MATHS M.C.Q

$\therefore \angle\text{AOB} = \angle\text{COD}$ (vertically opposite angle)
$\therefore \angle\text{COD}=115^\circ$

$\angle\text{A} = 60^\circ, \ \angle\text{C} = 60^\circ,\ \angle\text{B} = 120^\circ$
Opposite angles are equal and sum of adjacent angles are supplementary.

$BC = 4 \times 1.8 = 7.2\ cm.$

Let $ABCD$ be the rhombus with diagonals $AB = 10\ cm$ and $Ac = 16\ cm.$
Since the diagonals of a rhombus are perpendicular bisectors of each other,
$\Rightarrow\text{OA}=8\text{cm},\text{BD = 2OB}$ and $\angle\text{AOB}=90^{\circ}$
In right $\triangle\text{AOB},$
By Pythagoras theorem,
$\text{AB}^2=\text{OA}^2+\text{OB}^2$
$\Rightarrow(10)^{2}=(8)^2+\text{OB}^2$
$\Rightarrow100=64+\text{OB}^2$
$\Rightarrow\text{OB}^2=36$
$\Rightarrow\text{OB}=\sqrt{36}$
$\Rightarrow\text{OB}=6\text{cm}$
$\Rightarrow\text{BD}=2\times\text{OB}$
$\Rightarrow\text{BD}=2\times6$
$\Rightarrow\text{BD}=12\text{cm}$
Hence, the lenght of the other diagonal is $12\ cm.$

Given,

$ABCD$ is a trapezium,
$\text{AB || DC}$
M, N are mid-points of $AD \& BC$
$AB = 12cm, MN = 14cm$
$∵ \text{AB || MN || CD}$ [$M, N$ are mid points of $AD \& BC]$
$MP = NP$
By mid-point theorem,
$\text{MP}=\frac{1}{2}\text{CD}$ and $\text{NP}=\frac{1}{2}\text{AB}$
$\therefore\ \text{MN}=\frac{1}{2}(\text{AB + CD})$
$\Rightarrow\ 14=\frac{1}{2}(12\ +\ \text{CD})$
$\Rightarrow \text{CD}=28\ -\ 12=16\text{cm}$

Let the smallest angle $=\angle\text{ADC}=\text{x}^\circ$
Other angle $\angle\text{BCD}$
$\Rightarrow\angle\text{BCD}=2\text{x}^\circ-24^\circ$
Also, $\angle\text{ACD}+\angle\text{BCD}=180^\circ$ (Sum of adjacent angles in ||$^{gram}$ = 180$^{\circ}$)
$\Rightarrow\text{x}^\circ+2\text{x}^\circ-24^\circ=180^\circ$
$\Rightarrow3\text{x}^\circ=204^\circ$
$\Rightarrow\text{x}=68^\circ$
⇒ Largest angle $=\angle\text{BCD}=2\times68^\circ-24^\circ=112^\circ$

Let $x$ be one angle. Then $\text{x}+\frac{2}{3\text{x}}+23\text{x} =180$ {sum of the adjacent angles of a parallelogram is $180^{\circ}$}
$x = 108$, adjacent angle $= 180 - 108 = 72^{\circ}$

$ABCD$ 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 $

A line $DG$ is drawn parallel to $E F$ to meet $A C$.
$FE \| DG$ and $FE \| GH$
Now, consider $\triangle ADG$.
$E$ is the mid-point of $A D$ and $E F$ is line from $E \|$ to Base $DG.$
So by property, it will meet $AG$ at its midpoint
i.e. $F$ is midpoint of $A G$.
$\Rightarrow A F=F G \ldots(1)$
Now, consider $\triangle\text{FBC}\ \&\ \triangle\text{GDC}$
$FE \| GH$ and $FE \| GD$
$D$ is mid-point of $BC$.
$\Rightarrow\frac{\text{DC}}{\text{BC}}=\frac{1}{2}\dots(2)$
Because $\triangle\text{FBC}\sim\triangle\text{GDC},$
$\Rightarrow\frac{\text{GC}}{\text{FC}}=\frac{1}{2}$
$\Rightarrow FC = 2GC$
or $FG = GC ...(3)$
From equation $(1)$ and $(3)$
$AF = FG = GC$
$\Rightarrow\text{AF}=\frac{\text{AC}}{3}=\frac{10.5}{3}=3.5\text{cm}$

