3 Marks Question

Question 13 Marks

Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisect each other.

Answer

Given: $A$ quadrilateral $ABCD$ in which $EG$ and $FH$ are the line-segments joining the mid-points of opposite sides of a quadrilateral.

To prove: $EG$ and $FH$ bisect each other.
Construction: Join $AC, EF, FG, GH$ and $HE.$
Proof: $\ln A B C, E$ and $F$ are the mid-points of respective sides $A B$ and $B C$.
$\therefore E F \| A C$ and $E F=\frac{1}{2} A C..........(i)$
Similarly, in $ADC,$
$G$ and $H$ are the mid-points of respective sides $C D$ and $A D$.
$\therefore HG \| AC$ and $HG =\frac{1}{2} AC$.
From eq. $(i)$ and $(ii),$ we get,
$EF \| HG$ and $EF = HG$
$\therefore EFGH$ is a parallelogram.
Since the diagonals of a parallelogram bisect each other, therefore line segments (i.e. diagonals) $EG$ and $FH$ (of parallelogram $EFGH$) bisect each other.
Hence Proved.

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Question 23 Marks

In a parallelogram $ABCD$, $E$ and $F$ are the mid-points of sides $AB$ and $CD$ respectively. Show that the line segments $AF$ and $EC$ trisect the diagonal $BD.$

Answer

Given: In a parallelogram $ABCD, E$ and $F$ are the mid-points of sides $AB$ and $CD$ respectively.
To Prove: Line segments $AF$ and $EC$ intersect the diagonal $BD$.

Proof: $A B || CD . . .$[Opp. sides of $\| gm ABCD]$
$\therefore A E \| F C \ldots(1)$
As $A B=D C \ldots$ [Opp. sides of $\| gm ABCD]$
$\therefore \frac{1}{2} A B=\frac{1}{2} D C \ldots$ [Halves of equals are equal]
$\therefore A E=C F \ldots(2)$
According to $(1)$ and $(2)$
$AECF$ is a parallelogram . . [A quadrilateral is a parallelogram if a pair of opp. sides is parallel and of equal length]
$\therefore EC || AF . .$ [Opp. sides of $\|$ gm $AECF] . . (3)$
In $\triangle D B C$,
As $F$ is the mid-point of $D C$ and $F P \| C Q \ldots[A s E C \| A F]$
$P$ is the mid-point of $D Q \ldots$ [By converse of mid-point theorem]
$\therefore D P=P Q \ldots(4)$
Similarly, In $\triangle B A P$,
$B Q=P Q \ldots(5)$
$D P=P Q=B Q \ldots \text { From (4) and (5) }$
$\therefore$ Line segments $AF$ and $EC$ trisect the diagonal $BD .$

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Question 33 Marks

$ABCD$ is a trapezium in which $AB || DC, BD$ is a diagonal and $E$ is the mid-point of $AD$. A line is drawn through $E$ parallel to $AB$ intersecting $BC$ at $F$. Show that F is the mid-point of $BC.$

Answer

Given: $A B C D$ is a trapezium in which $A B \| D C, B D$ is a diagonal and $E$ is the mid-point of $A D . A$ line is drawn through $E$ parallel to $A B$ intersecting $B C$ at $F$.
To prove: $F$ is the mid-point of $B C$.

Proof: Let $DB$ intersect $EF$ at $G .$
In $\triangle D A B$,
As $E$ is the mid-point of $D A$ and $E G \| A B$
$\triangle G$ is the mid-point of $D B \ldots$ [By converse of mid-point theorem]
Again in $\triangle B D C$,
As $G$ is the mid-point of $B D$ and $G F\|A B\| D C$
$\triangle F$ is the mid point of $B C \ldots$ [By converse of mid-point theorem]

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Question 43 Marks

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer

Given: Diagonals of quadrilateral intersect each other at right angles.

