Exercise 8.1 (Revised) - Chapter 8 - Quadrilaterals - Ncert Solutions class 9 - Maths

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Chapter 8 - Quadrilaterals - NCERT Solutions for Class 9 Maths

Ex 8.1 Question 1.

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

Answer.

Given: $A B C D$ is a parallelogram with diagonal $A C=$ diagonal $B D$

To prove: $A B C D$ is a rectangle.

Proof: In triangles $\mathrm{ABC}$ and $\mathrm{ABD}$,
$\mathrm{AB}=\mathrm{AB}[$ Common $]$
$\mathrm{AC}=\mathrm{BD}[$ Given $]$
$\mathrm{AD}=\mathrm{BC}\left[\right.$ opp. Sides of $\left.\mathrm{a} \|_{\mathrm{gm}}\right]$
$\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{BAD}[\mathrm{By}$ SSS congruency]
$\Rightarrow \angle \mathrm{DAB}=\angle \mathrm{CBA}[\mathrm{By}$ C.P.C.T. $]$

But $\angle \mathrm{DAB}+\angle \mathrm{CBA}=180^{\circ}$
$\left[\because A D \|_{\mathrm{BC}}\right.$ and $\mathrm{AB}$ cuts them, the sum of the interior angles of the same side of transversal is $\left.180^{\circ}\right]$

From eq. (i) and (ii),
$
\angle \mathrm{DAB}=\angle \mathrm{CBA}=90^{\circ}
$

Hence $A B C D$ is a rectangle.

Ex 8.1 Question 2.

Show that the diagonals of a square are equal and bisect each other at right angles.

Answer.

Given: $A B C D$ is a square. $A C$ and $B D$ are its diagonals bisect each other at point $O$.

To prove: $\mathrm{AC}=\mathrm{BD}$ and $\mathrm{AC} \perp \mathrm{BD}$ at point $\mathrm{O}$.

Proof: In triangles $\mathrm{ABC}$ and $\mathrm{BAD}$,
$A B=A B[$ Common $]$
$\angle \mathrm{ABC}=\angle \mathrm{BAD}=90^{\circ}$
$\mathrm{BC}=\mathrm{AD}[$ Sides of a square $]$
$\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{BAD}$ [By SAS congruency]
$\Rightarrow \mathrm{AC}=\mathrm{BD}[\mathrm{By}$ C.P.C.T.]Hence proved.

Now in triangles $\mathrm{AOB}$ and $\mathrm{AOD}$,
$\mathrm{AO}=\mathrm{AO}[$ Common $]$
$\mathrm{AB}=\mathrm{AD}[$ Sides of a square $]$
$\therefore \triangle \mathrm{AOB} \cong \triangle \mathrm{AOD}[\mathrm{By}$ SSS congruency]
$\angle \mathrm{AOB}=\angle \mathrm{AOD}[\mathrm{By}$ C.P.C.T. $]$

But $\angle \mathrm{AOB}+\angle \mathrm{AOD}=180^{\circ}$ [Linear pair]

But $\angle \mathrm{AOB}+\angle \mathrm{AOD}=180^{\circ}$ [Linear pair]
$
\therefore \angle \mathrm{AOB}=\angle \mathrm{AOD}=90^{\circ}
$
$\Rightarrow \mathrm{OA} \perp \mathrm{BD}$ or $\mathrm{AC} \perp \mathrm{BD}$ Hence proved.

Ex 8.1 Question 3.

$\text {Diagonal AC of a parallelogram ABCD bisects } \angle \mathbf{A} \text { (See figure). Show that: }$

(i) It bisects $\angle \mathbf{C}$ also.
(ii) ABCD is a rhombus.

Answer.

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

(i) Since $\mathrm{AB} \|_{\mathrm{DC}}$ and $\mathrm{AC}$ intersects them.
$\therefore \angle 1=\angle 3$ [Alternate angles] $\qquad$
Similarly $\angle 2=\angle 4$ $\qquad$
But $\angle 1=\angle 2$ [Given] $\qquad$
$\therefore \angle 3=\angle 4$ [Using eq. (i), (ii) and (iii)]

Thus AC bisects $\angle \mathrm{C}$.
(ii) $\angle 2=\angle 3=\angle 4=\angle 1$
$\Rightarrow \mathrm{AD}=\mathrm{CD}[$ Sides opposite to equal angles $]$
$\therefore A B=C D=A D=B C$

Hence $A B C D$ is a rhombus.