In $\triangle\text{ABD},$
$\angle\text{BDA}+\angle\text{ABD}+\angle\text{DAB}=180^\circ$
$\angle\text{ABD}=50^\circ$ and $\angle\text{DAB}=90^\circ$
$\Rightarrow\angle\text{BDA}=180^\circ-90^\circ-50^\circ=40^\circ$
Consider $\triangle\text{ABD}\ \&\ \triangle\text{BAC}$
$\text{AD}=\text{BC},\ \angle\text{DAB}=\angle\text{ABC}=90^\circ,\text{BD}=\text{AC}$
Hence, by RHS property $\triangle\text{ABD}\cong\triangle\text{BAC}$
$\Rightarrow\angle\text{ABD}=\angle\text{BAC}=50^\circ$
Now, consider $\triangle\text{ABP}$
$\angle\text{PAB}+\angle\text{PBA}+\angle\text{APB}=180^\circ$
$\angle\text{PAB}=\angle\text{BAC}=50^\circ$
$\angle\text{PAB}=\angle\text{ABD}=50^\circ$
$\Rightarrow\angle\text{APB}=180^\circ-50^\circ-50^\circ=80^\circ$
Now, $\angle\text{APB}=\angle\text{DPC}$ (Opposite angles)
$\Rightarrow\angle\text{DPC}=80^\circ$

Given, $ABCD$ is a parallelogram having $\angle\text{BAO}=35^\circ, \ \angle\text{DAO}=40^\circ$ and $\angle\text{COD}=105^\circ.$
Now, $\angle\text{COD} = \angle\text{AOB} = 105^\circ$ [vertically opposite angles]
In $\angle\text{ABO},$ by angle sum property of triangle,
$\Rightarrow \angle\text{AOB} + \angle\text{OAB} + \angle\text{ABO} = 180^\circ$
$\Rightarrow 105^\circ + 35^\circ + \angle\text{ABO} = 180^\circ$
$\Rightarrow \angle\text{ABO} = 40^\circ$

The bisectors of $\angle\text{APQ},\angle\text{BPQ},\angle\text{CQP}$ and $\angle\text{PQD}$ enclose a rectengle.

The diagonals of a rhombus bisect each other at right angles.
$\Rightarrow\text{OA}=\frac{1}{2}\text{AC}$ and $\text{OB}=\frac{1}{2}\text{BD}$
Also, $\angle\text{AOB}=90^{\circ}$
In right $\triangle\text{AOB},$
$\therefore\text{AB}^2=\text{OA}^2+\text{OB}^2$ ...(By Pythagoras theorem)
$\Rightarrow\text{AB}^2=\frac{1}{4}\text{AC}^2+\frac{1}{4}\text{BD}^2$
$\Rightarrow\text{AC}^2+\text{BD}^2=4\text{AB}^2$

$\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$ (angle sum property)
$\angle\text{A} + 2\angle\text{A} + \angle\text{A} + \frac{1}{2}$ of $\angle\text{A} = 360^\circ$
$\frac{9}{2}$ of $\angle\text{A} = 360^\circ$
$\angle\text{A} = 80^\circ$
So, $\angle\text{B}=2(80^\circ)=160^\circ,\ \angle\text{C}=80^\circ$ and $\angle\text{D} = \frac{1}{2}$ of $80^\circ=40^\circ$

In $\triangle\text{APB},$ by angle sum property,
$\angle\text{APB}+\angle\text{PAB}+\angle\text{PBA}=180^{\circ}$
$\Rightarrow\angle\text{APB}+\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}=180^{\circ}$
$\Rightarrow\angle\text{APB}=180^{\circ}-\Big(\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}\Big)$
In $\triangle\text{CRD},$ by angle sum property,
$\angle\text{CRD}+\angle\text{RDC}+\angle\text{RCD}=180^{\circ}$
$\Rightarrow\angle\text{CRD}+\frac{1}{2}\angle\text{D}+\frac{1}{2}\angle\text{C}=180^{\circ}$
$\Rightarrow\angle\text{CRD}=180^{\circ}-\Big(\frac{1}{2}\angle\text{D}+\frac{1}{2}\angle\text{C}\Big)$
Now, $\angle\text{SPQ}+\angle\text{SRQ}=\angle\text{APB}+\angle\text{CRD}$
$=360^{\circ}-\Big(\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}+\frac{1}{2}\angle\text{D}\Big)$
$=360^{\circ}=\frac{1}{2}(\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D})$
$=360^{\circ}-\frac{1}{2}\times360^{\circ}$
$=360^{\circ}-180^{\circ}$
$=180^{\circ}$
Now, $\angle\text{PSR}+\angle\text{PQR}=360^{\circ}-(\angle\text{SPQ}+\angle\text{SRQ})$
$=360^{\circ}-180^{\circ}$
$=180^{\circ}$
Hence, $PQRS$ is a quadrilateral whose opposite angles are supplementary.

