Ex. 8.1 question NO 6 maths …

GRAMMAR CBSE NOTES CLS 6 7 8 9 10 11 12 MCQ QUESTIONS NCERT BOOKS STUDY MATERIAL NUMBERS CALCULATOR Learn CBSE NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12 NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1. NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 Ex 8.1 Class 9 Maths Question 1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. Solution: Let the angles of the quadrilateral be 3x, 5x, 9x and 13x. ∴ 3x + 5x + 9x + 13x = 360° [Angle sum property of a quadrilateral] ⇒ 30x = 360° ⇒ x = 360∘30 = 12° ∴ 3x = 3 x 12° = 36° 5x = 5 x 12° = 60° 9x = 9 x 12° = 108° 13a = 13 x 12° = 156° ⇒ The required angles of the quadrilateral are 36°, 60°, 108° and 156°. Ex 8.1 Class 9 Maths Question 2. If the diagonals of a parallelogram are equal, then show that it is a rectangle. Solution: Let ABCD is a parallelogram such that AC = BD.  In ∆ABC and ∆DCB, AC = DB [Given] AB = DC [Opposite sides of a parallelogram] BC = CB [Common] ∴ ∆ABC ≅ ∆DCB [By SSS congruency] ⇒ ∠ABC = ∠DCB [By C.P.C.T.] …(1) Now, AB || DC and BC is a transversal. [ ∵ ABCD is a parallelogram] ∴ ∠ABC + ∠DCB = 180° … (2) [Co-interior angles] From (1) and (2), we have ∠ABC = ∠DCB = 90° i.e., ABCD is a parallelogram having an angle equal to 90°. ∴ ABCD is a rectangle. Ex 8.1 Class 9 Maths Question 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. Solution: Let ABCD be a quadrilateral such that the diagonals AC and BD bisect each other at right angles at O.  ∴ In ∆AOB and ∆AOD, we have AO = AO [Common] OB = OD [O is the mid-point of BD] ∠AOB = ∠AOD [Each 90] ∴ ∆AQB ≅ ∆AOD [By,SAS congruency ∴ AB = AD [By C.P.C.T.] ……..(1) Similarly, AB = BC .. .(2) BC = CD …..(3) CD = DA ……(4) ∴ From (1), (2), (3) and (4), we have AB = BC = CD = DA Thus, the quadrilateral ABCD is a rhombus. Alternatively : ABCD can be proved first a parallelogram then proving one pair of adjacent sides equal will result in rhombus. Ex 8.1 Class 9 Maths Question 4. Show that the diagonals of a square are equal and bisect each other at right angles. Solution: Let ABCD be a square such that its diagonals AC and BD intersect at O.  (i) To prove that the diagonals are equal, we need to prove AC = BD. In ∆ABC and ∆BAD, we have AB = BA [Common] BC = AD [Sides of a square ABCD] ∠ABC = ∠BAD [Each angle is 90°] ∴ ∆ABC ≅ ∆BAD [By SAS congruency] AC = BD [By C.P.C.T.] …(1) (ii) AD || BC and AC is a transversal. [∵ A square is a parallelogram] ∴ ∠1 = ∠3 [Alternate interior angles are equal] Similarly, ∠2 = ∠4 Now, in ∆OAD and ∆OCB, we have AD = CB [Sides of a square ABCD] ∠1 = ∠3 [Proved] ∠2 = ∠4 [Proved] ∴ ∆OAD ≅ ∆OCB [By ASA congruency] ⇒ OA = OC and OD = OB [By C.P.C.T.] i.e., the diagonals AC and BD bisect each other at O. …….(2) (iii) In ∆OBA and ∆ODA, we have OB = OD [Proved] BA = DA [Sides of a square ABCD] OA = OA [Common] ∴ ∆OBA ≅ ∆ODA [By SSS congruency] ⇒ ∠AOB = ∠AOD [By C.P.C.T.] …(3) ∵ ∠AOB and ∠AOD form a linear pair. ∴∠AOB + ∠AOD = 180° ∴∠AOB = ∠AOD = 90° [By(3)] ⇒ AC ⊥ BD …(4) From (1), (2) and (4), we get AC and BD are equal and bisect each other at right angles. Ex 8.1 Class 9 Maths Question 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. Solution: Let ABCD be a quadrilateral such that diagonals AC and BD are equal and bisect each other at right angles.  Now, in ∆AOD and ∆AOB, We have ∠AOD = ∠AOB [Each 90°] AO = AO [Common] OD = OB [ ∵ O is the midpoint of BD] ∴ ∆AOD ≅ ∆AOB [By SAS congruency] ⇒ AD = AB [By C.P.C.T.] …(1) Similarly, we have AB = BC … (2) BC = CD …(3) CD = DA …(4) From (1), (2), (3) and (4), we have AB = BC = CD = DA ∴ Quadrilateral ABCD have all sides equal. In ∆AOD and ∆COB, we have AO = CO [Given] OD = OB [Given] ∠AOD = ∠COB [Vertically opposite angles] So, ∆AOD ≅ ∆COB [By SAS congruency] ∴∠1 = ∠2 [By C.P.C.T.] But, they form a pair of alternate interior angles. ∴ AD || BC Similarly, AB || DC ∴ ABCD is a parallelogram. ∴ Parallelogram having all its sides equal is a rhombus. ∴ ABCD is a rhombus. Now, in ∆ABC and ∆BAD, we have AC = BD [Given] BC = AD [Proved] AB = BA [Common] ∴ ∆ABC ≅ ∆BAD [By SSS congruency] ∴ ∠ABC = ∠BAD [By C.P.C.T.] ……(5) Since, AD || BC and AB is a transversal. ∴∠ABC + ∠BAD = 180° .. .(6) [ Co – interior angles] ⇒ ∠ABC = ∠BAD = 90° [By(5) & (6)] So, rhombus ABCD is having one angle equal to 90°. Thus, ABCD is a square. Ex 8.1 Class 9 Maths Question 6. Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus.  Solution: We have a parallelogram ABCD in which diagonal AC bisects ∠A ⇒ ∠DAC = ∠BAC  (i) Since, ABCD is a parallelogram. ∴ AB || DC and AC is a transversal. ∴ ∠1 = ∠3 …(1) [ ∵ Alternate interior angles are equal] Also, BC || AD and AC is a transversal. ∴ ∠2 = ∠4 …(2) [ v Alternate interior angles are equal] Also, ∠1 = ∠2 …(3) [ ∵ AC bisects ∠A] From (1), (2) and (3), we have ∠3 = ∠4 ⇒ AC bisects ∠C. (ii) In ∆ABC, we have ∠1 = ∠4 [From (2) and (3)] ⇒ BC = AB …(4) [ ∵ Sides opposite to equal angles of a ∆ are equal] Similarly, AD = DC ……..(5) But, ABCD is a parallelogram. [Given] ∴ AB = DC …(6) From (4), (5) and (6), we have AB = BC = CD = DA Thus, ABCD is a rhombus.