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
- If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Sol. Given : ABCD is a parallelogram with diagonal AC= diagonal BD To prove : ABCD is a rectangle. Proof : In triangle ABC and ABD, AB=AB [Common] AC=BD [Given] AD=BC[0pp. Sides of a ∥gm] ∴ΔABC≅△BAD [By SSS congruency] ⇒∠DAB=∠CBA [By C.P.C.T.] [∵AD∥BC and AB cuts them, the sum of the interior angle of the same side of transversal is 180∘ ] ∠DAB+∠CBA=180∘ From eq. (i) and (ii), ∠DAB=∠CBA=90∘ So, ABCD is a parallelogram with one of the angles equal to 90∘, Hence, ABCD is a rectangle.
- 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 △ABC and △BAD, AB=BA [Common] BC=AD [Opp. sides of square ABCD] ∠ABC=∠BAD[ Each =90∘(∵ABCD is a square )] ∴△ABC≅△BAD [SAS Rule] ∴AC=BD... (i) [C.P.C.T.] In △AOD and △BOC AD=CB [Opp. sides of square ABCD ] ∠OAD=∠OCB [Alternate angles as AD∥BC and transversal AC intersects them] ∠ODA=∠OBC [Alternate angles as AD∥BC and transversal BD intersects them] △AOD≅△COB [ASA Rule] ∴OA=OC and OB=OD ...(ii)[C.P.C.T.] So, 0 is the midpoint of AC and BD. Now, In △AOB and △COB AB=BCOA=OCOB=OB∴△AOB≅△COB∴∠AOB=∠BOC [Given] [from (ii)] [ Common] [ By SSS Rule] [ C.P.C.T] But ∠AOB+∠BOC=180∘ [Linear pair ] ∠AOB+∠AOB=180∘ [ ∠AOB=∠BOC proved earlier] ⇒2∠AOB=180∘ ⇒∠AOB=2180∘=90∘ ∴∠AOB=∠BOC=90∘ ∴AC and BD bisect each other at right angles.
- Diagonal AC of a parallelogram ABCD bisects ∠A. Show that
(i) it bisects ∠C also
(ii) ABCD is a rhombus.
- Diagonal AC bisects ∠A of the parallelogram ABCD. To prove: (i) AC bisects ∠C (ii) ABCD is a rhombus Proof: (i) Since AB∣∣DC and AC intersects them. ∴∠1=∠3 [Alternate angles] ...(i) Similarly, ∠2=∠4 But ∠1=∠2 [Given] ∴∠3=∠4 [Using eq. (i), (ii) and (iii)] Thus, AC bisects ∠C. (ii) since ∠2=∠3 (using (i) and (iii)) ⇒AD=CD [Sides opposite to equal angles] Also, ABCD is a parallelogram. ⇒AD=BC and AB=CD ∴AB=CD=AD=BC Hence, ABCD is a rhombus.
- ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that
(i) ABCD is a square
(ii) diagonal BD bisects ∠B as well as ∠D.
Sol. Given : ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.
To prove : (i) ABCD is a square (ii) diagonal BD bisects ∠B as well as ∠D. Proof: (i) ∵AB∥DC and transversal AC intersects them. So, ∠BAC=∠DCA [ Alternate angles] But ∠BAC=∠DAC[∵AC bisects ∠A] ∴∠DCA=∠DAC ⇒DA=CD [Sides opposite to equal angles of a triangle are equal] But AB=CD and DA=BC [Opposite side of a rectangle] ∴AB=BC=CD=DA Also, ∠A=∠B=∠C=∠D=90∘ [∵ABCD is a rectangle] Hence, ABCD is a square. (ii) In△BAD and △BCD, BA=BC[∵ABCD is a square ]AD=CD[∵ABCD is a square ]BD=BD[Common]∴ΔBAD≅△BCD[By SSS congruence rule ]∴∠ABD=∠CBD [By C.P.C.T.] ∠ADB=∠CDB[ By C.P.C.T. ] Hence, diagonal BD bisects ∠B as well as ∠D
- In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP= BQ. Show that:
(i) △APD≅△CQB
(ii) AP=CQ
(iii) △AQB≅△CPD
(iv) AQ=CP
(v) APCQ is a parallelogram
Sol. (i) In △APD and △CQB, we have DP=BQ [Given] AD=CB [Opposite sides of parallelogram ABCD] ∠ADP=∠CBQ [Pair of alternate angles] ⇒△APD≅△CQB [SAS congruence criteria] (ii) Then, by CPCT, we have AP = CQ (iii) We can prove ΔAQB≅△CPD [as we have done in (i)] (iv) By CPCT, we have AQ=CP (v) Now, we have AP=CQ and AQ=CP Hence, APCQ is a parallelogram.
- ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that
(i) △APB≅△CQD
(ii) AP = CQ
Sol. Given : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively. To prove : (i) △APB≅△CQD (ii) AP=CQ Proof: (i) In △APB and △CQD, AB=CD[ Opposite side of ॥ gm ABCD] ∠ABP=∠CDQ [∵AB∣∣DC and transversal BD intersect them] ∠APB=∠CQD [ Each =90∘] ∴△APB≅△CQD [AAS Rule] (ii) AP=CQ [C.P.C.T.]
- ABCD is a trapezium in which AB∥CD and AD=BC. Show that (figure)
(i) ∠A=∠B
(ii) ∠C=∠D
(iii) △ABC≅△BAD
(iv) diagonal AC= diagonal BD
Sol. Given :ABCD is a trapezium. AB∥CD and AD=BC To Prove: (i) ∠A=∠B (ii) ∠C=∠D (iii) ΔABC≅△BAD (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] ∴AD=EC But AD = BC [Given] ∴BC=EC ⇒∠3=∠4 [Angles opposite to equal sides are equal] Now, ∠1+∠4=180∘ [Co-Interior angles] And ∠2+∠3=180∘ [Linear pair] ⇒∠1+∠4=∠2+∠3 ⇒∠1=∠2[∵∠3=∠4] ⇒∠A=∠B (ii) ∠3=∠BCD [Alternate interior angles] ∠ADC=∠4 [Opposite angles of a parallelogram] But ∠3=∠4 [ △BCE is an isosceles triangle] ∴∠BCD=∠ADC ∴∠C=∠D (iii) In △ABC and △BAD, AB=AB [Common] ∠1=∠2 [Proved] AD=BC[ Given ]∴△ABC≅△BAD[ By SAS congruency] ⇒AC=BD
- The angles of quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral. Sol. Let the four angles of the quadrilateral be 3x,5x,9x and 13x. 3x+5x+9x+13x=360∘ [Sum of all the angles of quadrilateral is 360∘ ] ⇒30x=360∘ ⇒x=12∘ Hence, the angles of the quadrilateral are 3×12∘=36∘,5×12∘=60∘, 9×12∘=108∘ and 13×12∘=156∘.
- 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 AO=OC,OB=OD and AC⊥BD
To Prove: Quadrilateral ABCD is a rhombus,
i.e., AB=BC=CD=DA
Proof: In △AOB and △AOD,
OB=OD
[Given]
AO=AO
[Common]
∠AOB=∠AOD[ Each =90∘]
∴△AOB≅△AOD [SAS Rule]
∴AB=AD
[C.P.C.T.]
Similarly, we can prove that AB=BC BC=CD CD=AD From (i), (ii), (iii) and (iv), we obtain, AB=BC=CD=DA ∴ Quadrilateral ABCD is a rhombus.
- Show that if the diagonals of a 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 ABCD is a square.
