MATHS M.C.Q

If diagonals of $PQRS $are perpendicular.
Hence, $(c) $ is the correct answer.

We know that the opposite angles of a parallelogram are equal.
Therefore, $\angle\text{BCD} = \angle\text{BAD} = 75^\circ...\ \text{(i)}$
$(i)$ Now, in $\triangle\text{BCD},$
We have,
$\angle\text{CDB} + \angle\text{DBC} + \angle\text{BCD} = 180^\circ$ [Since, sum of the angles of a triangle is $180^\circ ]$
$\Rightarrow \angle\text{CDB} + 60^\circ + 75^\circ = 180^\circ$
$\Rightarrow \angle\text{CDB} + 135^\circ = 180^\circ$
$\Rightarrow \angle\text{CDB} = (180^\circ - 135^\circ) = 45^\circ$

Let $ABCD$ be a rhombus and $P,Q,R$ and $S$ be the mid-points of sides $AB, BC, CD$ and $DA$ respectively.
In $\triangle {ABD} $ and $\triangle {BDC} $
we have,
$ {SP || BD}$ and $ {SP}=\frac{1}{2} {BD}\ ...\ {(1)}$ [By mid-point and theorem]
$ {RQ || BD}$ and $ {RQ}=\frac{1}{2} {BD}\ ...\ {(2)}$ [By mid-point and theorem]
rom $(1)$ and $(2)$ we get,
$ {SP || RQ}$
$PQRS$ is a parallelogram
As diagonals of a rhombus bisect each other at right angles.
$\therefore {AC\perp BD}$
Since, $ {SP || BD, PQ || AC}$ and $ {AC\perp BD}$
$\therefore {SP\perp PQ}$
$\therefore \angle {QPS} = 90^\circ$
$\therefore {PQRS}$ is a rectangle.

Let $ABCD$ be a quadrilateral with $\angle\text{A} = 3\text{x}, \ \angle\text{B} = 5\text{x},\ \angle\text{C} = 9\text{x}$ and $\angle\text{D} = 13\text{x}$
$\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$ (angle sum property)
$3\text{x} + 5\text{x} + 9\text{x} + 13\text{x} = 360^\circ$
$30\text{x} = 360^\circ$
$\text{x} = 12^\circ$
$\angle\text{A} = 3 (12^\circ) = 36^\circ$
$\angle\text{B} = 5 (12^\circ) = 60^\circ$
$\angle\text{B} = 9 (12^\circ) = 1800^\circ$
$\angle\text{B} = 13 (12^\circ) = 156^\circ$

As $\angle\text{A} = 80^\circ$ opposite angles of a parallelogram are equal and $\angle\text{DGF} = 80^\circ$ as $GF$ is parallel to $AB$, corresponding angles are equal.

$A R, B R, C P, D P$ are the bisectors of angles of parallelogram.
Because two bisectors of adjacent angles make $90^{\circ}$ between them So $PQRS$ is a Rectangle
Because $DP$ and $BR$ are acute angle bisectors so the distance between them $PQ < PS$ (The distance between other two bisectors) So $PQ \neq PS$ (So $PQRS$ is not a square, but only a rectangle)

Given: A parallelogram $ABCD$ such that the bisectors of adjacent angles $A$ and $B$ intersect at $P.$
To prove: $\angle\text{APB} = 90^\circ$
Proof: Since $ABCD$ is a ||gm
$\therefore \ \text{AD || BC}$
$\Rightarrow \angle\text{A} + \angle\text{B} = 180^\circ$ [sum of consecutive interior angle]
$\Rightarrow \frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}=90^\circ$
$\Rightarrow\angle1+\angle2=90^\circ\ ...\ \text{(i)}$
[$\because$ AP is the bisector of $\angle\text{A}$ and BP is the bisector of $\angle\text{B}$]
$\angle1=\frac{1}{2}\angle\text{A}$ and $\angle2=\frac{1}{2}\angle\text{B}$
Now, $\triangle\text{APB},$ we have
$\angle1 + \angle\text{APB} + \angle2 = 180^\circ$ [sum of three angles of a $\triangle$]
$\Rightarrow 90^\circ + \angle\text{APB} + \angle2 = 180^\circ$ [$\because \angle1 + \angle2 = 90^\circ$ from (i)]
Hence, $\angle\text{APB} = 90^\circ$

Fourth angle of the quadrilateral
$= 360^\circ - (75^\circ + 90^\circ + 75^\circ )$
$= 360^\circ - 240^\circ $
$= 120^\circ $
Hence, $(d)$ is the corrent answer.

