorname{ar}(ABC) = \operatorname{ar}(ADC)$.
The area of the parallelogram is the sum of the areas of these two triangles:
$\operatorname{ar}(ABCD) = \operatorname{ar}(ABC) + \operatorname{ar}(ADC)$.
Since $\operatorname{ar}(ABC) = 42 \, \text{cm}^2$,then $\operatorname{ar}(ADC) = 42 \, \text{cm}^2$.
Thus,$\operatorname{ar}(ABCD) = 42 \, \text{cm}^2 + 42 \, \text{cm}^2 = 84 \, \text{cm}^2$.