To Prove: Quadrilateral is a rhombus.
Proof: In $\triangle A O B$ and $\triangle A O D$,
$A O=A O \ldots$ [Common]
$O B=O D \ldots$ [Given]
$\angle A O B=\angle A O D \ldots$ [Each $\left.90^{\circ}\right]$
$\therefore \angle A O B \cong \triangle A O D \ldots$ [By SAS property]
$\therefore A B=A D \ldots$ [c.p.c.t.] $\ldots (1)$
Similarly, we can prove that
$A B=B C \ldots \text { (2) }$
$B C=C D \ldots \text { (3) }$
$C D=A D \ldots \text { (4) }$
From $(1), (2), (3)$ and $(4)$
$A B=B C=C D=D A$
Since opposite sides of quadrilateral $A B C D$ are equal, it can be said that $A B C D$ is a parallelogram. Since all sides of a parallelogram $A B C D$ are equal, it can be said that $A B C D$ is a rhombus.

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Question 53 Marks

If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer

Given: The diagonals of a parallelogram are equal.

To prove: Parallelogram is a rectangle.
Proof: In $\triangle A C B$ and $\triangle B D A$,
$A C=B D \ldots$ [Given]
$AB = BA \ldots$ [Common]
$B C=A D \ldots$ [Opposite sides of parallelogram]
$\therefore \triangle ACB \cong \triangle B D A \ldots$ [By SSS property]
$\therefore \angle A B C=\angle B A D \ldots$ [c.p.c.t.] $\ldots (1)$
As $A D \| B C \ldots$ [Opposite sides of parallelogram]
transversal $A B$ intersects them.
$\therefore \angle B A D+\angle A B C=180^{\circ} \ldots$ [Sum of interior angle on the same side of a transversal] $\ldots (2)$.
$\angle B A D=\angle A B C=90^{\circ} \ldots[$ [From $(1)$ and $(2)]$
$\therefore \angle A=90^{\circ}$
$\therefore$ Parallelogram $A B C D$ is a rectangle.

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Question 63 Marks

$\text{ABCD}$ is a parallelogram and $AP$ and $CQ$ are perpendiculars from vertices $A$ and $C$ on diagonal $BD$ respectively.

Show that :
$i. \triangle APB \cong \triangle CQD$
$ii. AP = CQ.$

Answer

Given: $\text{ABCD}$ is a parallelogram and $A P$ and $C Q$ are perpendicular from vertices $A$ and $C$ on diagonal $B D$ respectively.
To Prove:
$\triangle APB \cong \triangle CQD$
$A P=C Q$
Proof:
In $\triangle A P B$ and $\triangle C Q D$
$A B=C D \ldots [$Opp. sides of $\| gm \text{ABCD}]$
$\angle A B P=\angle C D Q \ldots [$Alternate interior angles for $A B \| C D]$
$\therefore \angle APB =\angle CQD \ldots\left[\right.$ Each $\left.90^{\circ}\right]$
$\triangle APB \cong \triangle C Q D \ldots [$By $\text{AAS}$ rule$]$
As $\triangle A P B \cong \triangle C Q D \ldots [$As proved above$]$
$\therefore A P=C Q \ldots [\text {c.p.c.t.}]$

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Question 73 Marks

The angles of a quadrilateral are in the ratio $3 : 5 : 9 : 13$. Find all angles of the quadrilateral.

Answer

Let the quadrilateral $ABCD, \angle A=3x, \angle B= 5x, \angle C = 9x$ and $\angle D = 13x.$
Since, sum of all the angles of a quadrilateral = $360^\circ $
$ \therefore \angle A + \angle B + \angle C + \angle D = 360^\circ$
$\Rightarrow 3x + 5x + 9x + 13x = 360^\circ $
$\Rightarrow $$ 30x = 360^\circ $$\Rightarrow $ $x = 12^\circ $
Now $\angle $ A = $3x = 3 \times 12 = 36^\circ $
$\angle $ B =$ 5x = 5 \times 12 = 60^\circ $
$\angle $ C =$ 9x = 9 \times 12 = 108^\circ $
And$\angle $ D = $13x = 13 \times 12 = 156^\circ $
Hence angles of given quadrilateral are$ 36^\circ ,60^\circ ,108^\circ and 156^\circ .$

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Question 83 Marks

Show that the bisectors of angles of a parallelogram form a rectangle.

