NCERT Solutions Class 9 Chapter 8 Quadrilaterals Exercise 8.2 (original) (raw)

Last Updated : 26 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 = ½ 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 ∆ACD

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

Hence, SR _|| AC and SR =_ **½ AC (NCERT Theorem 8.9)............................(1)

****(ii)** So here, taking ∆ACB

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

Hence, PQ || AC and PQ = ½ 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 ∆ABD

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

Hence, PS || BD and PS = ½ BD (**NCERT Theorem 8.9)............................(1)

Similarly, is we take ∆CBD

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

Hence, RQ || BD and RQ = ½ 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 ∆ACD

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

Hence, SR || AC and RS = ½ AC (**NCERT Theorem 8.9)

from (2) RQ || BD and RQ = ½ BD (**NCERT Theorem 8.9)

Hence, **OGSH is a parallelogram.

∠HOG = 90°(**Diagonal of rhombus intersect at 90°)

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

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

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 ∆ACD

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

Hence, SR || AC and SR = ½ AC (NCERT Theorem 8.9)............................(1)

Now, taking ∆ACB

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

Hence, PQ || AC and PQ = ½ 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 ∆QBP and ∆QCR

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

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

∠QCR = ∠QBP (Each 90°)

**∆QBP ≅ ∆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 ∆ADB

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

Hence, **OD = ½ BD ...........(NCERT Theorem 8.10)

Now, taking ∆BCD

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

Hence, CF = ½ 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 ∆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 ∆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 ⊥ AC**

****(iii) CM = MA = ½ 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 ∆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, ∠ACB = ∠ADM = 90°

**Hence, MD ⊥AC

****(iii)** Considering ∆ADM and ∆CDM

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

∠CDM = ∠ADM (proved, MD ⊥AC)

DM = DM (common)

∆ADM ≅ ∆CDM (**By SAS congruency)

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

**CM = AM = ½ AB (M is the mid point of AB)