Maths M.C.Q

Two diagonals of quadrilateral form four triangles. Out of these four triangles two triangles of opposite to each other are congruent by $SAS$. By using $CPCT$ property we can prove that both pair of opposite sides in a quadrilateral are parallel. A quadrilateral with both pair of opposite sides parallel is called parallelogram.

Let the smallest angle be $x^\circ $
$\Rightarrow $ its adjacent angle $= (2x - 24)$
Since sum of the adjacent angles $= 180^\circ $
$\Rightarrow x + 2x - 24 = 180$
$\Rightarrow 3x = 204$
$\Rightarrow x = 68^\circ $
So, its adjacent angle $= 2(68) - 24 = 136 - 24 = 112^\circ .$
Hence, the largest angle of the parallelogram is $112^\circ .$

In $\triangle\text{ABC},$ P and $Q$ are the mid-point of sides $AB$ and $BC$ respectively.
$\therefore\text{PQ || AC}$ and $\text{PQ}=\frac{1}{2}\text{AC }...(\text{i})$
In $\triangle\text{ADC},$ R and $S$ are the mid-point of sides $CD$ and $AD$ respectively.
$\therefore\text{RS || AC}$ and $\text{RS}=\frac{1}{2}\text{AC }...(\text{ii})$
From $(i)$ and $(ii),$
$\text{PQ ∥ RS}$ and $\text{PQ = RS}$
Thus, in quadrilateral $PQRS$, a pair of opposite sides are equal are parallel.
So, $PQRS$ is a parallelogram.
Let the diagonals $AC$ and $BD$ intersect at $O.$
Now, in $\triangle\text{ABD},$ $P$ and S are the mid-points of sides $AB$ and $AD$ respectively.
$\therefore\text{PS || BD}$
$\Rightarrow\text{PN || MO}$
Also, from $(i)$, $\text{PQ || AC}$
$\Rightarrow\text{PM || NO}$
Thus, in quadrilateral $PMON$, $\text{PM || NO}$ and $\text{PN || MO},$
$⇒ PMON$ is a parallelogram.
$\Rightarrow\angle\text{MPN}=\angle\text{MON}$ (opposite angles of a parallelogram are equal)
$\Rightarrow\angle\text{MPN}=\angle\text{BOA}$ $(\text{Since }\angle\text{BOA}=\angle\text{MON})$
$\Rightarrow\angle\text{MPN}=90^{\circ}$ $(\text{Since }\text{AC}\perp\text{BD},\angle\text{BOA}=90^{\circ})$
$\Rightarrow\angle\text{QPS}=90^{\circ}$
Thus, $PQRS$ is a parallelogram whose one angle, i.e. $\angle\text{QPS}=90^{\circ}.$
Hence, $PQRS$ is a rectangle if $\text{AC}\perp\text{BD}.$

We know that The sum of angles of quad $= 360^\circ $
$4$th angle $= 360^\circ - (75^\circ + 90^\circ + 75^\circ )$
$= 360^\circ - 240^\circ $
$= 120^\circ $

$\angle\text{ABC}=\angle\text{ADC}$ ...(Opposite angles of a parallelogram are equal)
$\Rightarrow\frac{1}{2}\angle\text{ABC}=\frac{1}{2}\angle\text{ADC}$
$\Rightarrow\angle\text{ABD}=\angle\text{ADB}$
So, $\text{AD = AB}$ ...(Sides opposite equal angles are equal.)
$\therefore\triangle\text{ABD}$ is isosceles
Also, $M$ is the mid-point of $BD.$
$\therefore\text{AM}\perp\text{BD}$
$\therefore\angle\text{AMB}=90^{\circ}$

$ABCD$ is a rectangle with $AC$ and $BD$ as its diagonals.
$ABCD$ is a rectangle
$\therefore \angle\text{A} = 90^\circ, \ \text{AD = BC}$
$\text{AD || BC}$ and AB is a transversal
$\therefore \angle\text{A} + \angle\text{B} = 180^\circ$
$\Rightarrow \angle\text{B} = 90^\circ$
In $\triangle\text{ABD}$ and $\triangle\text{BAC}$
$AB = BA$
$\angle\text{A} = \angle\text{B}$
$AD = BC$
$\therefore \triangle\text{ABD}≅\triangle\text{BAC} (SAS)$
$\Rightarrow BD = AC (c.p.c.t)$
Hence, the diagonals of a rectangle are equal.

