MATHS M.C.Q

Join $PD$ and Produce it to meet $BA$ at $G.$

ln $\triangle\text{PCD}$ and $\triangle\text{APG},$
$\angle\text{DPC}= \angle\text{GPA},$
$\angle\text{PDC}= \angle\text{AGP},$
$\therefore\triangle\text{PDC}\cong\triangle\text{AGP}$
$CD = AG$ and $PD = PG$
ln $\triangle\text{BGD},$
$P$ is the mid-point of $GD$
$Q$ is the mid-point of $BD$
Therefore, By mid-point theorem, $\text{PQ || AB}$
and $\text{PQ} = \frac{1}{2}\text{(GB)}$
but $GB = GA + AB = CD + AB$
$\therefore\text{PQ}=\frac{1}{2}(\text{AB + CD})$

In triangle $ABC, P, Q$ and $R$ are the midpoints.
By midpoint theorem, $PQ$ is parallel to $BC$ and $\text{PQ} = \frac{1}{2}$ of $BC$
$QR$ is parallel to $AB$ and $\text{QR} = \frac{1}{2}$ of $AB$
$PR$ is parallel to $AC$ and $\text{PR} = \frac{1}{2}$ of $AC.$
So, perimeter of triangle $\text{PQR} = \text{PQ} + \text{QR} + \text{PR} = \frac{1}{2}$ of $\text{(AB + BC + AC)}$
$= \frac{1}{2}$ of $(10 + 8 + 12) =\frac{1}{2} $ of $30=15.$

By mid-point theorem of a triangle $E$ is the midpoint of $AC,$
Hence $AE = EC$

Let $ABCD$ be any quadrilateral.
Join $A$ and $C$.
Let $P$ and $Q$ be midpoints of sides $AB$ and $BC$
In $\triangle\text{ABC},$
$\text{PQ || AC}$ and $\text{PQ}=\frac{1}{2}\text{AC}\ ...\ \text{(i)}$ [By mid-point theorem]
S and R be the mid-points of AD and DC respectively
In $\triangle\text{ACD},$
$\text{SR || AC}$ and $\text{SR}=\frac{1}{2}\text{AC}\ ...\ \text{(ii)}$ [By mid-point theorem]
From $(i)$ and $(ii),$
$\text{PQ || SR}$ and $PQ = SR.$
$∵$ One pair of opposite sides are parallel and equal
$∴ ABCD$ is a parallelogram.

Given: $ABCD$ is a trapezium with $\text{AB || CD}$
Construction: Draw $DE$ and $CF \perp $ to $AB$​​​​​​​

Then in $\triangle\text{ABC}$
$\angle\text{BAC}$ is acute
$\therefore \mathrm{BC}^2=\mathrm{AC}^2+\mathrm{AB}^2-2 \mathrm{AF}: \mathrm{AB} \ldots(1)$
and In $\triangle\text{BDA}$
$\angle\text{DBA}$ is acute
$\therefore \mathrm{AD}^2=\mathrm{BD}^2+\mathrm{AB}^2-2 \mathrm{BE} \cdot \mathrm{AB} \ldots(2)$
Adding $(1)$ and $(2)$ we get
$B C^2+A D^2=A C^2+B D^2+2 A B^2-2 A F \cdot A B-2 B E \cdot A B$
$\Rightarrow A C^2+B D^2=B C^2+A D^2-2 A B[A B-A F-B E]$
$=B C^2+A D^2-2 A B[A B-(A E+E F)-(B F+E F)]$
$=B C^2+A D^2-2 A B[A B-(A E+E F+B F+E F)]$
$=B C^2+A D^2-2 A B[A B-(A B+C D)](\$ \therefore S E F=D C)$
$=B C^2+A D^2-2 A B[(-C D)]$
$=A D^2+B C^2+2 A B \times C D$

In parallelogram $ABCD,$
$\angle\text{A}+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{D}=180^\circ-75^\circ=105^\circ$
$\angle\text{ADB}=\angle\text{DBC}$ (Alternate angles)
$\Rightarrow\angle\text{ADB}=60^\circ$
$\angle\text{BDC}=\angle\text{ADC}-\angle\text{ADB}=105^\circ-60^\circ=45^\circ$