Proof:
In △AOD and △BOC, OA=OC [Given] OD=OB [Given] ∠AOD=∠COB [Vertically Opposite Angles] ∴△AOD≅△COB [SAS Rule] ∴AD=BC [C.P.C.T.] ∠ODA=∠OBC [C.P.C.T.] ∴AD∥BC Now, AD=CB and AD∥CB ∴ Quadrilateral ABCD is a |gm. In △AOB and △AOD, AO=AO[ Common ]OB=OD[ Given ]∠AOB=∠AOD[ Each =90∘( Given )]∴ΔAOB≅△AOD[ SAS Rule ]∴AB=AD Now, ∴ABCD is a parallelogram and AB=AD Again, in △ABC and △BAD, AC=BD [Given] BC=AD[∵ABCD is a ∥gm] AB=BA[ Common ] ∴△ABC≅△BAD[SSS rule] ∴∠ABC=∠BAD [C.P.C.T.] ∵AD∣∣BC [Opposite sides of ||gm ABCD] and transversal AB intersects them. ∴∠ABC+∠BAD=180∘ [Sum of consecutive interior angles on the same side of the transversal is 180∘ ] ∴∠ABC=∠BAD=90∘ Similarly, ∠BCD=∠ADC=90∘ ∴ABCD is a square.
- ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Sol Given : ABCD is a rhombus and AC and BD are its diagonals.
To prove :
(i) Diagonal AC bisects ∠A as well as ∠C.
(ii) Diagonal BD bisects ∠B as well as ∠D.
Proof: (i) In △ABC AB=BC (sides of Rhombus) So, ∠2=∠4 (Angle opposite to equal sides are equal) But ∠2=∠3 (Alternate angles as AB∥CD ) So, ∠2=∠3=∠4 But ∠1=∠4 (Alternate angles as AD∥BC ) So, ∠1=∠2=∠3=∠4 ∠1=∠2 by (1) So, AC bisect ∠A ∠3=∠4 by (1) So, AC bisect ∠C (ii) In △ABDAB=AD (Sides of Rhombus) So, ∠5=∠7 (Angle opposite to equal sides are equal) But ∠7=∠6 (Alternate angles as AD∥BC ) So, ∠5=∠6=∠7 ∠5=∠8 (Alternate angles as AB∥CD ) So, ∠5=∠6=∠7=∠8 ∠5=∠6 by (2) So, BD bisect ∠B ∠7=∠8 by (2) So, BD bisect ∠D
- In △ABC and △DEF,AB=DE,AB∥DE,BC= EF and BC | EF. Vertices A, B and C are joined to vertices D,E and F respectively. Show that :
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) AD∥CF and AD=CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC=DF
(vi) △ABC≅△DEF
Sol. Given : AB=DE,AB∥DE,BC=EF & BC∥EF To Prove (i) ABED is a parallelogram. (ii) BEFC is a parallelogram. (iii) AD∥CF and AD=CF (iv) ACFD is a parallelogram (v) AC=DF (vi) △ABC≅△DEF Proof (i) In quad. ABED AB=DE And AB || DE [Given] ∴ ABED is a parallelogram (ii) In quad. BEFC BC=EF [Given] And BC ∣EF [Given] ∴BEFC is a parallelogram. (iii) As ABED is a parallelogram. ∴AD∥BE and AD=BE Also, BEFC is a parallelogram ∴CF∥BE and CF=BE From (i) and (ii), we get ∴AD∥CF and AD=CF (iv) AsAD∥CF and AD=CF ⇒ ACFD is a parallelogram. (v) As ACFD is a parallelogram. ∴AC=DF (vi) In △ABC and △DEF, AB =DE BC=EF [Given] AC=DF [ Given] [Proved] ∴△ABC≅△DEF [By SSS congruency]
4.0Exercise: 8.2
- ABCD is a quadrilateral in which P,Q,R and S are mid points of the sides AB,BC, CD and DA (fig.) and AC is a diagonal. Show that
(i) SR∥AC and
SR =21AC
(ii) PQ=SR
(iii) PQRS is a parallelogram.
