NCERT Solutions
Class 9
Maths
Chapter 8 Quadrilaterals

NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals discusses quadrilateral, a four-sided polygon with four vertices and four angles. It has several types, such as trapezium, kite, rhombus, square, parallelogram, and rectangle. It helps students grasp the idea of quadrilaterals and their fundamental properties. This chapter effectively goes over all the important equations and theorems related to angles, diagonals, and sums. 

This article will provide students with high-quality NCERT solutions for Class 9 Maths Chapter 8 Quadrilaterals exercises designed specifically to help students overcome their challenges, improve their problem-solving skills, and boost their confidence in tackling complex mathematical problems. These solutions are developed by ALLEN's subject experts and include the entire chapter concepts as per the latest CBSE curriculum.

1.0Download NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals 

ALLEN'S Experts lucidly curated the solutions so that students can explore some practical applications of quadrilaterals in detail through various examples and sample questions. For a more precise idea about Quadrilaterals, and NCERT Solutions, students can download the NCERT Class 9 Maths Chapter 8 pdf solution below.


NCERT Class 8 Maths Chapter 9: Quadrilaterals

Topics Covered in NCERT Class 7 Maths Chapter 1 - Quadrilaterals

The following is a list of the subjects addressed in CBSE Class 7 Maths Chapter 1: Quadrilaterals.

  • Introduction to Quadrilaterals
  • Angle Sum Property of a Quadrilateral
  • Types of Quadrilaterals
  • Properties of a Parallelogram
  • Another Condition for a Quadrilateral to be a Parallelogram
  • The Mid-point Theorem
  • Summary

2.0Quadrilaterals in NCERT Solutions for Class 9 Maths, Chapter 8: All Exercises

Exercise

Total Questions

Exercise 8.1

7 Questions

Exercise 8.2

6 Questions

3.0NCERT Questions with Solutions for Class 9 Maths Chapter 8 - Detailed Solutions

