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Class 9 NCERT Solutions- Chapter 8 Quadrilaterals - Exercise 8.2

Last Updated : 11 Sep, 2024
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In Chapter 8 of Class 9 Mathematics, students explore quadrilaterals their properties, and various theorems related to them. Exercise 8.2 focuses on solving problems based on theorems like the mid-point theorem and others related to parallelograms. This exercise provides a strong foundation for understanding quadrilaterals which will be built upon in the higher classes. The solutions below are designed to help students understand the correct approach to solving these problems.

Quadrilaterals

A quadrilateral is a four-sided polygon with four vertices and four angles. The Common types of quadrilaterals include parallelograms, rectangles, squares, trapeziums, and rhombuses. Each type of quadrilateral has distinct properties based on its sides, angles, and diagonals. Understanding these properties is key to solving the exercises in this chapter.

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. 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 ∆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)

And since the diagonal of parallelogram bisects each other

so, QS and PR bisects each other.

Question 7. 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)

Conclusion

Exercise 8.2 of Chapter 8 provides essential practice with the theorems related to quadrilaterals reinforcing the student's understanding of different types of the quadrilaterals and their properties. Mastery of these concepts is critical for the solving problems involving the parallelograms and other quadrilateral shapes. This foundation will be beneficial as students progress to the more complex geometry topics.


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