Sol. Given : ABCD is a quadrilateral in which P,Q,R and S are mid-points of AB,BC,CD and DA.AC is a diagonal. To prove : (i) SR∥AC and SR=21AC (ii) PQ=SR (iii) PQRS is a parallelogram. Proof: (i) In△DAC, ∵S is the mid-point of DA and R is the mid-point of DC ∴SR∥AC and SR=21AC [By Mid-point theorem] (ii) In △BAC, ∵P is the mid-point of AB and Q is the mid-point of BC ∴PQ∥AC and PQ=21AC [By Mid-point theorem] But from (i) SR=21AC&(ii) PQ=21AC ⇒PQ=SR (iii) PQ∥AC[ From (ii)] SR | AC [From (i)] ∴PQ∥SR [Two lines parallel to the same line are parallel to each other] Also, PQ=SR [From (ii)] ∴ PQRS is a parallelogram. [A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length]
- ABCD is a rhombus and P,Q,R and S are the mid points of sides AB,BC,CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
Sol.
Given : P,Q,R and S are the mid-points of respective sides AB,BC,CD and DA of rhombus. PQ,QR,RS and SP are joined. To prove : PQRS is a rectangle. Construction : Join A and C. Proof: In △ABC,P is the mid-point of AB and Q is the mid-point of BC. ∴PQ∥AC and PQ=21AC (By midpoint theorem) In △ADC,R is the mid-point of CD and S is the mid-point of AD. ∴SR∥AC and SR=21AC. (By midpoint theorem) From eq. (i) and (ii), PQ∥SR and PQ=SR ∴PQRS is a parallelogram. Now ABCD is a rhombus [Given] ∴AB=BC ⇒21AB=21BC⇒PB=BQ ∴∠1=∠2 [Angles opposite to equal sides are equal] Now in triangles APS and CQR, we have, AP=CQ [ P and Q are the mid-points of AB and BC and AB=BC] Similarly, AS = CR and PS = QR [Opposite sides of a parallelogram] ∴ΔAPS≅△CQR [By SSS congruency] ⇒∠3=∠4 [By C.P.C.T.] Now, we have ∠1+∠SPQ+∠3=180∘ And ∠2+∠PQR+∠4=180∘ ∴∠1+∠SPQ+∠3=∠2+∠PQR+∠4 Since ∠1=∠2 and ∠3=∠4 [Proved above] ∴∠SPQ=∠PQR Now PQRS is a parallelogram [Proved above] ∴∠SPQ+∠PQR=180∘ [Co-Interior angles] Using eq. (iii) and (iv), ∠SPQ+∠SPQ=180∘ ⇒2∠SPQ=180∘ ⇒∠SPQ=90∘ Hence, PQRS is a rectangle.
- 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.
Sol. Given : A rectangle ABCD in which P,Q,R and S are the mid-points of the sides AB, BC,CD and DA respectively. PQ, QR, RS and SP are joined.
To prove : PQRS is a rhombus.
Construction : Join AC. Proof : In △ABC,P and Q are the midpoints of sides AB,BC respectively. ∴PQ∥AC and PQ=21AC [Mid- Point Theorem] In △ADC,R and S are the mid-points of sides CD,AD respectively. ∴SR∥AC and SR=21AC [Mid-Point Theorem] From eq.(i) and (ii), PQ∥SR and PQ=SR ∴PQRS is a parallelogram. Now ABCD is a rectangle. ∴AD=BC ⇒21AD=21BC ⇒AS=BQ In triangles APS and BPQ, AP=BP[P is the mid-point of AB ] ∠PAS=∠PBQ[ Each 90∘] And AS = BQ [From eq. (iv)] ∴△APS≅△BPQ [By SAS congruency] ⇒PS=PQ [By C.P.C.T.] From eq.(iii) and (v), we get that PQRS is a parallelogram. ⇒PS=PQ ⇒ Two adjacent sides are equal. Hence, PQRS is a rhombus.
