Maths M.C.Q

Let the half of the $\angle\text{A} , \angle\text{B}, \angle\text{C}$ and $\angle\text{D}$ are denoted by $a, b, c$ and $d$ respectively.
$\text{a + b} + \angle\text{APB} = 180$ and $\text{c + d} + \angle\text{DRC} = 180($angle sum property$).$
$\angle\text{APB} = \angle\text{QPS} $ and $\angle\text{DRC} = \angle\text{QRS} ($vertically opposite angles$).$
So, $\angle\text{QPS} + \angle\text{QRS} = (180 - \text{a} - \text{b)} + (180 - \text{c} - \text{d)}$
$= 360\ -\ (\text{a + b + c + d)}$
$= 360 - \frac{1}{2} $ of $(\angle\text{ A} + \angle\text{B} + \angle\text{C} + \angle\text{D})$
$= 360 - \frac{1}{2} $ 0f $360 = 180$
Therefore, $\angle\text{QPS}$ and $\angle\text{QRS}$ are supplementary.

 A quadrilateral formed by joining the mid points of a square is a square. So, $ABCD$ is a square. In Square, diagonals are equal and perpendicular.

Given,
$ABCD$ is a parallelogram

$P$ is mid-point of side $BC$
$\angle\text{BAP} = \angle\text{DAP}$
$\text{AD} = 10\text{cm}$
$∵ \text{AD || BC}$
$\angle\text{DAP} = \angle\text{APB} [$alternate angles$]$
$\angle\text{DAP} = \angle\text{BAP}$
$\text{AB = BP} ($side opposite to equal angles$)$
$\Rightarrow\ \text{BP}=\frac{1}{2}\text{BC}$
$\therefore\ \text{AB}=\frac{1}{2},\ \text{BC}=\frac{1}{2}\times10=5\text{cm}$
$\text{AB} = \text{CD} = 5\text{cm} ($sides of parallelogram$)$
Hence, $\text{CD} = 5\text{cm}.$

Let a line $BP$ is drawn $||$ to $AD$ to meet $DC$ at $ P.$
$ABPD$ is a parallelogram.
$AB || PD, AD || BP$
So $AB = DP$
Let $BP$ cuts $MN$ at $Q$.
$MQ$ is also $||$ to $AB || PD$
So $AB = MQ = PD = 12\ cm ...(1)$
$QN = MN - MQ = 14 - 12 = 2\ cm$
Consider $\triangle\text{BPC}.$
Q and N are the mid-points of $BP$ & $BC,$ and the line joining them $QN || PC.$
Then by property, $\frac{\text{QN}}{\text{PC}}=\frac{1}{2}$
$⇒ PC = 2QN = 2 × 2 = 4\ cm$
Now, $DC = DP + PC$
$DP = 12\ cm [$From $(1)]$
$⇒ DC = 12 + 4 = 16\ cm$

In a parallelogram, opposite corner angles are equal and sum of adjacent angles $= 108^\circ $
Hence, in quadrrilateral $PQRS,$
$\Rightarrow\angle\text{P}=\angle\text{R}$ and $\angle\text{Q}=\angle\text{S}$
Also, $\angle\text{P}+\angle\text{Q}=\angle\text{Q}+\text{R}=180^\circ$
Hence, if $\angle\text{P}=100^\circ$ and $\angle\text{Q}=80^\circ,$ then
$\angle\text{P}+\angle\text{Q}=100^\circ+80^\circ=180^\circ$
Also, if $\angle\text{Q}+=80^\circ$ and $\angle\text{R}=100^\circ$ then
$\angle\text{Q}+\angle\text{R}=80^\circ+100^\circ=180^\circ$

$ E$ and $F$ are midpoints of sides $AB$ and $AC.$ By midpoint theorem, $EF$ is parallel to $BC$ and $EF$ is $\frac{1}{2}$ of $BC.$
So, $\text{EF} = {1}{2}$ of $(7.2) = 3.6\text{cm}.$

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

$40^\circ\ \angle\text{C} = 60^\circ$ as opposite angles of a parallelogram are equal and $\angle\text{CDB} = 40^\circ$ angle sum property of a triangle. [In $\triangle\text{CDB},\ \angle\text{C} + \angle\text{CDB} + \angle\text{DBC} = 180^\circ$]