Exercise: 8.1

  1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
    Sol. Given : ABCD is a parallelogram with diagonal diagonal BD To prove : ABCD is a rectangle. Proof : In triangle ABC and ABD, [Common] [Given] . Sides of a [By SSS congruency] [By C.P.C.T.] and cuts them, the sum of the interior angle of the same side of transversal is ] From eq. (i) and (ii), So, ABCD is a parallelogram with one of the angles equal to , Hence, ABCD is a rectangle.
  2. Show that the diagonals of a square are equal and bisect each other at right angles. Sol. Given: ABCD is a square. To Prove : (i) AC = BD (ii) AC and BD bisect each other at right angles.
    Proof: In and , [Common] [Opp. sides of square ABCD] Each is a square [SAS Rule] ... (i) [C.P.C.T.] In and [Opp. sides of square ABCD ] [Alternate angles as and transversal AC intersects them] [Alternate angles as and transversal BD intersects them] [ASA Rule] and ...(ii)[C.P.C.T.] So, 0 is the midpoint of and . Now, In and But [Linear pair [ proved earlier] and BD bisect each other at right angles.
  3. Diagonal of a parallelogram bisects . Show that (i) it bisects also (ii) ABCD is a rhombus.
  4. Diagonal bisects of the parallelogram ABCD. To prove: (i) AC bisects (ii) ABCD is a rhombus Proof: (i) Since and AC intersects them. [Alternate angles] ...(i) Similarly, But [Given] [Using eq. (i), (ii) and (iii)] Thus, AC bisects . (ii) since (using (i) and (iii)) [Sides opposite to equal angles] Also, ABCD is a parallelogram. and Hence, ABCD is a rhombus.
  5. ABCD is a rectangle in which diagonal AC bisects as well as . Show that (i) ABCD is a square (ii) diagonal BD bisects as well as . Sol. Given : ABCD is a rectangle in which diagonal AC bisects as well as .
    To prove : (i) ABCD is a square (ii) diagonal BD bisects as well as . Proof: (i) and transversal AC intersects them. So, [ Alternate angles] But bisects [Sides opposite to equal angles of a triangle are equal] But and [Opposite side of a rectangle] Also, is a rectangle] Hence, ABCD is a square. (ii) and , Hence, diagonal BD bisects as well as
  6. In parallelogram , two points and Q are taken on diagonal BD such that BQ. Show that: (i) (ii) (iii) (iv) (v) APCQ is a parallelogram
    Sol. (i) In and , we have [Given] [Opposite sides of parallelogram ABCD] [Pair of alternate angles] [SAS congruence criteria] (ii) Then, by CPCT, we have AP = CQ (iii) We can prove [as we have done in (i)] (iv) By CPCT, we have (v) Now, we have and Hence, APCQ is a parallelogram.
  7. is a parallelogram and and are perpendiculars from vertices A and C on diagonal BD. Show that (i) (ii) AP = CQ
    Sol. Given : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices and on diagonal respectively. To prove : (i) (ii) Proof: (i) In and , Opposite side of ॥ gm ABCD] and transversal BD intersect them] Each [AAS Rule] (ii) [C.P.C.T.]
  8. ABCD is a trapezium in which and . Show that (figure) (i) (ii) (iii) (iv) diagonal diagonal BD
    Sol. Given :
    ABCD is a trapezium. and To Prove: (i) (ii) (iii) (iv) Diagonal AC = Diagonal BD Construction : Draw CE || AD and extend AB to intersect CE at E. Proof: (i) As AECD is a parallelogram. [By construction] But AD = BC [Given] [Angles opposite to equal sides are equal] Now, [Co-Interior angles] And [Linear pair] (ii) [Alternate interior angles] [Opposite angles of a parallelogram] But [ is an isosceles triangle] (iii) In and ,
  9. The angles of quadrilateral are in the ratio . Find all the angles of the quadrilateral. Sol. Let the four angles of the quadrilateral be and . [Sum of all the angles of quadrilateral is ] Hence, the angles of the quadrilateral are , and .
  10. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. Sol. Given : ABCD is a quadrilateral where diagonals AC and BD meet at 0 , such that and To Prove: Quadrilateral ABCD is a rhombus, i.e., Proof: In and , [Given] [Common] Each [SAS Rule] [C.P.C.T.]
    Similarly, we can prove that From (i), (ii), (iii) and (iv), we obtain, Quadrilateral ABCD is a rhombus.
  11. Show that if the diagonals of quadrilateral are equal and bisect each other at right angles, then it is a square. Sol. Given : The diagonals AC and BD of a quadrilateral ABCD are equal and bisect each other at right angles. To prove : Quadrilateral is a square. Proof:
    In and , [Vertically Opposite Angles] [SAS Rule] [C.P.C.T.] [C.P.C.T.] Now, and Quadrilateral ABCD is a |gm. In and , Now, is a parallelogram and Again, in and , [Given] is a Common rule] [C.P.C.T.] [Opposite sides of ||gm ABCD] and transversal AB intersects them. [Sum of consecutive interior angles on the same side of the transversal is ] Similarly, is a square.
  12. is a rhombus. Show that diagonal AC bisects as well as and diagonal BD bisects as well as . Sol Given : ABCD is a rhombus and AC and BD are its diagonals. To prove : (i) Diagonal AC bisects as well as . (ii) Diagonal BD bisects as well as .
    Proof: (i) In (sides of Rhombus) So, (Angle opposite to equal sides are equal) But (Alternate angles as ) So, But (Alternate angles as ) So, by (1) So, AC bisect by (1) So, AC bisect (ii) In
    (Sides of Rhombus) So, (Angle opposite to equal sides are equal) But (Alternate angles as ) So, (Alternate angles as ) So, by (2) So, BD bisect by (2) So, BD bisect
  13. In and EF and BC | EF. Vertices A, B and C are joined to vertices and F respectively. Show that : (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) and (iv) quadrilateral ACFD is a parallelogram (v) (vi)
    Sol. Given : & To Prove (i) ABED is a parallelogram. (ii) BEFC is a parallelogram. (iii) and (iv) ACFD is a parallelogram (v) (vi) Proof (i) In quad. ABED And AB || DE [Given] ABED is a parallelogram (ii) In quad. BEFC [Given] And [Given] is a parallelogram. (iii) As ABED is a parallelogram. and Also, BEFC is a parallelogram and From (i) and (ii), we get and (iv) and ACFD is a parallelogram. (v) As ACFD is a parallelogram. (vi) In and , AB [Given] Given] [Proved] [By SSS congruency]