- 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 (fig.). Show that F is the mid-point of BC .
Sol. Line ℓ∥AB and passes through E.Line ℓ meets BC at F and BD at G . In △ABD,E is mid-point of AD and EG∥AB. ⇒G is mid-point of BD. [Converse of Mid Point Theorem] Also, ℓ∥AB and AB∥CD ⇒ℓ∥CD ⇒ Fis mid-point of BC. [∵G is mid-point of BD] [Converse of Mid Point Theorem]
- In a parallelogram ABCD,E 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 AB and DC respectively.To prove: DP=PQ=QB Proof: Since E and F are the mid-points of AB and CD respectively. ∴AE=21AB and CF=21CD But ABCD is a parallelogram. ∴AB=CD and AB∥DC ⇒21AB=21CD and AB∥DC ⇒AE=FC and AE∥FC [From eq. (i)] ∴ AECF is a parallelogram. ⇒FA∥CE ⇒FP∥CQ [FP is a part of FA and CQ 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 △DCQ,F is the mid-point of CD and ⇒FP∥CQ ∴P is the mid-point of DQ . ⇒DP=PQ Similarly, In △ABP,E is the mid-point of AB and ⇒EQ∥AP ∴Q is the mid-point of BP . ⇒BQ=PQ From eq.(iii) and (iv), DP=PQ=BQ Now, BD=BQ+PQ+DP=BQ+BQ+BQ=3BQ ⇒BQ=31BD From eq (v) and (vi), DP=PQ=BQ=31BD ⇒ Points P and Q trisects BD . So, AF and CE trisects BD. Hence, AF and CE trisect the diagonal BO.
- 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=21AB
Sol. (i) Through M, we draw line ℓ∥BC. ℓ intersects AC at D.
M is a midpoint of AB
⇒D is mid-point of AC.
[By converse of mid-point theorem]
(ii) ∠ADM=∠ACB=90∘ [Corresponding angles] ⇒∠ADM=90∘ ⇒MD⊥AC. (iii) In △ CMD and △AMD; CD=AD,MD=MD and ∠CDM=∠ADM [ Each =90∘ ] Therefore, △CMD≅△AMD [SAS congruence rule] ⇒CM=AM; Also AM=21AB So, CM=21AB AM=CM=21AB
- Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Sol. P,Q,R and S are the mid-points of the sides AB,BC,CD and AD of the quadrilateral ABCD.
We have to prove that, PR and QS bisects each other. Now, join PQ, QR, RS and PS. Here, we can prove that PQRS is a parallelogram (as in solution 1). Now, PR and QS are the diagonals of the parallelogram PQRS. ∴PR and QS bisect each other as diagonals of parallelogram bisect each other.
NCERT Solutions for Class 9 Science Other Chapters:- |
Chapter 8: Quadrilaterals |
CBSE Notes for Class 9 Science - All Chapters:- |
Class 9 Maths Chapter 4 - Linear Equation In Two Variables Notes |
Class 9 Maths Chapter 5 - Introduction To Euclids Geometry Notes |
Frequently Asked Questions:
What are Quadrilaterals?
A quadrilateral is a plane figure that has four sides or edges and also has four corners or vertices. Quadrilaterals typically have standard shapes with four sides like rectangle, square, trapezoid, kite, or irregular and uncharacterized shapes.
What is the total number of questions in Quadrilaterals, Chapter 8 of NCERT Math for Class 9?
NCERT Class 9 Chapter 8 Quadrilaterals contains 13 questions in total. In addition to the exercise questions, students can practice the solved examples to gain a conceptual understanding.
What are the key concepts covered in NCERT Solutions Class 9 Maths Chapter 8?
The key concepts of NCERT Solution for Class 9 Maths Chapter 8 include: 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 of the chapter.
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