Since the line segment joiing the mid-poients of any two sides of a triangle is parallel to third side and is half to it, so

$\therefore\ \text{DE}=\frac{1}{2}\text{BC}$ and $\text{DE}||\text{BC}$
Similarly, $\text{DP}=\frac{1}{2}\text{AO}$ and $\text{DP||AO}$
And $\text{EQ}=\frac{1}{2}\text{AO}$ and $\text{EQ||AO}$
$\therefore\ \text{DP}=\text{EQ}[\therefore\text{Each}=\frac{1}{2}\text{AO}]$
And $\text{DP||EQ}[\therefore\text{Each}=\frac{1}{2}\text{AO}]$
Now, $DEQP$ is quadrilateral in which one pair of its opposite is equal and parallel.
Therefore, quadrilateral $DEQP$ is a parallelogram.
Hence, $(d)$ is the correct answer.

 Given that $ABCD$ is a parallelogram.
We konw that, opposite sides of a parallelogram are parallel.
$\Rightarrow\angle\text{A}+\angle\text{B}=180^{\circ} ...($interior angles$)$
Also, $\angle\text{A}=\angle\text{B}=90^{\circ} ...($Given$)$
Since opposite angles of a parallelogram are equal,
$\angle\text{A}=\angle\text{C}$ and $\angle\text{B}=\angle\text{D}$
So, $\angle\text{A}=\angle\text{C}=\angle\text{B}=\angle\text{D}=90^{\circ}$
$\therefore ABCD$ is a rectangle.

The figure will be a rectangle.
Hence, $(b)$ is the correct answer.

$ABCD$ is a rhombus and a rhombus is also a parallelogram. A rhombus has four equal sides.
The diagonals of a rhombus are perpendicular bisector of each other.
So, in $\triangle\text{AOB}, \ \angle\text{OAB} = 40^\circ, \angle\text{AOB} = 90^\circ $ and $\angle\text{ABO} = 180^\circ- (40^\circ + 90^\circ) = 50^\circ$
$\therefore \text{x} = 50^\circ$
In $\triangle\text{ABD, AB = AD}$
So, $\angle\text{ABD} = \angle\text{ADB} = 50^\circ$
Hence, $x = 50^\circ $ and $y = 50^\circ $

$\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360 ($angle sum property$)$
$3x + 5x + 20x + 8x = 360$
$36x = 360$
$x = 10$

In a parallelogram $ABCD,$
$\angle\text{OAB}=\angle\text{OCB}$
In $\triangle\text{OCB}$
$\angle\text{OCD}+\angle\text{COD}+\angle\text{ODC}=180^\circ$
$\angle\text{COD}=90^\circ$
$\angle\text{ODC}=50^\circ$ (given)
$\angle\text{OCD}=180^\circ-90^\circ-50^\circ=40^\circ$
$\Rightarrow\angle\text{OAB}=\angle\text{OCD}=40^\circ$

$ABCD$ is a rhombus. $AB = BC = CD = DA$
In Rhombus, diagonals bisect each other at right angles.
So, $AO= CO$ and $BO = DO$
In triangle $\mathrm{AOB} \times \mathrm{AO} \mathrm{F}^2+\mathrm{BO}^2=\mathrm{AB}^2$ (Pythagoras theorem)
$\Big(\frac{1}{2}\text{AC}\Big)^2 + \Big(\frac{1}{2}\text{BD}\Big)^2 = \text{AB}^2$
$\frac{\text{AC}^2}{4} + \frac{\text{BD}^2}{4} = \text{AB}^2$
$=A C^2+B D^2=4 A B^2$

In Rhombus, diagonals bisect each other right angle. By using angle sum property in any of the four triangles formed by intersection of diagonals, we get $\angle\text{CBD} = 50$ and $\angle\text{CBD} = \angle\text{ADC}$ (alternate angles).
So, $\angle\text{ADC} = 50$

A quadrilateral formed by joining the mid-points of a square is a square. So, $ABCD$ is a square. In Square, diagonals are equal and perpendicular.