By $SSS$ congruence condition, opposite sides are equal and common diagonal.

We know that, in a $\| gm$ opposite sides are equal, opposite angles are equal and also the diagonals bisect each other.
So, opposite angles are bisected by the diagonals is not true.

In $\triangle ABD$ and $\triangle CBD$
$P S|\mid B D$ and $Q R| \mid B D$
\{A line joining mid-points of two sides of triangle is parallel to third side}
$\Rightarrow PS \| QR$
Similiarly PQ || SR
Because $S R \| A C$ and $Q R \| B D$,
And angle between the diagonals of a Rhombus $A C$ and $B D=90^{\circ}$,
Angle between $S R$ and $Q R=90^{\circ}$
$\Rightarrow PQRS$ is a rectangle.

Let us assume a rhombus $A B C D$ where,
$A B=B C=C D=D A$
Now, in triangle $O B C$ by using Pythagoras theorem we get:
$\mathrm{BC}^2=O B^2+O C^2$
$\mathrm{BC}^2=6^2+8^2$
$\mathrm{BC}^2=36+64$
$\mathrm{BC}^2=100$
$\mathrm{BC}=\sqrt{100}$
$\mathrm{BC}=10 \mathrm{~cm}$
$\therefore A B=B C=C D=D A=10 \mathrm{~cm}$

Diagonals of a square are perpendicular bisectors, hence angle $AOB = 90^\circ $. So triangle $AOB$ is right-angled.

Let the common multiple be $x .$
$\therefore$ The angle measure $3 x, 7 x, 6 x$ and $4 x$.
Since the sum of the angles of a quadrilateral is $360^{\circ}$, we have
$3x + 7x + 6x + 4x = 360$
$⇒ 20x = 360$
$⇒ x = 18°$
$\therefore$ The angles of the quadrilateral are
$3x = 3(18) = 54°$
$7x = 7(18) = 126°$
$6x = 6(18) = 108°$ and
$4x = 4(18) = 72°$Now, $54+126=180^{\circ}$ and $108+72=180^{\circ}$
So, the angles are interior angles and hence we get one pair of parallel sides of $A B C D$. Hence, $A B C D$ is a trapezium.

Let ABCD be a parallelogram. $\angle\text{A} + \angle\text{D} = 180$ (co-interior angles)
$\frac{1}{2}$ of $(\angle\text{A} + \angle\text{D}) = 90.$
Triangle formed by bisectors of $\angle\text{A} $ and $\angle\text{D} ,$ have sum of two angles equals to $90$ therefore, remaining angle is of $90$. similarly, we can prove that other angles formed are of $90$ each by bisectors of angles of $ABCD. $The quadrilateral formed by angle bisectors of $ABCD$ has all angles equal to $90$ (Vertically opposite angles). $A $quadrilateral with all right angles is a Rectangle.

Area of a parallelogram = base ⨯ height
Since two parallelogram stand on equal bases and between the same parallel lines,
their heights are same.
$\therefore$ Areas are also same.
$\therefore$ The ratio of their area is $1 : 1.$

