# Class 9 NCERT Solutions- Chapter 8 Quadrilaterals – Exercise 8.2

Last Updated : 04 Apr, 2024

### Question 1. 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. Show that:

(i) SR || AC and SR = Â½ AC

(ii) PQ = SR

(iii) PQRS is a parallelogram

Solution:

Given that, P, Q, R and S are the mid points of quadrilateral ABCD

Theorem 8.9: The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

(i) So here, taking âˆ†ACD

we can see S and R are the mid points of side AD and DC respectively.    [Given]

Hence, SR || AC and SR = Â½ AC (NCERT Theorem 8.9)……………………….(1)

(ii) So here, taking âˆ†ACB

we can see P and Q are the mid points of side AB and BC respectively.    [Given]

Hence, PQ || AC and PQ = Â½ AC (NCERT Theorem 8.9)…………………………(2)

From (1) and (2) we can say,

PQ = SR

(iii) so from (i) and (ii) we can say that

PQ || AC and SR || AC

so, PQ || SR and PQ = SR

If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Hence, PQRS is a parallelogram.

### Question 2. 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 that, P, Q, R and S are the mid points of Rhombus ABCD.

Theorem 8.9: The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Construction: Join AC and BD

So here, taking âˆ†ABD

We can see P and S are the mid points of side AB and AD respectively.    [Given]

Hence, PS || BD and PS = Â½ BD (NCERT Theorem 8.9)……………………….(1)

Similarly, is we take âˆ†CBD

We can see R and Q are the mid points of side CD and CB respectively.    [Given]

Hence, RQ || BD and RQ = Â½ BD (NCERT Theorem 8.9)……………………….(2)

So from (1) and (2), we conclude that

PS || RQ and PS = RQ

If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Hence, PQRS is a parallelogram.

Now in âˆ†ACD

We can see S and R are the mid points of side AD and CD respectively.    [Given]

Hence, SR || AC and RS = Â½ AC (NCERT Theorem 8.9)

from (2) RQ || BD and RQ = Â½ BD (NCERT Theorem 8.9)

Hence, OGSH is a parallelogram.

âˆ HOG = 90Â° (Diagonal of rhombus intersect at 90Â°)

So âˆ HSG = 90Â° (opposite angle of a parallelogram are equal)

As, PQRS is a parallelogram having vertices angles equal to 90Â°.

Hence, PQRS is a Rectangle.

### Question 3. 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 that, P, Q, R and S are the mid points of Rectangle ABCD.

Construction: Join AC

Theorem 8.9 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

So here, taking âˆ†ACD

we can see S and R are the mid points of side AD and DC respectively.    [Given]

Hence, SR || AC and SR = Â½ AC (NCERT Theorem 8.9)……………………….(1)

Now, taking âˆ†ACB

we can see P and Q are the mid points of side AB and BC respectively.    [Given]

Hence, PQ || AC and PQ = Â½ AC (NCERT Theorem 8.9)…………………………(2)

So from (1) and (2), we conclude that

SR || PQ and SR = PQ

Hence, PQRS is a parallelogram. (If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram)

Now in âˆ†QBP and âˆ†QCR

QC = QB (Q is the mid point of BC)

RC = PB (opposite sides are equal, hence half length is also equal)

âˆ QCR = âˆ QBP (Each 90Â°)

âˆ†QBP â‰… âˆ†QCR (By SAS congruency)

QR = QP (By C.P.C.T.)

As PQRS is a parallelogram and having adjacent sides equal

Hence, PQRS is a rhombus

### Question 4. ABCD is a trapezium in which AB || 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). Show that F is the mid-point of BC.

Solution:

Let O be the point of intersection of lines BD and EF

Theorem 8.10 : The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

So here, taking âˆ†ADB

we can see S is the mid point of side AD and ED || AB   [Given]

Hence, OD = Â½ BD ………..(NCERT Theorem 8.10)

Now, taking âˆ†BCD

we can see O is the mid point of side BD and OF || AB   [Proved and Given]

Hence, CF = Â½ BC…….. (NCERT Theorem 8.10)

Hence proved!!

### Question 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD.

Solution:

Theorem 8.10 : The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

Given, E and F are the mid points of side AB and CD of parallelogram ABCD.

As, AB || CD and AB = CD  (opposite sides of parallelogram)………..(1)

AE = CF  (halves of opposite sides of parallelogram)………………………(2)

from (1) and (2)

AECF is a parallelogram

Hence, AF || EC

Now taking âˆ†APB

we can see E is the mid point of side AB and EF || AP   [Given and proved]

Hence, BQ = PQ………..(NCERT Theorem 8.10)………………….(1)

Now taking âˆ†CQD

we can see F is the mid point of side CD and CQ || FP   [Given and proved]

Hence, DP = PQ………..(NCERT Theorem 8.10)………………..(2)

From (1) and (2) we conclude that,

BQ = PQ = DQ

Hence, we can say that line segments AF and EC trisect the diagonal BD

### Question 6. 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

(i) D is the mid-point of AC

(ii) MD âŠ¥ AC

(iii) CM = MA = Â½ AB

Solution:

Theorem 8.10 : The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

(i) while taking âˆ†ABC

we can see M is the mid point of side AB and DM || BC   [Given]

This implies, DC= AD ……….. (NCERT Theorem 8.10)

Hence, D is the mid-point of AC.

(ii) As we know MD || BC and AC is transversal

This implies, âˆ ACB = âˆ ADM = 90Â°

Hence, MD âŠ¥AC

(iii) Considering âˆ†ADM and âˆ†CDM

AD = CD (D is the mid point of AC (Proved))

âˆ CDM = âˆ ADM (proved, MD âŠ¥AC)

DM = DM (common)

âˆ†ADM â‰… âˆ†CDM (By SAS congruency)

CM = AM (By C.P.C.T.)

CM = AM = Â½ AB (M is the mid point of AB)

## Deleted Questions

### Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Solution:

Theorem 8.9 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

So here, taking âˆ†ACD

we can see S and R are the mid points of side AD and DC respectively.    [Given]

Hence, SR || AC and SR = Â½ AC (NCERT Theorem 8.9)……………………….(1)

Now, taking âˆ†ACB

we can see P and Q are the mid points of side AB and BC respectively.    [Given]

Hence, PQ || AC and PQ = Â½ AC (NCERT Theorem 8.9)…………………………(2)

So from (1) and (2), we conclude that

SR || PQ and SR = PQ

Hence, PQRS is a parallelogram. (If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram)

And since the diagonal of parallelogram bisects each other

so, QS and PR bisects each other.

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