$E$ and $F$ are the given to be the mid-points of $AD$ and $BC$ respectively.
$\therefore\text{EF} = \frac{1}{2}(\text{AB + DC})$
$=\frac{1}{2}(\text{a + b})$

If in any triangle, all the mid-points (of each sides) are joined to form a triangle, then that triangle is similian to a parent triangle.
i.e.$\triangle\text{QPR}\sim\triangle\text{ABC}$
So angles of $\triangle\text{PQR}$ will be same as angles of $\triangle\text{ABC}.$
Thus, angles are $30^\circ , 40^\circ , 110^\circ $

since $E$ and $F$ are the mid points of sides $AB$ and $AC$ respectively.
according to mid-point theorem of triangle;
$\text{EF} = \frac{1}{2}×\text{BC}$
$\text{EF} = \frac{1}{2}×5$

​​​​​Let $ABCD$ be a quadrilateral in which $P, Q, R$ and $S$ are the mid-points of $AB, BC, CD$ and $DA$ respectively.
Join $AC$
In $\triangle\text{ABC},$ the points $P$ and $Q$ are the mid-points of sides $AB$ and $BC$ respectively.
$\therefore\ \text{PQ || AC}$ and $\text{PQ}=\frac{1}{2}\text{AC}\ ...\ \text{(i)}$ [By mid-point theorem]
Again, in $\triangle\text{DAC},$ the points S and R are the mid-points of $AD$ and $DC$.
$\therefore\ \text{SR || AC}$ and $\text{SR}=\frac{1}{2}\text{AC}\ ...\ \text{(ii)}$

Given,
$ABCD$ is a parallelogram
Diagonal AC bisects $\angle\text{BAD}$

$\angle\text{BAC} = 35^\circ$
$∵ \angle\text{A} + \angle\text{B} = 180^\circ ...\text{(i)}$ [angle sum property of quadrilateral]
$\angle\text{A} = 2\angle\text{BAC} = 2 × 35^\circ = 70^\circ$
Putting value of $\angle\text{A}$ in equation (i)
$70^\circ + \angle\text{B} = 180^\circ$
$\angle\text{B} = 180^\circ - 70^\circ = 110^\circ$
$\angle\text{ABC} = 110^\circ$

$ABCO$ is a rectangle The diagonals of a rectangle are congruent and bisect each other. Therefore, in
$\triangle\text{AOB},$ we have:
$OA = OB$
$\angle\text{OAB} = \angle\text{OBA} = 35^\circ$
$\text{x} = 90^\circ - 35^\circ = 55^\circ$ and $\angle\text{AOB} = 180^\circ - (35^\circ + 35^\circ) = 110^\circ$
$\text{y} = \angle\text{AOB} = 110^\circ$ [Vertically opposite angles]
Hence, $x = 55^\circ $ and $y = 110^\circ $

We can look at a quadrilateral as:



The opposite sides of the above quadrilateral $AB$ and $CD$ have no point in common.

Centroid is the point where all medians of a meet.
In $\triangle\text{ABD}, E$ is the centroid,
And in $\triangle\text{BCD}, F$ is the centroid.
By the property of centroid, centroid divides a median in $2 : 1$
So from figure,
$\frac{\text{AE}}{\text{EO}}=\frac{2}{1}\Rightarrow\text{EO}=\frac{\text{AE}}{2}\ ...(1)$
Also $\frac{\text{CF}}{\text{FO}}=\frac{2}{1}\Rightarrow\text{FO}=\frac{\text{CF}}{2}\ ...(2)$
Because $AC$ is a digonal of a parallelogram, $O$ is its midpoint.
$\Rightarrow OA = OC$
$\Rightarrow AE = CF$
Adding equations $(1) \& (2),$
$\text{EO + FO} =\frac{\text{AE}+\text{CF}}{2}=\frac{\not2\text{AE}}{\not2}$
$\Rightarrow\text{EF}=\frac{\not2\text{AE}}{\not2}$
$\Rightarrow\text{EF}=\text{AE}$