Answer

Let $P , Q , R$ and S be the points of intersection of the bisectors of $\angle A$ and $\angle B, \angle B$ and $\angle C, \angle C$ and $\angle D$, and $\angle D$ and $\angle A$ respectively of parallelogram $A B C D$.
$\text { In } \triangle A S D$
Since DS bisects $\angle D$ and AS bisects $\angle A$, therefore,
$\angle D A S+\angle A D S=\frac{1}{2} \angle A+\frac{1}{2} \angle D$
$=\frac{1}{2}(\angle A+\angle D)$
$=\frac{1}{2} \times 180^{\circ} \text { ( } \angle A \text { and } \angle D$ are interior angles on the same side of the transversal)
$=90^{\circ}$
Also, $\angle D A S+\angle A D S+\angle D S A=180^{\circ}$ (Angle sum property of a triangle)
$\text { or, } 90^{\circ}+\angle D S A=180^{\circ}$
or, $\angle D S A=90^{\circ}$
So, $\angle P S R=90^{\circ}$ (Being vertically opposite to $\angle D S A$ )
Similarly, it can be shown that $\angle A P B=90^{\circ}$ or $\angle S P Q=90^{\circ}$ (as it was shown for $\angle D S A$ ).
Similarly, $\angle P Q R=90^{\circ}$ and $\angle S R Q=90^{\circ}$.
So, PQRS is a quadrilateral in which all angles are right angles.
We have shown that $\angle P S R=\angle P Q R=90^{\circ}$ and $\angle S P Q=\angle S R Q=90^{\circ}$.
So both pairs of opposite angles are equal.
Therefore, PQRS is a parallelogram in which one angle (in fact all angles) is $90^{\circ}$ and so, PQRS is a rectangle.

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Question 93 Marks

In Figure $\text{ABC}$ is an isosceles triangle in which $AB = AC. AD$ bisects exterior $\angle PAC$ and $CD \| AB$. Show that
$i.\angle DAC = \angle BCA$
$ii. \text{ABCD}$ is a parallelogram

Answer

$i. \triangle A B C$ is isosceles in which $A B=A C ($Given$)$
So, $\angle A B C=\angle A C B ($Angles opposite to equal sides$)$
Also, $\angle P A C=\angle A B C+\angle A C B ($Exterior angle of a triangle$)$
or, $\angle P A C=2 \angle A C B \ldots (1)$
Now, $A D$ bisects $\angle P A C$.
So, $\angle P A C=2 \angle D A C \ldots (2)$
Therefore, $2 \angle D A C=2 \angle A C B$ [From $(1)$ and $(2)]$
or, $\angle D A C=\angle A C B$
$ii.$ Now, these equal angles $\angle D A C=\angle A C B$ form a pair of alternate angles when line segments $BC$ and $AD$ are intersected by a transversal $A C$.
So, $B C \| A D$
Also, $BA \| CD ($Given$)$
Now, both pairs of opposite sides of quadrilateral $\text{ABCD}$ are parallel.

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Question 103 Marks

If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Answer

Note that in Fig. 8.5, it is given that $\mathrm{OA}=\mathrm{OC}$ and $\mathrm{OB}=\mathrm{OD}$.
So,
$\triangle \mathrm{AOB} \cong \triangle \mathrm{COD} \text { (Why?) }$
Therefore, $\angle \mathrm{ABO}=\angle \mathrm{CDO}$ (Why?)
From this, we get $A B \| C D$
Similarly, $\quad B C \| A D$
Therefore ABCD is a parallelogram.
Let us now take some examples.

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Question 113 Marks

If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.

Answer

There is yet another property of a parallelogram. Let us study the same. Draw a parallelogram $A B C D$ and draw both its diagonals intersecting at the point $O$ (see Fig. 8.4).
Measure the lengths of $\mathrm{OA}, \mathrm{OB}, \mathrm{OC}$ and $\mathrm{OD}$.
What do you observe? You will observe that
$\mathrm{OA}=\mathrm{OC} \text { and } \mathrm{OB}=\mathrm{OD} \text {. }$
or, $\mathrm{O}$ is the mid-point of both the diagonals.
Repeat this activity with some more parallelograms.
Each time you will find that $\mathrm{O}$ is the mid-point of both the diagonals.
So, we have the following theorem :

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