NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 - Quadrilaterals

Access Quadrilaterals Class 9 Chapter 8 Exercise: 8.2

Q1 (i) ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $8.20$ ). AC is a diagonal. Show that : $SR\parallel AC$ and $SR=\frac{1}{2}AC$

Solution:

Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $8.29$ ). AC is a diagonal.

To prove:

$SR\parallel AC$ and $SR=\frac{1}{2}AC$

Proof: In $\mathrm{△}$ ACD,

S is the midpoint of DA. (Given)

R is the midpoint of DC. (Given)

By the midpoint theorem,

$SR\parallel AC$ and $SR=\frac{1}{2}AC$

Q1 (ii) ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $8.20$ ). AC is a diagonal. Show that : $PQ=SR$

Solution:

Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $8.29$ ). AC is a diagonal.

To prove : $PQ=SR$

Proof: In $\mathrm{△}$ ACD,

S is the midpoint of DA. (Given)

R is the midpoint of DC. (Given)

By the mid-point theorem,

$SR\parallel AC$ and $SR=\frac{1}{2}AC$ ...........(1)

In $\mathrm{△}$ ABC,

P is the midpoint of AB. (Given)

Q is the midpoint of BC. (Given)

By the mid-point theorem,

$PQ\parallel AC$ and $PQ=\frac{1}{2}AC$ ..............(2)

From 1 and 2, we get

$PQ\parallel SR$ and $PQ=SR=\frac{1}{2}AC$

Thus, $PQ=SR$

Q1 (iii) ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $8.20$ ). AC is a diagonal. Show that : PQRS is a parallelogram.

Solution:

Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $8.29$ ). AC is a diagonal.

To prove: PQRS is a parallelogram.

Proof: In PQRS,

Since,

$PQ\parallel SR$ and $PQ=SR$ .

So, PQRS is a parallelogram.

Q2 ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Solution:

Given: ABCD is a rhombus in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC, BD are diagonals.

To prove: the quadrilateral PQRS is a rectangle.

Proof: In $\mathrm{△}$ ACD,

S is the midpoint of DA. (Given)

R is the midpoint of DC. (Given)

By the midpoint theorem,

$SR\parallel AC$ and $SR=\frac{1}{2}AC$ ........(1)

In $\mathrm{△}$ ABC,

P is the midpoint of AB. (Given)

Q is the midpoint of BC. (Given)

By the mid-point theorem,

$PQ\parallel AC$ and $PQ=\frac{1}{2}AC$ ...........(2)

From (1) and (2), we get

$PQ\parallel SR$ and $PQ=SR=\frac{1}{2}AC$

Thus, $PQ=SR$ and $PQ\parallel SR$

So, the quadrilateral PQRS is a parallelogram.

Similarly, in $\mathrm{△}$ BCD,

Q is the midpoint of BC. (Given)

R is the midpoint of DC. (Given)

By the midpoint theorem,

$QR\parallel BD$

So, QN || LM .........(5)

LQ || MN ..........(6) (Since, PQ || AC)

From (5) and (6), we get

LMPQ is a parallelogram.

Hence, $\mathrm{\angle }$ LMN= $\mathrm{\angle }$ LQN (opposite angles of the parallelogram)

But, $\mathrm{\angle }$ LMN= 90° (Diagonals of a rhombus are perpendicular)

So, $\mathrm{\angle }$ LQN = 90°

Thus, a parallelogram whose one angle is a right angle is a rectangle. Hence, PQRS is a rectangle.

Q3 ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Solution:

Given: ABCD is a rectangle and P, Q, R, and S are midpoints of the sides AB, BC, CD and DA, respectively.

To prove: the quadrilateral PQRS is a rhombus.

Proof:

In $\mathrm{△}$ ACD,

S is the midpoint of DA. (Given)

R is the midpoint of DC. (Given)

By the midpoint theorem,

$SR\parallel AC$ and $SR=\frac{1}{2}AC$ ..........(1)

In $\mathrm{△}$ ABC,

P is the midpoint of AB. (Given)

Q is the midpoint of BC. (Given)

By the midpoint theorem,

$PQ\parallel AC$ and $PQ=\frac{1}{2}AC$ ..........(2)

From (1) and (2), we get

$PQ\parallel SR$ and $PQ=SR=\frac{1}{2}AC$

Thus, $PQ=SR$ and $PQ\parallel SR$

So, the quadrilateral PQRS is a parallelogram.