Let $ABCD$ is the parallelogram, in which $AC$ and $BD$ are diagonals that intersect at $O.$
We have, $\angle\text{DAC} = 32^\circ; \ \angle\text{AOB} = 70^\circ$ [given]
Since $ABCD$ is a ||gm, so $\text{AB || CD}$ and $\text{AD || BC}.$
Since $\text{AD || BC},$ and AC is a transversal,
Then $\angle\text{ACB} = \angle\text{DAC}$ [alternate interior angles]
So, $\angle\text{ACB} = 32^\circ \Rightarrow \angle\text{OCB} = 32^\circ$
Since, $\angle\text{AOB}$ and $\angle\text{OBC}$ form a linear pair, then $\angle\text{AOB} + \angle\text{BOC}=180^\circ$
$\Rightarrow 70^\circ+\angle\text{BOC}=180^\circ\Rightarrow \angle\text{BOC} = 180^\circ − 70^\circ \Rightarrow \angle\text{BOC}=110^\circ$
In $\angle\text{OBC},$
We have, $\angle\text{BOC} + \angle\text{OCB} + \angle\text{OBC} =180^\circ$ [angle sum property]
$\Rightarrow 110^\circ + 32^\circ + \angle\text{OBC} = 180^\circ$
$\Rightarrow 142^\circ + \angle\text{OBC}=180^\circ$
$\Rightarrow \angle\text{OBC} = 180^\circ−142^\circ$
$\Rightarrow \angle\text{OBC} = 38^\circ$
$\Rightarrow \angle\text{DBC} = 38^\circ$

Here, $\angle\text{BAC}=50^\circ$ (angles in same segment are equal)

In $\triangle\text{BCD},$ we have
$\angle\text{BCD}=180^\circ−(\angle\text{BDC}+\angle\text{DBC})$
$=180^\circ−(50^\circ+60^\circ)$
$= 70^\circ$

Consider parallelogram $ABCD,$
We know that, the sum if the adjacent angles of a parallelogram is $180^\circ .$
$\Rightarrow\angle\text{DAB}+\angle\text{CBA}=180^{\circ}$
$\Rightarrow\frac{1}{2}\angle\text{DAB}+\frac{1}{2}\angle\text{CBA}=\frac{1}{2}(180^{\circ})$
$\Rightarrow\angle\text{OAB}+\angle\text{OBA}=90^{\circ} ...(\text{i})$
Now,
In $\triangle\text{OAB},$
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^{\circ}$ ...(Angle sum property)
$\Rightarrow90^{\circ}+\angle\text{AOB}=180^{\circ}$ ...(from $(i))$
$\Rightarrow\angle\text{AOB}=90^{\circ}$
The bisectore of any adjacent angles of a parallelogram intersects at $90^\circ .$

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

As $D$ and $E$ are the midpoints of $AB$ and $AC.$
So, by mid-point theorem
$\text{DE} = \frac{\text{BC}}{2} = \frac{12}{2} = 6\text{cm}$
$\text{AD} = \frac{\text{AB}}{2} = \frac{9}{2} = 4.5\text{cm}$
Area of $\triangle\text{ADE} = 0.5 × \text{DE} × \text{AD}$
$= 0.5 × 6 × 4.5 = 13.5\text{cm}^2$

$PQRS$ is a quadrilateral whose opposite angles are supplementary.


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

$3x - 10^\circ = x + 80^\circ $ [opposite angles of a parallelogram are equal]
$3x - x = 80^\circ + 10^\circ $
$2x = 90^\circ $
$\text{x} = \frac{90^\circ}{2}$
$x = 45^\circ $
$3x - 10^\circ + y + 25^\circ = 180^\circ $[In a parallelogram co-interior angles are supplementary]
$3 \times 45^\circ - 10^\circ + y + 25^\circ = 180^\circ $
$135^\circ + 25^\circ - 10^\circ + y = 180^\circ $
$150^\circ + y = 180^\circ $
$y = 180^\circ - 150^\circ $
$y = 30^\circ $

$ABCD$ is a rhombus and a rhombus is also a parallelogram. A rhombus has four equal sides.
In $\triangle\text{ABC},\ \angle\text{BAC} = \angle\text{BCA} = \text{x}$
In $\triangle\text{ABC}$
$x + x + 110^{\circ} = 180^{\circ}$(angle sum property of triangle)
$\Rightarrow 2 x=180^{\circ}-110^{\circ}=70^{\circ}$
$\Rightarrow x=35^{\circ}$
Now, $\angle\text{B} + \angle\text{C} = 180^\circ$ (Adjacent angles are supplementary)
But, $\angle\text{C} = \text{x} + \text{y} = 70^\circ$
$\Rightarrow y=70^{\circ}-x$
$\Rightarrow y=70^{\circ}-35^{\circ}=35^{\circ}$
Hence, $x=35^{\circ}$ and $y=35^{\circ}$

$PNQM$ is a rectangle.