4.0Exercise: 8.2

  1. is a quadrilateral in which and are mid points of the sides , CD and DA (fig.) and AC is a diagonal. Show that (i) and SR (ii) (iii) is a parallelogram.
    Sol. Given : ABCD is a quadrilateral in which and are mid-points of and is a diagonal. To prove : (i) and (ii) (iii) is a parallelogram. Proof: (i) , is the mid-point of and is the mid-point of DC and [By Mid-point theorem] (ii) In , is the mid-point of and is the mid-point of and [By Mid-point theorem] But from (i) (iii) From (ii)] SR | AC [From (i)] [Two lines parallel to the same line are parallel to each other] Also, [From (ii)] PQRS is a parallelogram. [A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length]
  2. is a rhombus and and are the mid points of sides and DA respectively. Show that the quadrilateral PQRS is a rectangle. Sol.
    Given : and are the mid-points of respective sides and of rhombus. and SP are joined. To prove : PQRS is a rectangle. Construction : Join A and C. Proof: In is the mid-point of and is the mid-point of . and (By midpoint theorem) In is the mid-point of and is the mid-point of . and . (By midpoint theorem) From eq. (i) and (ii), and is a parallelogram. Now ABCD is a rhombus [Given] [Angles opposite to equal sides are equal] Now in triangles APS and CQR, we have, [ P and Q are the mid-points of AB and BC and Similarly, AS = CR and PS = QR [Opposite sides of a parallelogram] [By SSS congruency] [By C.P.C.T.] Now, we have And Since and [Proved above] Now PQRS is a parallelogram [Proved above] [Co-Interior angles] Using eq. (iii) and (iv), Hence, is a rectangle.
  3. is a rectangle and and are mid-points of the sides and DA respectively. Show that the quadrilateral is a rhombus. Sol. Given : A rectangle ABCD in which and are the mid-points of the sides , and DA respectively. PQ, QR, RS and SP are joined. To prove : PQRS is a rhombus.
    Construction : Join AC. Proof : In and Q are the midpoints of sides respectively. and [Mid- Point Theorem] In and S are the mid-points of sides respectively. and [Mid-Point Theorem] From eq.(i) and (ii), and is a parallelogram. Now ABCD is a rectangle. In triangles APS and BPQ, is the mid-point of AB ] Each And AS = BQ [From eq. (iv)] [By SAS congruency] [By C.P.C.T.] From eq.(iii) and (v), we get that PQRS is a parallelogram. Two adjacent sides are equal. Hence, PQRS is a rhombus.
  4. ABCD is a trapezium in which is a diagonal and E is the mid-point of AD . A line is drawn through E parallel to intersecting BC at F (fig.). Show that F is the mid-point of BC .
    Sol. Line and passes through E.
    Line meets BC at F and BD at G . In is mid-point of and . is mid-point of . [Converse of Mid Point Theorem] Also, and Fis mid-point of . is mid-point of [Converse of Mid Point Theorem]
  5. In a parallelogram and F are the mid-points of sides AB and CD respectively (fig.). Show that the line segments AF and EC trisect the diagonal BD .
    Sol. Given : ABCD is a parallelogram. E and F are midpoints of and respectively.
    To prove: Proof: Since E and F are the mid-points of and respectively. and But ABCD is a parallelogram. and and and [From eq. (i)] AECF is a parallelogram. [FP is a part of and is a part of CE] ... (ii) Since the line segment drawn through the mid-point of one side of a triangle and parallel to the other side bisects the third side. In is the mid-point of and is the mid-point of DQ . Similarly, In is the mid-point of AB and is the mid-point of BP . From eq.(iii) and (iv), Now, From eq (v) and (vi), Points P and Q trisects BD . So, AF and CE trisects BD. Hence, AF and CE trisect the diagonal BO.
  6. ABC is a triangle right angled at C . A line through the mid-point M of hypotenuse and parallel to intersects at . Show that (i) D is the mid-point of AC (ii) (iii) Sol. (i) Through M, we draw line . intersects AC at D. is a midpoint of is mid-point of . [By converse of mid-point theorem]
    (ii) [Corresponding angles] . (iii) In CMD and ; and Each ] Therefore, [SAS congruence rule] ; Also So,
  7. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. Sol. P,Q,R and are the mid-points of the sides and of the quadrilateral ABCD.
    We have to prove that, and bisects each other. Now, join PQ, QR, RS and PS. Here, we can prove that is a parallelogram (as in solution 1). Now, PR and QS are the diagonals of the parallelogram PQRS. and QS bisect each other as diagonals of parallelogram bisect each other.

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