Similarly, in $\mathrm{△}$ BCD,

Q is the midpoint of BC. (Given)

R is the midpoint of DC. (Given)

By the midpoint theorem,

$QR\parallel BD$ and $QR=\frac{1}{2}BD$ ...........(3)

AC = BD........(4) (diagonals )

From (2), (3) and (4), we get

PQ = QR

Thus, a parallelogram whose adjacent sides are equal is a rhombus. Hence, PQRS is a rhombus.

Q4 ABCD is a trapezium in which $AB\parallel DC$ , 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 (see Fig. $8.21$ ). Show that F is the mid-point of BC.

Solution:

Given: ABCD is a trapezium in which $AB\parallel DC$ , 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 (see Fig. $8.30$ ).

To prove: F is the midpoint of BC.

In $\mathrm{△}$ ABD,

E is the midpoint of AD. (Given)

EG || AB (Given)

By the converse of the midpoint theorem,

G is the midpoint of BD.

In $\mathrm{△}$ BCD,

G is the midpoint of BD. (Proved above)

FG || DC (Given)

By the converse of the midpoint theorem,

F is the midpoint of BC.

Q5 In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. $8.22$ ). Show that the line segments AF and EC trisect the diagonal BD.

Solution:

Given: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD, respectively

To prove: the line segments AF and EC trisect the diagonal BD.

Proof: In quadrilateral ABCD,

AB = CD (Given)

$\frac{1}{2}AB=\frac{1}{2}CD$

$\Rightarrow AE=CF$ (E and F are midpoints of AB and CD)

In quadrilateral AECF,

AE = CF (Given)

AE || CF (Opposite sides of a parallelogram)

Hence, AECF is a parallelogram.

In $\mathrm{△}$ DCQ,

F is the midpoint of DC. (given )

FP || CQ (AECF is a parallelogram)

By the converse of the midpoint theorem,

P is the midpoint of DQ.

DP = PQ........(1)

In $\mathrm{△}$ ABP,

E is the midpoint of AB. (given )

EQ || AP (AECF is a parallelogram)

By the converse of the midpoint theorem,

Q is the midpoint of PB.

PQ = QB........(2)

From (1) and (2), we have

DP = PQ = QB.

Hence, the line segments AF and EC trisect the diagonal BD.

Q6 (i) ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that D is the mid-point of AC .

Solution:

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

To prove: D is the midpoint of AC.

Proof: In $\mathrm{△}$ ABC,

M is the midpoint of AB. (Given)

DM || BC (Given)

By the converse of the midpoint theorem,

D is the midpoint of AC.

Q6(ii) ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that $MD\perp AC$

Solution:

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

To prove : $MD\perp AC$

Proof : $\mathrm{\angle }$ ADM = $\mathrm{\angle }$ ACB (Corresponding angles)

$\mathrm{\angle }$ ADM= ${90}^{\circ }$ . ( $\mathrm{\angle }$ ACB = ${90}^{\circ }$ )

Hence, $MD\perp AC$ .

Q6 (iii) ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that $CM=MA=\frac{1}{2}AB$

Solution:

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

To prove : $CM=MA=\frac{1}{2}AB$

Proof : In $\mathrm{△}$ ABC,

M is the midpoint of AB. (Given)

DM || BC (Given)

By the converse of the midpoint theorem,

D is the midpoint of AC, i.e. AD = DC.

In $\mathrm{△}$ AMD and $\mathrm{△}$ CMD,

AD = DC (proved above)

$\mathrm{\angle }$ ADM = $\mathrm{\angle }$ CDM (Each right angle)

DM = DM (Common)

$\mathrm{△}$ AMD $\cong$ $\mathrm{△}$ CMD (By SAS)

AM = CM (CPCT)

But , $AM=\frac{1}{2}AB$

Hence, $CM=MA=\frac{1}{2}AB$ .