Given: $ABCD$ is a trapezium with $\text{AB || CD}$
Construction: Draw $DE$ and $CF$ $\bot$ to $AB.$
Then in $\triangle\text{ABC}$
$\angle\text{BAC}$ is acute
$\therefore B C^2=A C^2+A B^2-2 A F: A B \ldots(1)$
and in $\triangle \mathrm{BDA}$
$\angle \mathrm{DBA}$ is acute
$\therefore A D^2=B D^2+A B^2-2 B E: A B \ldots(2)$
Adding $(1)$ and $(2)$ we get
$B C^2+A D^2=A C^2+B D^2+2 A B^2-2 A F \cdot A B-2 B E \cdot A B$
$\Rightarrow A C^2+B D^2=B C^2+A D^2-2 A B[A B-A F-B E)$
$=B C^2+A D^2-2 A B[A B-(A E+E F)-(B F+E F)]$
$=B C^2+A D^2-2 A B[A B-(A E+E F+B F+E F)]$
$=B C^2+A D^2-2 A B[A B-(A B+C D)](\therefore E F=D C)$
$=B C^2+A D^2-2 A B[-(C D)]$
$=A D^2+B C^2+2 A B \times C D$

Given: $ABCD$ is a parallelogram in which $AO$ and $BO$ are angle bisectors of $\angle\text{A}$ and $\angle\text{B}.$
Now since $ABCD$ is a parallelogram
$\therefore\text{AD || BC}$
Now $\text{AD || BC}$ and transversal $AB$ intersect them
$\therefore\angle\text{A}+ \angle\text{B} = 180^\circ  ( \therefore$ sum of consecutive interior angle is $180º)$
$\Rightarrow\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}=90^\circ$
$\therefore\angle\text{1}+ \angle\text{2} = 90^\circ ( \therefore  AO$ and $BO$ are angle bisectors$) ... (1)$
In $\triangle\text{AOB}$ we have
$\angle1 + \angle\text{AOB} + \angle\text{2} = 180^\circ$
$\Rightarrow90^\circ +\angle\text{AOB} = 180^\circ$ from $(1))$
$\Rightarrow \angle\text{AOB} = 180^\circ - 90^\circ = 90^\circ$

$ABCD$ is a rhombus.
$\Rightarrow\text{AD || BC}$ and $\text{AC}$ is the transversal.
$\Rightarrow\angle\text{DAC}=\angle\text{ACB}$ (alternate angles)
$\Rightarrow\angle\text{DAC}=50^{\circ}$
In $\triangle\text{AOD},$ by angle sum property,
$\angle\text{AOD}+\angle\text{DAO}+\angle\text{ADO}=180^{\circ}$
$\Rightarrow90^{\circ}+\angle\text{50}^{\circ}+\angle\text{ADO}=180^{\circ}$
$\Rightarrow\angle\text{ADO}=40^{\circ}$
$\Rightarrow\angle\text{ADB}=40^{\circ}$

 Given,
$ABCD$ is a parallelogram
$\angle\text{A} + \angle\text{C} = 2(\angle\text{B} + \angle\text{D})$
$\angle\text{A} = 40^\circ$
$∵ \angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$ [angle sum property of quadrilateral]
$\Rightarrow \angle\text{A} + \angle\text{C} + \angle\text{B} + \angle\text{D} = 360^\circ$
$⇒ 2(\angle\text{B} + \angle\text{D})+ \angle\text{B} + \angle\text{D} = 360^\circ$
$⇒ 3(\angle\text{B} + \angle\text{D})= 360^\circ$
$\Rightarrow\angle\text{B}+\angle\text{D}=\frac{360^\circ}{3}=120^\circ$
$\because\ \angle\text{B}=60^\circ$ [given]
$\therefore\ \angle\text{B}=120^\circ-60^\circ=60^\circ$

We know that, sum of the angles of a quadrilateral is $360^\circ .$
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^{\circ}$
$\Rightarrow\angle\text{A}+\angle\text{B}+70^{\circ}+30^{\circ}=360^{\circ}$
$\Rightarrow\angle\text{A}+\angle\text{B}=260^{\circ}$
$\Rightarrow\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}=\frac{1}{2}(260)^{\circ}$
$\Rightarrow\angle\text{BAO}+\angle\text{ABO}=130^{\circ}...(\text{i})$
In $\triangle\text{AOB},$
$\angle\text{BAO}+\angle\text{ABO}+\angle\text{AOB}=180^{\circ} ...($Angle sum Property$)$
$\Rightarrow130^{\circ}+\angle\text{AOB}=180^{\circ} ...($from $(i))$
$\Rightarrow\angle\text{AOB}=50^{\circ}$