Ex 8.1 Question 4.

ABCD is a rectangle in which diagonal $\mathbf{A C}$ bisects $\angle \mathbf{A}$ as well as $\angle \mathbf{C}$. Show that:
(i) ABCD is a square.
(ii) Diagonal BD bisects both $\angle$ B as well as $\angle$ D.

Answer.

$A B C D$ is a rectangle. Therefore $A B=D C \ldots . . . . .$. (i)

And $B C=A D$

Also $\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=\angle \mathrm{D}=90^{\circ}$

(i) In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{ADC}$
$
\angle 1=\angle 2 \text { and } \angle 3=\angle 4
$
[AC bisects $\angle \mathrm{A}$ and $\angle \mathrm{C}$ (given)]
$\mathrm{AC}=\mathrm{AC}[$ Common $]$
$\therefore \Delta \mathrm{ABC} \cong \triangle \mathrm{ADC}[\mathrm{By}$ ASA congruency]
$
\Rightarrow \mathrm{AB}=\mathrm{AD}
$

From eq. (i) and (ii), $A B=B C=C D=A D$

Hence $A B C D$ is a square.

(ii) In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{ADC}$
$A B=B A[$ Since $A B C D$ is a square $]$
$A D=D C[$ Since $A B C D$ is a square $]$
$
\begin{aligned}
& \mathrm{BD}=\mathrm{BD} \text { [Common] } \\
& \therefore \Delta \mathrm{ABD} \cong \Delta \mathrm{CBD} \text { [By SSS congruency] } \\
& \Rightarrow \angle \mathrm{ABD}=\angle \mathrm{CBD} \text { [By C.P.C.T.].........(iii) }
\end{aligned}
$

And $\angle \mathrm{ADB}=\angle \mathrm{CDB}[\mathrm{By}$ C.P.C.T.] $\qquad$ (iv)

From eq. (iii) and (iv), it is clear that diagonal $\mathrm{BD}$ bisects both $\angle \mathrm{B}$ and $\angle \mathrm{D}$.

Ex 8.1 Question 5.

In parallelogram ABCD, two points $P$ and $Q$ are taken on diagonal BD such that $D P=B Q$ (See figure). Show that:

(i) $\triangle$ APD $\cong \triangle$ CQB
(ii) $\mathbf{A P}=\mathbf{C Q}$
(iii) $\triangle \mathrm{AQB} \cong \triangle \mathrm{CPD}$
(iv) $A Q=C P$
(v) APCQ is a parallelogram.

Answer.

(i) In $\triangle \mathrm{APD}$ and $\triangle C Q B$,
$D P=B Q[$ Given $]$
$\angle \mathrm{ADP}=\angle \mathrm{QBC}$ [Alternate angles (AD $\|_{\mathrm{BC}}$ and $\mathrm{BD}$ is transversal)]
$\mathrm{AD}=\mathrm{CB}$ [Opposite sides of parallelogram]
$\therefore \triangle \mathrm{APD} \cong \triangle \mathrm{CQB}$ [By SAS congruency]
(ii) Since $\triangle \mathrm{APD} \cong \triangle \mathrm{CQB}$
$
\Rightarrow \mathrm{AP}=\mathrm{CQ}[\mathrm{By} \text { C.P.C.T.] }
$
(iii) In $\triangle \mathrm{AQB}$ and $\triangle \mathrm{CPD}$,
$
\mathrm{BQ}=\mathrm{DP}[\text { Given }]
$

$\angle \mathrm{ABQ}=\angle \mathrm{PDC}$ [Alternate angles (AB $\|^{\mathrm{CD}}$ and $\mathrm{BD}$ is transversal)]
$A B=C D[$ Opposite sides of parallelogram $]$
$\therefore \triangle \mathrm{AQB} \cong \triangle \mathrm{CPD}[$ By SAS congruency]
(iv) Since $\triangle A Q B \cong \triangle C P D$
$\Rightarrow \mathrm{AQ}=\mathrm{CP}[\mathrm{By}$ C.P.C.T. $]$
(v) In quadrilateral APCQ,
$\mathrm{AP}=\mathrm{CQ}[$ proved in part (i)]
$\mathrm{AQ}=\mathrm{CP}[$ proved in part (iv)]

Since opposite sides of quadrilateral APCQ are equal.

