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

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Chapter 8 - Quadrilaterals - NCERT Solutions for Class 9 Maths

Example 1 :

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

Let us recall what a rectangle is.
A rectangle is a parallelogram in which one angle is a right angle.
Let $\mathrm{ABCD}$ be a rectangle in which $\angle \mathrm{A}=90^{\circ}$.
We have to show that $\angle \mathrm{B}=\angle \mathrm{C}=\angle \mathrm{D}=90^{\circ}$
We have, $A D \| B C$ and $A B$ is a transversal (see Fig. 8.6).
So, $\angle \mathrm{A}+\angle \mathrm{B}=180^{\circ}$ (Interior angles on the same side of the transversal)

But, $\quad \angle \mathrm{A}=90^{\circ}$

So, $\quad \angle \mathrm{B}=180^{\circ}-\angle \mathrm{A}=180^{\circ}-90^{\circ}=90^{\circ}$
Now, $\quad \angle \mathrm{C}=\angle \mathrm{A}$ and $\angle \mathrm{D}=\angle \mathrm{B}$
(Opposite angles of the parallellogram)

So, $\angle \mathrm{C}=90^{\circ}$ and $\angle \mathrm{D}=90^{\circ}$.

Therefore, each of the angles of a rectangle is a right angle.

Example 2 :

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

Consider the rhombus ABCD (see Fig. 8.7). You know that $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$ (Why?) Now, in $\triangle A O D$ and $\triangle$ COD,
$
\begin{aligned}
& \mathrm{OA}=\mathrm{OC} \text { (Diagonals of a parallelogram } \\
& \text { bisect each other) } \\
& \mathrm{OD}=\mathrm{OD} \\
& \text { (Common) } \\
& \mathrm{AD}=\mathrm{CD} \\
&
\end{aligned}
$

Therefore, $\triangle \mathrm{AOD} \cong \triangle \mathrm{COD}$
(SSS congruence rule)

This gives, $\angle \mathrm{AOD}=\angle \mathrm{COD}$
(CPCT)

But, $\angle \mathrm{AOD}+\angle \mathrm{COD}=180^{\circ}$ (Linear pair)
So, $\quad 2 \angle \mathrm{AOD}=180^{\circ}$
or, $\angle \mathrm{AOD}=90^{\circ}$
So, the diagonals of a rhombus are perpendicular to each other.

Example 3 :

ABC is an isosceles triangle in which $\mathrm{AB}=\mathrm{AC}$. $\mathrm{AD}$ bisects exterior angle $\mathrm{PAC}$ and $\mathrm{CD} \| \mathrm{AB}$ (see Fig. 8.8). Show that
(i) $\angle \mathrm{DAC}=\angle \mathrm{BCA}$ and
(ii) $\mathrm{ABCD}$ is a parallelogram.

Solution :

(i) $\triangle \mathrm{ABC}$ is isosceles in which $\mathrm{AB}=\mathrm{AC}$ (Given)
So, $\angle \mathrm{ABC}=\angle \mathrm{ACB} \quad$ (Angles opposite to equal sides)
Also, $\quad \angle \mathrm{PAC}=\angle \mathrm{ABC}+\angle \mathrm{ACB}$
(Exterior angle of a triangle)
or, $\quad \angle \mathrm{PAC}=2 \angle \mathrm{ACB}$
Now, AD bisects $\angle \mathrm{PAC}$.
So, $\quad \angle \mathrm{PAC}=2 \angle \mathrm{DAC}$
Therefore,
$
\begin{aligned}
2 \angle \mathrm{DAC} & =2 \angle \mathrm{ACB} \quad[\text { From (1) and (2)] } \\
\text { or, } \quad \angle \mathrm{DAC} & =\angle \mathrm{ACB}
\end{aligned}
$

(ii) Now, these equal angles form a pair of alternate angles when line segments BC and $\mathrm{AD}$ are intersected by a transversal $\mathrm{AC}$.
So, $\quad \mathrm{BC} \| \mathrm{AD}$
Also, BA $\|$ CD
(Given)

Now, both pairs of opposite sides of quadrilateral $\mathrm{ABCD}$ are parallel.
So, ABCD is a parallelogram.
Example 4 :

Two parallel lines $l$ and $m$ are intersected by a transversal $p$ (see Fig. 8.9). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.

Solution :

It is given that $\mathrm{PS} \| \mathrm{QR}$ and transversal $p$ intersects them at points $\mathrm{A}$ and C respectively.
The bisectors of $\angle \mathrm{PAC}$ and $\angle \mathrm{ACQ}$ intersect at $\mathrm{B}$ and bisectors of $\angle \mathrm{ACR}$ and $\angle \mathrm{SAC}$ intersect at D.
We are to show that quadrilateral $\mathrm{ABCD}$ is a rectangle.
Now, $\quad \angle \mathrm{PAC}=\angle \mathrm{ACR}$
(Alternate angles as $l \| m$ and $p$ is a transversal)
So, $\quad \frac{1}{2} \angle \mathrm{PAC}=\frac{1}{2} \angle \mathrm{ACR}$
i.e., $\quad \angle \mathrm{BAC}=\angle \mathrm{ACD}$

These form a pair of alternate angles for lines $\mathrm{AB}$ and $\mathrm{DC}$ with $\mathrm{AC}$ as transversal and they are equal also.