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 $180^\circ )$
$\Rightarrow 70^\circ + \angle\text{B} = 180^\circ$
$\Rightarrow \angle\text{B} = 180^\circ − 70^\circ$
$\Rightarrow \angle\text{B} = 110^\circ$
$\Rightarrow \text{y} + 15 = 110^\circ$
$\Rightarrow \text{y} = 95^\circ$
Hence, $x = 30^\circ $ and $y = 95^\circ $

Since $ABCD$ is a parallelogram, $AB \| CD$ since opposite angles of a parallelogram are equal.
$\Rightarrow\angle\text{ABD}=\angle\text{BCD}=45^{\circ}$ ...(Alternate angles)
In $\triangle\text{ADB},$
$\angle\text{ABD}+\angle\text{BDA}+\angle\text{DAB}=180^{\circ}$ ...(Angle sum property)
$\Rightarrow45+\angle\text{BDA}+75=180$
$\Rightarrow\angle\text{BDA}+120=180$
$\Rightarrow\angle\text{BDA}=60^{\circ}$
$\Rightarrow\angle\text{CBD}=\angle\text{BDA}=60^{\circ}$ ...(Alternate angles)
$\Rightarrow\angle\text{CBD}=60^{\circ}$

If three or more points lie on same line then they are called collinear points. $A, B$ and $C$ are collinear means they form a line. Point $D$ is outside the line. Point $D$ can be join with the end points of the line formed by points $A, B$, and $C$. For this we need total three lines to form a closed figure. A closed figure with three sides is called triangle.

$P, Q, R \& S$ are the mid-points of $AB, BC, CD \& AD$ respectively.
Consider $\triangle\text{ADB},$
If in a triangle, the mid-points of two sides are joint by a line then the line is parallel to the third side.
$\Rightarrow\text{PS}||\text{DB}$ in $\triangle\text{ADB}$
Similarly in $\triangle\text{CDB},$
$RQ \| DB$
Hence $PS \| RQ ...(1)$
Similarly in $\triangle\text{ABC}$ and $\triangle\text{ADC}$
$S R\|A C, P Q\| A C$
$\Rightarrow S R \| P Q \ldots(2)$
From eq. $(1)$ and $(2), PQRS$ is a parallelogram.

Consider $\triangle\text{AOD} \ \&\ \triangle\text{COB}$ $$
$\angle\text{AOD}=\angle\text{COB}=90^\circ$
$AD = BC$ (Sides of Rhombus)
$AO = CO$ (Diagonals bisects each other)
So by RHS property, $\triangle\text{AOD}\cong\triangle\text{COB}$
$\Rightarrow\angle\text{OAD}=\angle\text{OCB}=40^\circ$
$\angle\text{ADB}=\angle\text{ADO}=180^\circ-90^\circ-40^\circ=50^\circ$

From the given choices, only in a square the diagonals bisect the opposite angles.
Let us prove it.
Take the following square $ABCD$ with diagonal $AD.$

In $\triangle\text{ABD}$ and $\triangle\text{CBD},$
$AD = BC$ (Opposite sides of a square are equal.)
$BD = BD$ (Common)
$AB = DC$ (Opposite sides of a square are equal.)
Thus.
$\triangle\text{ABD≅ΔCBD}$ (By $SSS$ Congruence Rule)
By Corresponging parts of congruent triangle property,
we have:
$\angle\text{ABD} =\angle\text{CBD}$
$\angle\text{ADB} =\angle\text{CDB}$
Therefore, in a square the diagonals bisect the opposite angles.

In $\triangle\text{PCD}$ and $\triangle\text{APG},$
$\angle\text{DPC} = \angle\text{GPA},$
$\angle\text{PDC} = \angle\text{AGP}$
$∴\triangle\text{PCD}≅\triangle\text{APG}$
$CD = AG$ and $PD = PG$
In $\triangle\text{BGD},$
$P$ is the mid-point of $GD$
$Q$ is the mid-point of $BD$
Therefore, By mid-point theorem, $\text{PQ || AB}$
and $\text{PQ} = \frac{1}{2}\text{(GB)}$
but $GB = GA + AB = CD + AB$
$∴\text{PQ}=\big(\frac{1}{2}\big)\text{(AB + CD)}$