Let $ABCD$ be a rhombus.
Join $BD$ which forms two triangles $ABD$ and $DCB$. In $\triangle\text{ABD, AB = AD}.$
So, $\angle\text{ADB} = \angle\text{ABD}$ (angles opposite to equal sides are equal) $...(i)$
But, $\angle\text{ABD} = \angle\text{BDC}$ and $\angle\text{ADB} = \angle\text{CBD}$ (alternate angles)$ ...(ii)$
So, from $(i)$ and $(ii)$
$\angle\text{ADB} = \angle\text{ABD}=\angle\text{BDC} = \angle\text{CBD}$
$\therefore$ diagonal $BD$ bisects $\angle\text{B}$ and $\angle\text{D}.$

$\angle\text{DAO}+\angle\text{OAB}=\angle\text{DAB}$
$\Rightarrow\angle\text{DAO}+35^{\circ}=90^{\circ}$
$\Rightarrow\angle\text{DAO}=55^{\circ}$
ABCD is a rectangle and diagonals of a rectangle are equal and bisect each other.
$\text{OA = OD}$
$\Rightarrow\angle\text{ODA}=\angle\text{DAO}$ (angles opposte to equal sides are equal)
$\Rightarrow\angle\text{ODA}=55^{\circ}$
In $DODA$, by angle sum property,
$\angle\text{ODA}+\angle\text{DAO}+\angle\text{AOD}=180^{\circ}$
$\Rightarrow55^{\circ}+\angle55^{\circ}+\angle\text{AOD}=180^{\circ}$
$\Rightarrow\angle\text{AOD}=70^{\circ}$

Given,
$ABCD$ is a parallelogram,
Angles of quadrilateral $4x, 7x, 9x, 10x$
$\Rightarrow 4x + 7x + 9x + 10x = 360^\circ $ [angle sum property of quadrilateral]
$\Rightarrow 30x = 360$°
$\Rightarrow\text{x}=\frac{360^\circ}{30}=12^\circ$
Hence, sum of smallest and largest angles $= 4x + 10x = 4 \times 12^\circ + 10 \times 12^\circ$
$= 48^\circ + 120^\circ = 168^\circ$

In $\triangle\text{ABC},$
$R$ and $P$ are the mid-points of $AB$ and $BC.$
$\therefore\ \text{RP || AC},\ \text{RP}=\frac{1}{2\text{AC}}$ [By mid-point theorem]
In a quadrilateral $RPQA$,
$\Rightarrow \text{RP || AQ}\Rightarrow \text{RP || AQ}$
$\therefore\ \text{RPQA}$ is a parallelogram
$\Rightarrow \text{AR}=\frac{1}{2\text{AB}}$
$\therefore\ \text{AR}=12\times 30=15\text{cm}$
$\Rightarrow \text{AR} = \text{PQ} = 15\text{cm}$ [Since, opposite sides are equal]
$\Rightarrow \text{RP}=\frac{1}{2\text{AC}}=\frac{1}{2}\times 21=10.5\text{cm}$ [Since, opposite sides are equal]
$\Rightarrow $ Perimeter of $ARPQ = AR + QP + RP + AQ$
$= 15 + 15 + 10.5 + 10.5$
$= 51cm$
$\therefore $ Perimeter of $ARPQ$ is $51cm.$

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 $\angle\text{B} = \angle\text{D} = 60^\circ$

$PS \| QR, PQ \| SR ...(1)$
{Because lines joining the mid-points of any two sides of a triangle are parallel to the third side}
$\text{AC } \bot \text{ BD}$ & $\text{BR } \bot \text{ QS}$ (From Figure)
$SR \| AC$ and $QR \| BD$
$\text{AC } \bot \text{ BD}$
$\Rightarrow \text{SR }\bot \text{ QR}$
Hence $\angle\text{SRQ}=90^\circ\ ...(2)$
Also $\triangle\text{APS}\cong\triangle\text{DSR}$
$\Rightarrow\text{PS} = \text{SR}\dots(3)$
From equations $(1), (2), (3)$
$PQRS$ is a square.

In quadrilateral, the sum of the all four angles equal to 360º. let $\angle\text{B} = 2\text{x},\ \angle\text{C} = 3\text{x}$ and $\angle\text{D} = 7\text{x}.$
$\angle\text{A} + \angle\text{B} + \angle\text{C} +\angle\text{D} = 360^\circ$
$60 + 2x + 3x + 7x = 360$
$12x = 300^{\circ}$
$x = 25^{\circ}$
So, $\angle\text{D} = 7\text{x} = 7(25^\circ) = 175^\circ$

Area of Parallelogram = Base $\times $ Height
If 'a' is the side and $'b'$ is the base the height will be less than $'a'$ using Pythagoras theorem, a as Hypotenuse, h as height, $A < B.$

In quadrilateral, the sum of the all four angles equal to $360^\circ .$
Let $\angle\text{B} = 2\text{x}, \ \angle\text{C} = 3\text{x}$ and $\angle\text{D} = 7\text{x}.$
$\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$
$60 + 2x + 3x + 7x = 360$
$12x = 300$
$x = 25$
So, $\angle\text{D} = 7\text{x} = 7 (25) = 175^\circ$