Exercise 8.2 (Revised) - Chapter 8 - Quadrilaterals - Ncert Solutions class 9 - Maths

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Chapter 8 - Quadrilaterals - NCERT Solutions for Class 9 Maths

Ex 8.2 Question 1.

ABCD is a quadrilateral in which $P, Q, R$ and $S$ are the mid-points of sides $A B, B C, C D$ and DA respectively (See figure). AC is a diagonal. Show that:

(i) SR AC and $S R=\frac{1}{2} \mathrm{AC}$
(ii) $\mathbf{P Q}=\mathbf{S R}$
(iii) PQRS is a parallelogram.

Answer.

In $\triangle A B C, P$ is the mid-point of $A B$ and $Q$ is the mid-point of $B C$.

Then $\mathrm{PQ} \| \mathrm{AC}$ and $\mathrm{PQ}=\frac{1}{2} \mathrm{AC}$
(i) In $\triangle A C D, R$ is the mid-point of $C D$ and $S$ is the mid-point of $A D$.

Then $S R \| A C$ and $S R=\frac{1}{2} \mathrm{AC}$
(ii) Since $P Q=\frac{1}{2} \mathrm{AC}$ and $S R=\frac{1}{2} \mathrm{AC}$

Therefore, $\mathrm{PQ}=\mathrm{SR}$
(iii) Since $P Q \| A C$ and $S R \| A C$

Therefore, PQ II SR [two lines parallel to given line are parallel to each other]

Now $P Q=S R$ and $P Q \| S R$

Therefore, PQRS is a parallelogram.

Ex 8.2 Question 2.

ABCD is a rhombus and $P, Q, R, S$ are mid-points of $A B, B C, C D$ and $D A$ respectively. Prove that quadrilateral PQRS is a rectangle.

Answer.

Given: $P, Q, R$ and $S$ are the mid-points of respective sides $A B, B C, C D$ and $D A$ of rhombus. $P Q$, $Q R, R S$ and $S P$ are joined.

To prove: PQRS is a rectangle.

Construction: Join A and C.

Proof: In $\triangle A B C, P$ is the mid-point of $A B$ and $Q$ is the mid-point of $B C$.
$\therefore P Q \| A C$ and $P Q=\frac{1}{2} \mathrm{AC}$

In $\triangle A D C, R$ is the mid-point of $C D$ and $S$ is the mid-point of $A D$.
$\therefore S R \| A C$ and $S R=\frac{1}{2} A C$ $\qquad$
From eq. (i) and (ii), $P Q{ }^{\|} \mathrm{SR}$ and $\mathrm{PQ}=\mathrm{SR}$
$\therefore P Q R S$ is a parallelogram.

Now $A B C D$ is a rhombus. [Given]
$\therefore A B=B C$
$\Rightarrow \frac{1}{2} \mathrm{AB}=\frac{1}{2} \mathrm{BC} \Rightarrow \mathrm{PB}=\mathrm{BQ}$
$\therefore \angle 1=\angle 2$ [Angles opposite to equal sides are equal]

Now in triangles APS and CQR, we have,

$A P=C Q[P$ and $Q$ are the mid-points of $A B$ and $B C$ and $A B=B C]$

Similarly, AS $=\mathrm{CR}$ and $\mathrm{PS}=\mathrm{QR}$ [Opposite sides of a parallelogram]
$
\begin{aligned}
& \therefore \triangle \mathrm{APS} \cong \triangle \mathrm{CQR} \text { [By SSS congreuancy] } \\
& \Rightarrow \angle 3=\angle 4 \text { [By C.P.C.T.] }
\end{aligned}
$

Now we have $\angle 1+\angle \mathrm{SPQ}+\angle 3=180^{\circ}$

And $\angle 2+\angle \mathrm{PQR}+\angle 4=180^{\circ}$ [Linear pairs]
$
\therefore \angle 1+\angle \mathrm{SPQ}+\angle 3=\angle 2+\angle \mathrm{PQR}+\angle 4
$