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

If $\angle\text{ACB} = 30^\circ$ then $\angle\text{CAD} = 30^\circ$ (The diagonals bisect the angles in a Rhombus) and in $\triangle\text{AOD}, \ \angle\text{AOB} = 90^\circ$ diagonals perpendicular to each other, so by angle sum property of a $\angle\text{ADB} = 60^\circ.$

Given:
$A, B, C$ is a triangle .
$P, Q, R$ are the mid-points of sides $BC , CA $ and $AB .$
$AC = 21cm$
$BC = 29cm$
$AB = 30cm$
To find:
Perimeter of quadrilateral $A R Q P$ ?
$Q$ is the mid-point of $A C$
$P$ is the mid-point of $B C$
$Q P$ is parallel to $A B$
$QP =$ half of AB (according to mid point theorem)
$A B=30 cm, Q P=15 CM (Q P$ is half of $B A)$ (proved above)
$R$ is the mid-point of side $A B$.
$Q P$ is also parallel to $A R$ (half of side $A B$ )
$P R$ is parallel to $A C$
$P R=$ half of $A C$ (according to mid point theorem)
$A C=21 cm, P R=10.5 cm$ ( $P R$ is half of $A C$ ) (proved above)
$P R$ is parallel to $A Q$ ( $A Q$ is half of $A C$ )
Since, in quadrilateral $ARQP$ both the opposite sides are parallel it is a parallelogram.
Therefore, $ARQP$ is a parallelogram.
We know that
In parallelogram, opp sides are equal.
Therefore,
$PR = AQ = 10.5cm$
$QP = AR = 15cm$
$10.5cm + 10.5cm + 15cm + 15cm = 51cm.$
Therefore the perimeter of quadrilateral,
$ARQP = 51cm.$

Let the measure of the fourth angle be $x^0$.
We know that, the sum of the angles of a quadrilateral is $360^{\circ}$.
$\text { So }, 80^{\circ}+95^{\circ}+112^{\circ}+x=360^{\circ}$
$\Rightarrow 287^{\circ}+x=360^{\circ}$
$\Rightarrow x=73^{\circ}$
$\therefore$  Its fourth angle is $73^{\circ}$.

By the congruency of triangles,
$BEF$ and $CED$ (AAS Rule) are congruent
So, $DC = BF (CPCT) ...(1)$
But $DC = AB ...(2)$
So, $AB = BF$
but $AF = AB + BF$
So, $AF = 2AB$

$\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ\dots(1)$
Now, $\angle\text{A}+\angle\text{C}=2(\angle\text{B}+\angle\text{D})$ (given)$ ...(2)$
Also, $\angle\text{A}=140^\circ,\ \angle\text{D}=60^\circ$
Putting value of $(\angle\text{A}+\angle\text{C})$ from eq. $(2)$ in eq. $(1)$
$2(\angle\text{B}+\angle\text{D})+\angle\text{B}+\angle\text{D}=360^\circ$
$3(\angle\text{B}+\angle\text{D})=360^\circ$
$\Rightarrow\angle\text{B}+\angle\text{D}=120^\circ$
$\Rightarrow\angle\text{B}+60^\circ=120^\circ$
$\Rightarrow\angle\text{B}=60^\circ$

According to the question,
$\text{Area of given quadrilateral}=\frac{1}{2}\times\text{Product of diagonals}$
$=\frac{1}{2}\times48\times32$
$=768\text{m}^2$