Let $ABCD$ be a parallelogram and $AP$ and $CQ$ are the altitudes drawn from vertex $A$ on sides $DC$ and $BC.$
In quadrilateral $APCQ$, sum of the all angles $= 360^\circ $
So, $60^\circ + 90^\circ + \angle\text{C} + 90^\circ = 360^\circ$
$\angle\text{C} = 360^\circ - 240^\circ = 120^\circ$
$\angle\text{C} + \angle\text{B} = 180^\circ$ (co-interior angles)
$\angle\text{B} = 180^\circ - 120^\circ = 60^\circ$
In parallelogram, opposite angles are equal.
So, $\angle\text{A} = \angle\text{C} = 120^\circ$ and $\angle\text{B} = \angle\text{D} = 60^\circ$

$\text{PR}||\text{AD}\Rightarrow\text{AB}\not\bot\text{AD}$
$\text{QS}||\text{AB}\Rightarrow\text{PR}\not\bot\text{QS}$
Since diagonals of $PQRS$ are not making $90^\circ $ between them,
$PQRS$ is not a Rhombus.

$P, Q, R$ and $S$ are the mid-points,
$P R$ and $Q S$ are diagonals of quadrilateral $P Q R S$.
$P R\|A D, Q S\| A B$
Because they are Formed by joning of mid-points of sides of Rhombus $ABCD.$
$AD$ is not $\perp$ to $AB$
$\Rightarrow P R$ will not be $\perp$ to $Q S$
i.e angle between diagonals $PR \& QS$ is not $90^{\circ}$.
So, $PQRS$ is not a Rhombus.

$PR$ and $QS$ are making $90^{\circ}$ with each - other.
Because $P R\|A D, Q S\| A B$ and $A D \perp A B$
So $P R$ and $Q S$ are diagonals of $P Q R S$ and are $\perp$ to each other.
​​​​​​​Hence, $PQRS$ is a Rhombus.​​​​​​​

By joining the mid-points of sides of a triangle, no quadrilateral is formed.

$\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$ (angle sum property of quadrilateral).
$110 + 75^\circ + 35^\circ + \angle\text{D} = 360^\circ$
$\angle\text{D} = 360^\circ - 220^\circ = 140^\circ$

Let $ABCD$ be a quadrilateral with $\angle\text{A} = 3\text{x},\ \angle\text{B} = 4\text{x} ,\ \angle\text{C} = 4\text{x}$ and $\angle\text{D} = 7\text{x}$
$\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$ (angle sum property)
$3\text{x} + 4\text{x} + 4\text{x} + 7\text{x} = 360^\circ$
$18\text{x} = 360^\circ$
$\text{x} = 20^\circ$
$\angle\text{A} = 3 (20^\circ) = 60^\circ$
$\angle\text{B} = \angle\text{C} = 4 (20^\circ) = 80^\circ$
$\angle\text{D} = 7 (20^\circ) = 140^\circ$

$AB$ is parallel to $DC.$
$\angle\text{A} + \angle\text{D} = 180^\circ$ (co-interior angle)
$\angle\text{D} = 180^\circ - 45^\circ = 135^\circ$
Similarly by following same argument, $\angle\text{C} = 135^\circ.$

Given,
$ABCD$ is a rhombus

$AC = 24\ cm, BD = 18\ cm$
$AB = BC = CD = DA$ [side of rhombus]
We know that diagonals of rhombus bisect each other at $90^{\circ}$.
In right $\triangle\text{AOB}$
$AB^2 = BO^2 + AO^2$
$AB^2 = 122 + 92 = 144 + 81 = 225$
$\text{AB} = \sqrt{225}$
$= 15cm$
Side of rhombus $= 15cm$

Construction: join $CF$ and extend it to meet $AB$ at $G.$
In $\triangle\text{CDF}$ and $\triangle\text{GFB},$
$\angle\text{CDF}=\angle\text{GFB}$ ...(Vertically opposite angles)
$\text{AB || CD},$
So, $\angle\text{DCF}=\angle\text{GFB,}$ ...(alternate angles)
$\text{DF = FB}$ $...(F$ is the mid-point of $BD)$
$\Rightarrow\triangle\text{CDF}\cong\triangle\text{GFB}$ $...(ASA$ congruence criterion)
So, $\text{CD = GB}$ and $\text{CF = GF}$ $...(C.P.C.T.)$
Since $E$ and $F$ are the mid-points of $CA$ and $CG$ respectively,
we have $\text{EF}=\frac{1}{2}\text{AG}$
$=\frac{1}{2}(\text{AB}-\text{GB})$
$=\frac{1}{2}(\text{AB}-\text{CD})$