Hence APCQ is a parallelogram.

Ex 8.1 Question 6.

ABCD is a parallelogram and AP and CQ are the perpendiculars from vertices $A$ and $C$ on its diagonal BD (See figure). Show that:

(i) $\triangle$ APB $\cong \triangle C Q D$
(ii) $\mathbf{A P}=\mathbf{C Q}$

Answer.

Given: $\mathrm{ABCD}$ is a parallelogram. $\mathrm{AP} \perp \mathrm{BD}$ and $\mathrm{CQ} \perp \mathrm{BD}$

To prove: (i) $\triangle \mathrm{APB} \cong \triangle C Q D$ (ii) $\mathrm{AP}=\mathrm{CQ}$

Proof: (i) In $\triangle \mathrm{APB}$ and $\triangle C Q D$,
$\angle 1=\angle 2$ [Alternate interior angles]
$\mathrm{AB}=\mathrm{CD}[$ Opposite sides of a parallelogram are equal]
$\angle \mathrm{APB}=\angle \mathrm{CQD}=90^{\circ}$
$\therefore \triangle \mathrm{APB} \cong \triangle \mathrm{CQD}[$ By ASA Congruency]
(ii) Since $\triangle \mathrm{APB} \cong \triangle C Q D$
$\therefore A P=C Q[B y ~ C . P . C . T$.

Ex 8.1 Question 7.

$\text {} A B C D \text { is a trapezium in which } A B \| C D \text { and } A D=B C \text { (See figure). Show that: }$

(i) $\angle \mathrm{A}=\angle \mathrm{B}$
(ii) $\angle \mathbf{c}=\angle \mathrm{D}$
(iii) $\triangle \mathrm{ABC} \cong \triangle \mathrm{BAD}$
(iv) Diagonal AC $=$ Diagonal BD

Answer.

Given: $A B C D$ is a trapezium.
$A B \| C D$ and $A D=B C$

To prove: (i) $\angle \mathrm{A}=\angle \mathrm{B}$
(ii) $\angle \mathrm{C}=\angle \mathrm{D}$
(iii) $\triangle \mathrm{ABC} \cong \triangle \mathrm{BAD}$
(iv) Diag. $\mathrm{AC}=$ Diag. $\mathrm{BD}$

Construction: Draw CE $\|$ AD and extend
$A B$ to intersect $C E$ at $E$.

Proof: (i) As AECD is a parallelogram.[By construction]
$
\therefore A D=E C
$

But $A D=B C$ [Given]
$
\therefore B C=E C
$
$\Rightarrow \angle 3=\angle 4$ [Angles opposite to equal sides are equal]

Now $\angle 1+\angle 4=180^{\circ}$ [Interior angles]

And $\angle 2+\angle 3=180^{\circ}$ [Linear pair]
$
\Rightarrow \angle 1+\angle 4=\angle 2+\angle 3
$

$
\begin{aligned}
& \Rightarrow \angle 1=\angle 2[\because \angle 3=\angle 4] \\
& \Rightarrow \angle \mathrm{A}=\angle \mathrm{B}
\end{aligned}
$
(ii) $\angle 3=\angle$ C[Alternate interior angles]

And $\angle \mathrm{D}=\angle 4$ [Opposite angles of a parallelogram]

But $\angle 3=\angle 4$ [ $\triangle \mathrm{BCE}$ is an isosceles triangle]
$
\therefore \angle C=\angle D
$
(iii) In $\triangle A B C$ and $\triangle B A D$,
$
\begin{aligned}
& \mathrm{AB}=\mathrm{AB} \text { [Common] } \\
& \angle 1=\angle 2 \text { [Proved] } \\
& \mathrm{AD}=\mathrm{BC} \text { [Given] } \\
& \therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{BAD}[\mathrm{By} \text { SAS congruency] }
\end{aligned}
$
(iv) We had observed that,
$
\therefore \triangle \mathrm{ABC} \cong \triangle B A D
$

$\Rightarrow \mathrm{AC}=\mathrm{BD} \text { [By C.P.C.T.] }$