Since $\angle 1=\angle 2$ and $\angle 3=\angle 4$ [Proved above]
$
\therefore \angle \mathrm{SPQ}=\angle \mathrm{PQR}
$

Now PQRS is a parallelogram [Proved above]
$
\therefore \angle \mathrm{SPQ}+\angle \mathrm{PQR}=180^{\circ}
$
$\qquad$ (iv) [Interior angles]

Using eq. (iii) and (iv),
$
\begin{aligned}
& \angle \mathrm{SPQ}+\angle \mathrm{SPQ}=180^{\circ} \Rightarrow 2 \angle \mathrm{SPQ}=180^{\circ} \\
& \Rightarrow \angle \mathrm{SPQ}=90^{\circ}
\end{aligned}
$

Hence $P Q R S$ is a rectangle.

Ex 8.2 Question 3.

ABCD is a rectangle and $P, Q, R$ and $S$ are the mid-points of the sides $A B, B C, C D$ and $D A$ respectively. Show that the quadrilateral PQRS is a rhombus.

Answer.

Given: $A$ rectangle $A B C D$ in which $P, Q, R$ and $S$ are the mid-points of the sides $A B, B C, C D$ and $D A$ respectively. $\mathrm{PQ}, \mathrm{QR}, \mathrm{RS}$ and $S P$ are joined.

To prove: $P Q R S$ is a rhombus.

Construction: Join AC.

Proof: In $\triangle \mathrm{ABC}, \mathrm{P}$ and $\mathrm{Q}$ are the mid-points of sides $\mathrm{AB}, \mathrm{BC}$ respectively.
$\therefore P Q \| A C$ and $P Q=\frac{1}{2} \mathrm{AC}$ $\qquad$
In $\triangle A D C, R$ and $S$ are the mid-points of sides $C D, A D$ respectively.
$\therefore S R \| A C$ and $S R=\frac{1}{2} \mathrm{AC}$ $\qquad$
From eq. (i) and (ii), $P Q \| S R$ and $P Q=S R$ $\qquad$
$\therefore$ PQRS is a parallelogram.

Now $A B C D$ is a rectangle. [Given]
$\therefore A D=B C$
$\Rightarrow \frac{1}{2} \mathrm{AD}=\frac{1}{2} \mathrm{BC} \Rightarrow \mathrm{AS}=\mathrm{BQ}$ $\qquad$
In triangles APS and BPQ,
$A P=B P[P$ is the mid-point of $A B]$

$
\angle \mathrm{PAS}=\angle \mathrm{PBQ}\left[\text { Each } 90^{\circ}\right]
$

And $A S=B Q[$ From eq. (iv)]
$\therefore \triangle \mathrm{APS} \cong \triangle \mathrm{BPQ}$ [By SAS congruency]
$
\Rightarrow P S=P Q[B y \text { C.P.C.T.] }
$

From eq. (iii) and (v), we get that PQRS is a parallelogram.
$
\Rightarrow P S=P Q
$
$\Rightarrow$ Two adjacent sides are equal.

Hence, $\mathrm{PQRS}$ is a rhombus.

Ex 8.2 Question 4.

ABCD 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 $F$ (See figure). Show that $F$ is the mid-point of BC.

Answer.

Let diagonal BD intersect line EF at point $P$.

In $\triangle \mathrm{DAB}$,
$E$ is the mid-point of $A D$ and $E P \| A B[\because E F \| A B$ (given) $P$ is the part of $E F]$
$\therefore P$ is the mid-point of other side, $B D$ of $\triangle D A B$.
[A line drawn through the mid-point of one side of a triangle, parallel to another side intersects the third side at the mid-point]

Now in $\triangle B C D$,
$P$ is the mid-point of $B D$ and $P F \| D C[\because E F \| A B$ (given) and $A B \| D C$ (given) $]$
$\therefore E F \| D C$ and PF is a part of EF.
$\therefore F$ is the mid-point of other side, $B C$ of $\triangle B C D$. [Converse of mid-point of theorem]

Ex 8.2 Question 5.