Diagonals of $PQRS$ are at right angles form all the internal angles as right angles. $[$according to angle property of rectangle, i.e., all the angles of a rectangle are right angle $(90^\circ )].$

Applying Pythagoras theorem in $\triangle\text{ABC}$
$A C^2=A B^2+B C^2$
$15^2=9^2+B C^2$
$225=81+\mathrm{BC}^2$
$225-81=\mathrm{BC}^2$
$B C^2=144$
$B C=12 \mathrm{~cm}$

$ AB = 7\ cm ($by mid-point theorm$)$
$AC = 5\ cm ($by mid-point theorm$)$
$BC = 6\ cm ($by mid-point theorm$)$

 $\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$ (angle sum property)
$\angle\text{C} = 3 ( 38) = 114^\circ$
$\angle\text{D} = 4 ( 38) = 152^\circ$
So, $38^\circ + \angle\text{B} + 114^\circ + 152^\circ = 360^\circ$
$\angle\text{B} = 360^\circ - 304^\circ = 56^\circ$

 Use pythagoras theorem in right triangle,
$102 -\Big[\frac{16}{2}\Big]^2 = 100 - 64 = 36 = [6]^2$
Hence, the other diagonal $= 6 × 2 = 12\ cm$

Given: $ABCD$ is a parallelogram in which $AO$ and $BO$ are angle bisectors of $\angle\text{A}$ and $\angle\text{B}.$
Now since $ABCD$ is a parallelogram
$∴ \text{AD || BC}$
Now $\text{AD || BC}$ and transversal $AB$ intersect them
$∴\ \angle\text{A} + \angle\text{B} = 180^\circ (∴$ sum of consecutive interior angle is $180º)$
$⇒\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}=90^\circ$
$⇒ \angle1 + \angle2 = 90^\circ (∴ AO$ and $BO$ are angle bisectors$) ...(i)$
In $\triangle\text{AOB}$ we have
$\angle1 + \angle\text{AOB} + \angle2 = 180^\circ$
$⇒ 90^\circ +\angle\text{AOB} = 180^\circ ($from $(i))$
$⇒ \angle\text{AOB} = 180^\circ – 90^\circ = 90^\circ$

 
Let $ABCD$ be a rhombus and $p, q, R$ and $S$ be the mid-point.
Let $ABCD$ be a rhombus and $P, Q, R$ and $S$ be the mid-points of sides $AB, BC, CD$ and $DA$ respectively.
In $\triangle\text{ABD}$ and BDC we have
$\text{SP || BD}$ and $\text{SP}=\frac{1}{2}\text{BD}\ ...\ \text{(i)}$
$\text{RQ || BD}$ and $\text{RQ}=\frac{1}{2}\text{BD}\ ...\ \text{(ii)}$
From $(i)$ and $(ii)$ we get
$PQRS$ is a $||gm$
As diagonals of a rhombus bisect each other at right angles.
$∴\ \text{AC⊥BD}$
Since $\text{SP || BD, PQ || AC}$ and $\text{AC⊥BD}$
$∴\ \text{SP⊥ PQ}$
$∴\ \angle\text{QPS}=90^\circ ... \text{(iii)}$
From above results,
we have,
$||gm\ PQRS$ is a rectangle.

 
Given $ABCD$ is a rhombus. Diagonals bisect each other perpendicularly.
Hence $\angle\text{BOC} = 90^\circ$
Given $\angle\text{OCB} = 40^\circ$
$\text{AD || BC}$ and BD is the transversal
$∴ \angle\text{ADB} = \angle\text{DBC}$ (Alternate angles)
Hence in right angled $\triangle\text{BOC},$
$\angle\text{BOC} + \angle\text{OCB} + \angle\text{OBC} = 180^\circ$
$⇒ 90^\circ + 40^\circ + \angle\text{OBC} = 180^\circ$
$⇒130^\circ + \angle\text{OBC} = 180^\circ$
$⇒ \angle\text{OBC} = 180^\circ - 130^\circ = 50^\circ$
But $\angle\text{OBC} = \angle\text{DBC}$
Therefore, $\angle\text{ADB} = 50^\circ.$