$\angle\text{DAC} = \angle\text{ACB} = 30$ (alternate angles)
$\angle\text{BOA} = \angle\text{BOC} = 180$ (linear pair)
$\angle\text{BOC}= 180 - 70 = 110$
In $\triangle\text{BOC}, \ \angle\text{BOC} + \angle\text{OCB} + \angle\text{CBO} = 180$ (angle sum property)
$110 + 30 + \angle\text{CBO} = 180$
$\angle\text{CBO} = 180 - 140 =40$
$\Rightarrow \angle\text{DBC}=40^\circ$

We need $\text{DE}=\text{EF}.$

Hence, $(c)$ is the correct answer.

In $\triangle\text{ABC}, P$ and $Q$ are the mid-points of sides $AB$ and $BC$ respectively.
$\therefore\text{PQ || AC}$ and $\text{PQ}=\frac{1}{2}\text{AC ...(i)}$
In $\triangle\text{BCD},$ Q and R are the mid-points of sides $BC$ and $CD$ respectively.
$\therefore\text{QR || BD}$ and $\text{QR}=\frac{1}{2}\text{BD ...(ii)}$
In $\triangle\text{ADC},$ S and R are the mid-points of sides $AD$ and $CD$ respectively.
$\therefore\text{RS || AC}$ and $\text{RS}=\frac{1}{2}\text{AC ...(iii)}$
In $\triangle\text{ABD},$ P and S are the mid-points of sides $AB$ and $AD$ respectively.
$\therefore\text{SP || BD}$ and $\text{SP}=\frac{1}{2}\text{BD ... (iv)}$
$\Rightarrow\text{PQ ∥ RS}$ and $\text{QR ∥ SP}$ [From $(i), (ii), (iii)$ and $(iv)]$
Thus, $PQRS$ is a parallelogram.
Now, $AC = BD$ (given)
$\Rightarrow\frac{1}{2}\text{AC}=\frac{1}{2}\text{BD}$
$\Rightarrow\text{PQ = QR = RS = SP }$ [From $(i), (ii), (iii)$ and $(iv)]$
Hence, $PQRS$ is a rhombus if diagonals of $ABCD$ are equal.

Here, length of rectangle $A B C D=8 cm$ and breadth of rectangle $A B C D=6 cm$.
Let $E , F , G$ and H are the mid-points of the sides of rectangle $ABCD$ , then $EFGH $is a rhombus.

Here, length of rectangle $A B C D=8 cm$
and breadth of rectangle $A B C D=6 cm$
Let $E, F, G$ and $H$ are the mid-points of the sides of rectangle $A B C D$, then $E F G H$ is a rhombus. Then, diagonal of rhombus EFGH are EG and HF.
Here, $EG = BC =8 cm$
And $HF = AB =6 cm$
$\therefore\ \text{Area of rhombus}=\frac{\text{Product of diagonals}}{2}$
$=\frac{8\times6}{2}=4\times6=24\text{cm}^2$
Hence, joining the mid-points of the adjacent sides of a rectangle forms a rhombus of area $24cm^2$.

If $DE= EF$ then $\triangle\text{AED}$ become congruent to $\triangle\text{CEF}$ by $SSS$ congruence rule.
By $CPCT,$ $\angle\text{ECF} = \angle\text{EAD}$ which forms a pair of an alternate angles.
Which proves that $AD$ is parallel to $CF$.

$ABCD$ is a Quadrilateral.
The opposite sides $AB$ and $DC, AD$ and $BC$ have no common point.

Construction: Join $DG$ and $G$ be the mid-point of $FC$.
Now,
In $\triangle\text{BCF, D}$ is the mid-point of $BC$ and $G$ is the mid-point of $FC$ and $F$ is the mid-point of $AG.$
$\Rightarrow\text{DG || BF}$
$\Rightarrow\text{DG || EF} ...(B - E - F)$
$\Rightarrow\text{AF}=\text{FG}=\text{GC}$ ...(Since $G$ is the mid-point.)
$\Rightarrow\text{AF}=\frac{1}{2}\text{AC}$

