2 Marks Questions

Question 12 Marks

$ABCD$ is a rhombus show that diagonal $AC$ bisects $\angle $A as well as $\angle C$ and diagonal $BD$ bisects $\angle B$ as well as $\angle D$

Answer

Given: $ABCD$ is a rhombus

In $\triangle ABC$ and $\triangle ADC,$
$AB = CD$ [Sides of a rhombus]
$BC = DA$ [Sides of a rhombus]
$AC = AC$ [Common]
$\therefore \triangle ABC \cong \triangle ADC$ [By $SSS$ Congruency]
$\therefore \angle C A B=\angle C A D \text { And } \angle A C B=\angle A C D$
Hence $A C$ bisects $\angle A$ as well as $\angle C$
Similarly, by joining $B$ to $D$, we can prove that $\triangle A B D \cong \triangle C B D$
Hence $B D$ bisects $\angle B$ as well as $\angle D$

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Question 22 Marks

Diagonal $AC$ of a parallelogram $\text{ABCD}$ bisects $\angle A ($See figure$).$ Show that:
$i.$ It bisects $\angle C$ also.
$ii. \text{ABCD}$ is a rhombus.

Answer

Diagonal $AC$ bisects $\angle A$ of the parallelogram $\text{ABCD}.$

$i.$ Since $A B \| D C$ and $A C$ intersects them.
$\therefore \angle 1=\angle 3 [$Alternate angles$] ...(i)$
Similarly $\angle 2=\angle 4 \ldots (ii)$
But $\angle 1=\angle 2 [$Given$]$
Thus $A C$ bisects $\angle C$.
$ii. \angle 2=\angle 3=\angle 4=\angle 1$
$\Rightarrow A D=C D [$Sides opposite to equal angles$]$
$\therefore A B=C D=A D=B C$
Hence $\text{ABCD}$ is a rhombus.

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Question 32 Marks

$I, m$ and $n$ are three parallel lines intersected by transversal $p$ and $q$ such that $I, m$ and $n$ cut-off equal intercepts $A B$ and $B C$ on $p$ Show that $I, m$ and $n$ cut-off equal intercepts $D E$ and $E F$ on $q$ also.

Answer

It is given that $A B=B C$ and we need to prove that $D E=E F$.
Construction: Join $A$ to $F$ which intersects m at $G$ as shown below.

The trapezium $ACFD$ is divided into two triangles; namely $\triangle A C F$ and $\triangle A F D$.
In $\triangle A C F$, it is given that $B$ is the mid-point of $A C(A B=B C)$
and $B G \| C F$ (Since $m \| n$ )
So, By the converse of Mid-point Theorem, $G$ is the mid-point of AF.
Now, in $\triangle A F D$, by applying the same argument as $G$ is the mid-point of $A F$, we have $G E \| A D$ so $E$ is the mid-point of $D F$,
$\text { i.e., } D E=E F$
In other words $/, m$ and n cut-off equal intercepts on q also.

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Question 42 Marks

In $\triangle ABC, D, E$ and $F$ are respectively the mid-points of sides $AB, BC$ and $CA$. Show that $\triangle ABC$ is divided into four congruent triangles by joining $D, E$ and $F.$

Answer

As $D$ and $E$ are mid-points of sides $A B$ and $B C$ of the triangle $A B C$ (Mid-points of two sides of a triangle is parallel to the third side),
$D E \| A C$
Similarly, $D F \| B C$ and $E F \| A B$
Therefore $ADEF, BDFE$ and $DFCE$ are all parallelograms.
Now $D E$ is a diagonal of the parallelogram $BDFE,$,
therefore, $\triangle BDE$ $\cong$ $\triangle FED$
Similarly $\triangle DAF$ $\cong$ $\triangle FED$
and $\triangle EFC$ $\cong$ $\triangle FED$
So, all the four triangles are congruent.

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Question 52 Marks

$\text{ABCD}$ is a parallelogram in which $P$ and $Q$ are the mid$-$points of opposite sides $A B$ and $C D$. If $A Q$ intersects $D P$ at $S$ and $B Q$ intersects $C P$ at $R$, show that
$i. \text{APCQ}$ is a parallelogram
$ii. \text{DPBQ}$ is a parallelogram
$iii. \text{PSQR}$ is a parallelogram

Answer

$i.$ Since $\text{ABCD}$ is a parallelogram, we have
$AB=CD$
$\frac{1}{2} AB=\frac{1}{2} DC$
Since $P$ and $Q$ are the mid$-$points of $A B$ and $C D$, we have
$AP=QC$
Also, $AP \| QC$
Therefore, $\text{APCQ}$ is a parallelogram.
$ii.$ Similarly, quadrilateral $\text{DPBQ}$ is a parallelogram,
$ \because D Q \| P B$ and $D Q=P B$
$iii.$ In quadrilateral $\text{PSQR},$
$S P \| Q R(S P$ is a part of $D P$ and $Q R$ is a part of $Q B)$
Similarly, $SQ \| PR$
So, $\text{PSQR}$ is a parallelogram.

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Question 62 Marks

Show that the diagonals of a rhombus are perpendicular to each other.

Answer

Consider the rhombus ABCD .
You know that $A B=B C=C D=D A$
Now, in $\triangle A O D$ and $\triangle C O D, O A=O C$ (Diagonals of a parallelogram bisect each other)
$O D=O D \text { (Common) }$
$A D=C D$
Therefore, $\triangle A O D \cong \triangle C O D$ ($SSS$ congruence rule)
This gives, $\angle A O D=\angle C O D( CPCT )$
But, $\angle A O D+\angle C O D=180^{\circ}$ (Linear pair)
So, $2 \angle A O D=180^{\circ}$
or, $\angle A O D=90^{\circ}$
So, the diagonals of a rhombus are perpendicular to each other.

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Question 72 Marks

Show that each angle of a rectangle is a right angle.

Answer

We know that rectangle is a parallelogram whose one angle is right angle.

Let ABCD be a rectangle. $\angle A = {90^0}$
To prove $\angle B = \angle C = \angle D = {90^0}$
Proof: $\because AD\parallel BC$ and AB is transversal
$\therefore \angle A + \angle B = {180^0}$
${90^0} + \angle B = {180^0}$
$\angle B = {180^0} - {90^0} = {90^0}$
$\angle C = \angle A$
$\therefore {\text{ }}\angle C = {90^0}$
$\angle D = \angle B$
$\therefore {\text{ }}\angle D = {90^0}$
Hence, all the angles of the rectangles is a right angle.

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