In a parallelogram $A B C D, E$ and $F$ are the mid-points of sides $A B$ and CD respectively (See figure). Show that the line segments AF and EC trisect the diagonal BD.

Answer.

Since $E$ and $F$ are the mid-points of $A B$ and $C D$ respectively.
$\therefore A E=\frac{1}{2} \mathrm{AB}$ and $C F=\frac{1}{2}$ CD..........(i)

But $A B C D$ is a parallelogram.
$\therefore A B=C D$ and $A B \|_{D C}$
$\Rightarrow \frac{1}{2} \mathrm{AB}=\frac{1}{2} \mathrm{CD}$ and $\mathrm{AB} \|_{\mathrm{DC}}$
$\Rightarrow A E=F C$ and $A E \|^{F C}[$ From eq. (i)]
$\therefore \mathrm{AECF}$ is a parallelogram.
$\Rightarrow$ FA \| CE $\Rightarrow$ FP \| CQ [FP is a part of FA and CQ is a part of CE]

Since the segment drawn through the mid-point of one side of a triangle and parallel to the other side bisects the third side.

In $\triangle \mathrm{DCQ}, \mathrm{F}$ is the mid-point of $\mathrm{CD}$ and $\Rightarrow \mathrm{FP} \| \mathrm{CQ}$
$\therefore P$ is the mid-point of $D Q$.
$
\Rightarrow D P=P Q
$
Similarly, In $\triangle A B P, E$ is the mid-point of $A B$ and $\Rightarrow E Q \| A P$

$\therefore Q$ is the mid-point of $B P$.
$
\Rightarrow B Q=P Q
$
From eq. (iii) and (iv),
$
D P=P Q=B Q \ldots \ldots . . .(v)
$

Now $B D=B Q+P Q+D P=B Q+B Q+B Q=3 B Q$
$
\Rightarrow \mathrm{BQ}=\frac{1}{3} \mathrm{BD}
$
From eq. (v) and (vi),
$D P=P Q=B Q=\frac{1}{3} B D$
$\Rightarrow$ Points $\mathrm{P}$ and $\mathrm{Q}$ trisects $\mathrm{BD}$.

So $\mathrm{AF}$ and $\mathrm{CE}$ trisects $\mathrm{BD}$.

Ex 8.2 Question 6.

ABC is a triangle right angled at $C$. A line through the mid-point $M$ of hypotenuse $A B$ and parallel to BC intersects AC at D.

Answer.

(i) In $\triangle A B C, M$ is the mid-point of $A B$ [Given]

$M D \| B C$
$\therefore \mathrm{AD}=\mathrm{DC}$ [Converse of mid-point theorem]

Thus $D$ is the mid-point of $A C$.
(ii) $l \| \mathrm{BC}$ (given) consider $\mathrm{AC}$ as a transversal.
$\therefore \angle 1=\angle$ C [Corresponding angles]
$
\Rightarrow \angle 1=90^{\circ}\left[\angle \mathrm{C}=90^{\circ}\right]
$

Thus MD $\perp \mathrm{AC}$.
(iii) In $\triangle \mathrm{AMD}$ and $\triangle \mathrm{CMD}$,
$\mathrm{AD}=\mathrm{DC}$ [proved above]
$\angle 1=\angle 2=90^{\circ}$ [proved above]
$M D=M D$ [common]
$\therefore \triangle \mathrm{AMD} \cong \triangle \mathrm{CMD}$ [By SAS congruency]
$\Rightarrow \mathrm{AM}=\mathrm{CM}$ [By C.P.C.T.]

Given that $M$ is the mid-point of $A B$.

$
\therefore A M=\frac{1}{2} \mathrm{AB}
$
From eq. (i) and (ii),
$
\mathrm{CM}=\mathrm{AM}=\frac{1}{2} \mathrm{AB}
$