In Rectangle, diagonals are equal and bisect each other.
In $\triangle\text{APD, AP = PD}$
$\Rightarrow \angle\text{ADP} = \angle\text{PAD} = \text{x}$ (angle opposite to equal sides are equal)
In $\triangle\text{APD}, \angle\text{APD} + \angle\text{PDA} + \angle\text{DAP} = 180^\circ$ (angle sum property)
$52^\circ + \text{x} + \text{x} = 180^\circ$
$2\text{x} = 180^\circ - 52^\circ = 128^\circ$
$\text{x} = 64^\circ$
$\angle\text{DAC} = \angle\text{BCA} = 64^\circ$ (alternate angles)
In $\triangle\text{ADB}, \angle\text{ADB} + \angle\text{DBA} + \angle\text{BAD} = 180^\circ$ (angle sum property)
$64^\circ + \angle\text{DBA} + 90^\circ = 180^\circ$
$\angle\text{DBA} = 180^\circ - 154^\circ = 26^\circ$

We know that, the opposite angles of a parallelogram are equal.
$\therefore\angle\text{C}=\angle\text{A}=75^{\circ}$
In $\triangle\text{BCD},$
$\angle\text{BCD}+\angle\text{BDC}+\angle\text{CBD}=180^{\circ}$ ...(Angle sum property)
$\Rightarrow75^{\circ}+\angle\text{BDC}+60^{\circ}=180^{\circ}$
$\Rightarrow\angle\text{BDC}=45^{\circ}$

Consider $\triangle\text{AOD}\ \&\ \triangle\text{COB},$
$AO = CO$ {Diagonals bisects each other}
$OD = OB$ {Diagonals bisects each other}
$\angle\text{AOD}=\angle\text{COB}$ (Opposite angles)
So by SAS property, $\triangle\text{AOD}\cong\triangle\text{COB},$
$\Rightarrow\angle\text{ADO}=\angle\text{CBO}\dots(1)$
$\angle\text{ABD}=180^\circ-\angle\text{A}-\angle\text{ADO}$ $($in $\triangle\text{ADB})$
$=180^\circ-45^\circ-\angle\text{ADO}$
$\angle\text{ABD}=135^\circ-\angle\text{ADO}\dots(2)$
$\angle\text{B}=\angle\text{ABD}+\angle\text{CBO}$
Putting values From eq $(1)$ and $(2)$
$\angle\text{B}=135^\circ-\angle\text{ADO}+\angle\text{ADO}$
$\angle\text{B}=135^\circ$

 One Diagonal divides it into two triangles and the sum of angles of one triangle is $180^\circ $ and $180 \times 2 = 360^\circ .$

$ ABCD$ is a parallelogram. $BD$ is the diagonal and M is the mid point of $BD.\ BD$ is a bisector of $\angle\text{B}.$
We know that, diagonals of the parallelogram bisect each other.
$\therefore $ M is the mid point of $AC.$
$\text{AB || CD}$ and $BD$ is the transversal,
$\therefore \angle\text{ABD} = \angle\text{BDC}\ ...(1)$ (Alternate interior angles)
$\angle\text{ABD} = \angle\text{DBC}\ ...(2)$ (Given)
From $(1)$ and $(2),$ we get
$\angle\text{BDC} = \angle\text{DBC}$
In $\triangle\text{BCD},$
$\angle\text{BDC} = \angle\text{DBC}$
$\Rightarrow BC = CD ...(3) ($In a triangle, equal angles have equal sides opposite to them$)$
$AB = CD$ and $BC = AD ...(4) ($Opposite sides of the parallelogram are equal$)$
From $(3)$ and $(4),$ we get
$AB = BC = CD = DA$
$\therefore ABCD$ is a rhombus.
$\Rightarrow \angle\text{AMB} = 90^\circ ($Diagonals of rhombus are perpendicular to each other$)$

Parallelograms on the same base and between the same parallels are equal in area. Hence the ration of two parallelograms will be $1 : 1.$

Let's assume our quadrilateral $ABCD$ as a parallelogram:

We know
$\angle\text{DCB} + \angle\text{ABC} = 180^\circ($ Co-interior angles of parallelogram are supplementary$)$
$\Rightarrow\frac{1}{2}\angle\text{DCB} + \angle\text{ABC} = 90^\circ ($Both sides divide by $2)$
$\Rightarrow \angle1+ \angle2 =90^\circ ... (1)$
In, $\triangle\text{CQB}$ we know
$\Rightarrow\angle1+ \angle2 +\angle\text{CQB} = 180^\circ ... (2)$
From eq. $(1)$ and eq. $(2),$ We get
$\Rightarrow\angle\text{CQB} = 180^\circ - 90^\circ$
$\Rightarrow\angle\text{CQB}= 90^\circ$
$\Rightarrow\angle\text{PQR}= 90^\circ$ (because $\angle\text{CQB} = \angle\text{PQR},$ vertically opposite angles)
Similarly, it can be shown
$\angle\text{QPS} = \angle\text{PSR} =\angle\text{SRQ} = 90^\circ$
So, quadrilateral $PQRS$ is a rectangle.

Given,
In $\angle\text{ABC}$

$\angle\text{A} = 30^\circ,\ \angle\text{B} = 40^\circ,\ \angle\text{C} = 110^\circ$
In figure, $BDEF, DCEF, DEAF$ are parallelograms so,
$\angle\text{B} = \angle\text{E} = 40^\circ [∵ \angle\text{B}$ & $\angle\text{E}$ are opposite angles of parallelogram $BDEF]$
$\angle\text{C} = \angle\text{F} = 110^\circ [∵ \angle\text{C}$ & $\angle\text{F}$ are opposite angles of parallelogram $DCEF]$
$\angle\text{A} = \angle\text{D} = 30^\circ [∵ \angle\text{A}$ & $\angle\text{D}$ are opposite angles of parallelogram $DEAF]$
Hence, $\angle\text{D} = 30^\circ$
$\angle\text{E} = 40^\circ$
$\angle\text{F} = 110^\circ$

 Given, $ABCD$ is a parallelogram.
So,
$\angle\text{A} =\angle\text{C} ($Opposite angles of parallelogram are equal in size$)$
$\Rightarrow 3x - 20 = x + 40$
$\Rightarrow 3x - x = 40 + 20$
$\Rightarrow 2x = 60$
$\Rightarrow x = 30^\circ $
Thus, $\angle\text{A} = 3 \times 30 - 20 = 90 - 20 = 70^\circ$
Now, $\angle\text{A}+\angle\text{B}=180^\circ ($Sum of interior angles of parallelogram is $180^\circ )$
$\Rightarrow 70^\circ + \angle\text{B} = 180^\circ$
$\Rightarrow \angle\text{B}= 180^\circ - 70^\circ$
$\Rightarrow \angle\text{B}= 110^\circ$
$\Rightarrow y + 15 = 110^\circ $
$\Rightarrow y = 95^\circ $
Hence, $x = 30^\circ $ and $y = 95^\circ $

 $\angle\text{OAC} = \angle\text{ACB} = 32^\circ$ (alternate angles)
$\angle\text{AOB} = \angle\text{COB} = 180^\circ$ (linear pair)
$\angle\text{COB} = 180 - 70 = 110^\circ$
In $\triangle\text{BOC}, \ \angle\text{BOC} + \angle\text{OCB} + \angle\text{CBO} = 180^\circ$ (angle sum property)
$110 + 32 + \angle\text{CBO} = 180^\circ$
$\angle\text{CBO} = 180 - 142 = 38^\circ$

$\angle\text{ABM}=\angle\text{CBM}\ ...(1) (BM$ bisects $\angle\text{B})$
$\angle\text{ABM}=\angle\text{MDC}\ ...(2)($Alternate angles$)$
$\angle\text{CBM}=\angle\text{ADM}\ ...(3)($Alternate angles$)$
From equations $(1), (2)$ & $(3)$
$\angle\text{MDC}=\angle\text{ADM}...(4)$
Now, consider $\triangle\text{ABM}$ & $\triangle\text{CBD}$
$\angle\text{CBD}=\angle\text{ABD} \{$from eq $(1)\}$
$DB = DB ($Common$)$
$\angle\text{ADB}=\angle\text{CDB} \{$from eq $(4)\}$
Hence, by ASA property,
$\triangle\text{ADB}\cong\triangle\text{CBD}$
$⇒ AB = CB, AD = CD$
Hence, it becomes a Rhombus.
So now diagonals of a Rhombus bisect each other at $90^\circ .$
$\Rightarrow\angle\text{AMB}=90^\circ$