In a parallelogram, sum of adjacent angles =$ 180^\circ $
$\Rightarrow \angle \text{A}+\angle\text{B}=180^\circ$
$\Rightarrow\frac{\angle\text{A}}{2}+\frac{\angle\text{B}}{2}=90^\circ\ ...(1)$
$\Rightarrow\angle\text{OAB}=\frac{\angle\text{A}}{2}$ and $\angle\text{OBA}=\frac{\angle\text{B}}{2}$
Thus, $\angle\text{OAB}+\angle\text{OBA}=90^\circ$ [From eq $(1)]$
$\Rightarrow\angle\text{AOB}=180^\circ-(\angle\text{OAB}+\angle\text{OBA})=180^\circ-90^\circ$
$\Rightarrow\angle\text{AOB}=90^\circ$

$PQ || AC$ (since in $\triangle\text{ABC}$ mid-points of $AB \& BC$ are meeting by $PQ)$
Similarly, $SR || AC$
$\Rightarrow PQ || SR$
Now in $\triangle\text{ABD}$ and $\triangle\text{CBD},$
$PS || BD$ and $QR || BD$
$\Rightarrow PS || QR$
Hence, PQRS is a parallelogram.
But $\text{PR }\bot \text{ QS}$
$\Rightarrow $ Diagonals cut at $90^\circ $
$\Rightarrow PQRS$ is a Rhomus.

Let $ABCD$ be a quadrilateral with $\angle\text{A} = 108^\circ$ and $\angle\text{B} = \angle\text{C} = \angle\text{D} = \text{x}$
$\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$ (angle sum property)
$108^\circ + x + x + x = 360^\circ$
$3x = 360^\circ - 108^\circ$
$3x = 252^\circ$
$x = 84^\circ$

Given $\angle\text{AOB}=110^\circ$
$\Rightarrow \angle\text{DOC}=110^\circ$ vertically opposite angles
$\triangle\text{DOC}$ we have:
$DO = OC$
Now, $\angle\text{ODC} = \angle\text{OCD} = \text{y}$
now in $\triangle\text{ODC}$
$y+y+110^{\circ}=180^{\circ} \text { (angle sum property of triangle) }$
$\Rightarrow 2 y=180^{\circ}-110^{\circ}=70^{\circ}$
$\Rightarrow y=35^{\circ}$
Also, $x=90^{\circ}-y$
$x=90^{\circ}-35^{\circ}=55^{\circ}$
Hence, $x=55^{\circ}$ and $y=35^{\circ}$

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

Consecutive sides of a Quadrilateral $ABCD$ are
$AB$ and $BC,$
$BC$ and $CD,$
$CD$ and $AD,$
$AD$ and $AB,$
Which have only one point in common
i.e the joint point of their ends.

$PQ \| SR \| AC$
$QR \| PS \| BD$
{Because line joining the mid-points of two sides of triangle is $\|$ to third side}
Now because $AC$ is not prependicular to $BD$ in parallelogram,
$\Rightarrow SR$ is not perpendicular to $QR$
Also $\triangle\text{ASP}\not\cong\triangle\text{DRS}$
$\Rightarrow \text{PS} \neq \text{SR}$
$\Rightarrow PQRS$ is just a parallelogram.

By theorm diagonals of quadrilateral bisect each other if and only if it is a parallelogram.For a quadrilateral to be parallelogram some other properties are required: Opposite sides are equal and parallel. Opposite angles are equal. Sum of any adjacent angles is $180.$

Let angles of parallelogram are $\angle\text{A}, \angle\text{B}, \angle\text{C}, \angle\text{D}$

Let smallest angle $= \angle\text{A}$
Let largest angle $= \angle\text{B}$
$= \angle\text{B} = 2\angle\text{A} – 24^\circ ...\text{(i)}$
$\angle\text{A} + \angle\text{B} = 180^\circ$ [adjacent angle of parallelogram]
So, $\angle\text{A} + 2\angle\text{A} -24^\circ = 180^\circ$
$= 3\angle\text{A} = 180^\circ + 24^\circ = 204^\circ$
$=\angle\text{A}=\frac{204^\circ}{3}=68^\circ$