So,
$
\mathrm{AB} \| \mathrm{DC}
$

Similarly,
$
\mathrm{BC} \| \mathrm{AD}
$
(Considering $\angle \mathrm{ACB}$ and $\angle \mathrm{CAD}$ )
Therefore, quadrilateral $\mathrm{ABCD}$ is a parallelogram.
Also,
$
\angle \mathrm{PAC}+\angle \mathrm{CAS}=180^{\circ}
$
(Linear pair)

So,
$
\frac{1}{2} \angle \mathrm{PAC}+\frac{1}{2} \angle \mathrm{CAS}=\frac{1}{2} \times 180^{\circ}=90^{\circ}
$
or,
$
\begin{aligned}
\angle \mathrm{BAC}+\angle \mathrm{CAD} & =90^{\circ} \\
\angle \mathrm{BAD} & =90^{\circ}
\end{aligned}
$
So, $\mathrm{ABCD}$ is a parallelogram in which one angle is $90^{\circ}$.
Therefore, $\mathrm{ABCD}$ is a rectangle.

Example 5 :

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

Let P, Q, R and S be the points of intersection of the bisectors of $\angle \mathrm{A}$ and $\angle \mathrm{B}, \angle \mathrm{B}$ and $\angle \mathrm{C}, \angle \mathrm{C}$ and $\angle \mathrm{D}$, and $\angle \mathrm{D}$ and $\angle \mathrm{A}$ respectively of parallelogram $\mathrm{ABCD}$ (see Fig. 8.10).
In $\triangle$ ASD, what do you observe?
Since $\mathrm{DS}$ bisects $\angle \mathrm{D}$ and $\mathrm{AS}$ bisects $\angle \mathrm{A}$, therefore,

$
\begin{aligned}
\angle \mathrm{DAS}+\angle \mathrm{ADS} & =\frac{1}{2} \angle \mathrm{A}+\frac{1}{2} \angle \mathrm{D} \\
& =\frac{1}{2}(\angle \mathrm{A}+\angle \mathrm{D}) \\
& =\frac{1}{2} \times 180^{\circ} \quad(\angle \mathrm{A} \text { and } \angle \mathrm{D} \text { are interior angles } \\
& \left.=90^{\circ} \quad \text { on the same side of the transversal }\right)
\end{aligned}
$

Similarly, it can be shown that $\angle \mathrm{APB}=90^{\circ}$ or $\angle \mathrm{SPQ}=90^{\circ}$ (as it was shown for $\angle \mathrm{DSA})$. Similarly, $\angle \mathrm{PQR}=90^{\circ}$ and $\angle \mathrm{SRQ}=90^{\circ}$.
So, $\mathrm{PQRS}$ is a quadrilateral in which all angles are right angles.
Can we conclude that it is a rectangle? Let us examine. We have shown that $\angle \mathrm{PSR}=\angle \mathrm{PQR}=90^{\circ}$ and $\angle \mathrm{SPQ}=\angle \mathrm{SRQ}=90^{\circ}$. So both pairs of opposite angles are equal.
Therefore, $\mathrm{PQRS}$ is a parallelogram in which one angle (in fact all angles) is $90^{\circ}$ and so, $\mathrm{PQRS}$ is a rectangle.

Example 6 :

In $\triangle \mathrm{ABC}, \mathrm{D}, \mathrm{E}$ and $\mathrm{F}$ are respectively the mid-points of sides $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{CA}$ (see Fig. 8.18). Show that $\triangle A B C$ is divided into four congruent triangles by joining $\mathrm{D}, \mathrm{E}$ and $\mathrm{F}$.
Solution :

As D and E are mid-points of sides $\mathrm{AB}$ and $\mathrm{BC}$ of the triangle $\mathrm{ABC}$, by Theorem 8.8 ,
$\mathrm{DE} \| \mathrm{AC}$

Similarly, $\quad \mathrm{DF} \| \mathrm{BC}$ and $\mathrm{EF} \| \mathrm{AB}$

Therefore $\mathrm{ADEF}, \mathrm{BDFE}$ and $\mathrm{DFCE}$ are all parallelograms.
Now DE is a diagonal of the parallelogram BDFE,
therefore, $\quad \Delta \mathrm{BDE} \cong \triangle \mathrm{FED}$
Similarly $\quad \triangle \mathrm{DAF} \cong \triangle \mathrm{FED}$
and $\quad \triangle \mathrm{EFC} \cong \triangle \mathrm{FED}$
So, all the four triangles are congruent.
Example 7:

l,m and $n$ are three parallel lines intersected by transversals $p$ and $q$ such that $l, m$ and $n$ cut off equal intercepts $\mathrm{AB}$ and $\mathrm{BC}$ on $p$ (see Fig. 8.19). Show that $l, m$ and $n$ cut off equal intercepts $\mathrm{DE}$ and $\mathrm{EF}$ on $q$ also.
Solution :

We are given that $\mathrm{AB}=\mathrm{BC}$ and have to prove that $\mathrm{DE}=\mathrm{EF}$.

Let us join $\mathrm{A}$ to $\mathrm{F}$ intersecting $m$ at $\mathrm{G}$.
The trapezium ACFD is divided into two triangles; namely $\triangle \mathrm{ACF}$ and $\triangle \mathrm{AFD}$.

In $\triangle \mathrm{ACF}$, it is given that $\mathrm{B}$ is the mid-point of $\mathrm{AC}(\mathrm{AB}=\mathrm{BC})$
and $\quad \mathrm{BG} \| \mathrm{CF} \quad$ (since $m \| n$ ).
So, $\mathrm{G}$ is the mid-point of AF (by using Theorem 8.9)
Now, in $\triangle \mathrm{AFD}$, we can apply the same argument as G is the mid-point of AF, $\mathrm{GE} \| \mathrm{AD}$ and so by Theorem $8.9, \mathrm{E}$ is the mid-point of $\mathrm{DF}$,
i.e., $\quad \mathrm{DE}=\mathrm{EF}$.

In other words, $l, m$ and $n$ cut off equal intercepts on $q$ also.