{"id":2015,"date":"2020-09-08T05:38:37","date_gmt":"2020-09-08T00:08:37","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=2015"},"modified":"2021-07-02T15:03:45","modified_gmt":"2021-07-02T09:33:45","slug":"kseeb-solutions-class-9-maths-chapter-7-ex-7-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-7-ex-7-1\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 7 Quadrilaterals Ex 7.1"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 7 Quadrilaterals Ex 7.1<\/strong> are part of KSEEB Solutions for Class 9 Maths<\/a>. Here we have given Karnataka Board Class 9 Maths Chapter 7 Quadrilaterals Exercise 7.1.<\/p>\n

Karnataka Board Class 9 Maths Chapter 7 Quadrilaterals Ex 7.1<\/h2>\n

Question 1.
\nThe angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
\n\"KSEEB
\nSolution:
\nThe angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13.
\nLet \u2220A : \u2220B : \u2220C : \u2220D = 3 : 5 : 9 : 13.
\nSum of ratio = 3x + 5x + 9x + 13x = 30x
\nSum of 4 angles of quadrilateral is 360\u00b0
\n\u2234 \u2220A + \u2220B + \u2220C + \u2220D = 360\u00b0
\n3x + 5x + 9x + 13x = 360
\n30x = 360
\n\u2234 x = \\(\\frac{360}{30}\\) = 12\u00b0
\n\u2220A = 3x = 3 \u00d7 12 = 36\u00b0
\n\u2220B = 5x = 5 \u00d7 12 = 60\u00b0
\n\u2220C = 9x = 9 \u00d7 12 = 108\u00b0
\n\u2220D = 13x = 13 \u00d7 12 = 156\u00b0<\/p>\n

Question 2.
\nIf the diagonals of a parallelogram are equal, then show that it is a rectangle.
\n\"KSEEB
\nSolution:
\nData: Diagonals of a parallelogram are equal.
\nTo Prove: ABCD is a rectangle.
\nProof: Now ABCD is a parallelogram and diagonal AC = Diagonal BD (Data)
\nIn \u2206ABC and \u2206ABD,
\nBC = AD (Opposite sides of a quadrilateral)
\nAC = BD (Data)
\nAB common.
\n\u2234 \u2206ABC \u2245 \u2206ABD (SSS postulate)
\n\u2220ABC = \u2220BAD
\nBut, \u2220ABC + \u2220BAD = 180\u00b0
\n\u2220ABC + \u2220ABC = 180\u00b0
\n2 \u2220ABC = 180\u00b0
\n\u2234 \u2220ABC = 90\u00b0
\nIf an angle of a parallelogram is a right angle, it is called a rectangle.
\n\u2234 ABCD is a rectangle.<\/p>\n

Question 3.
\nShow that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
\n\"KSEEB
\nSolution:
\nData: ABCD is a parallelogram and diagonals AC and BD bisect at right angles at O’.
\nTo Prove: ABCD is a rhombus.
\nProof: Here, AC and BD bisect each other at right angles.
\n\u2234 AO = OC
\nBO = OD
\nand \u2220AOB = \u2220BOC = \u2220COD = \u2220AOD = 90\u00b0
\nIf sides are equal to each other, then ABCD is said to be a rhombus.
\nNow, \u2206AOD and \u2206COD,
\nAO = OC (Data)
\n\u2220AOD = \u2220COD = 90\u00b0 (Data)
\nOD is common.
\n\u2234 \u2206AOD \u2245 \u2206COD (SAS Postulate)
\n\u2234 AD = CD …………… (i)
\nSimilarly,
\n\u2206AOD = \u2206AOB
\nAD = AB ………… (ii)
\n\u2206AOB \u2245 \u2206COB
\n\u2234 AB = BC ……….. (iii)
\n\u2206COB \u2245 \u2206COD
\n\u2234 BC = CD ……………. (iv)
\nFrom (i), (ii), (iii) and (iv),
\nAB = BC = CD = AD
\nAll 4 sides of parallelogram ABCD are equal, then it is a rhombus.<\/p>\n

Question 4.
\nShow that the diagonals of a square are equal and bisect each other at right angles.
\n\"KSEEB
\nSolution:
\nData: ABCD is a parallelogram. Its diagonals are AC and BD. They meet at ‘O’.
\nTo Prove: i) AO = OC
\nBO = OD
\nii) \u2220AOB = \u2220BOC = \u2220COD = \u2220DOA = 90\u00b0.
\nProof: In \u2206ABC and \u2206ABD,
\nBC = AD (diagonals of square is equal)
\n\u2220ABC = \u2220BAD = 90\u00b0 (angles of square)
\nAB is common.
\n\u2234 \u2206ABC \u2245 \u2206ABD (SAS Postulate.)
\nIn \u2206AOB and \u2206COD,
\nAB = DC (sides of square)
\n\u2220OAB = \u2220OCD (alternate angles)
\n\u2220OBA = \u2220ODC (alternate angles)
\n\u2234 \u2206AOB \u2245 \u2206COD (ASA Postulate)
\nAO = OC
\nBO = OD …………….. (i)
\nSimilarly,
\n\u2206AOB \u2245 \u2206BOC.
\n\u2234 \u2220AOB = \u2220BOC = 90\u00b0
\nNow, \u2206COD \u2245 \u2206AOD, then
\n\u2234 \u2220COD = \u2220DOA = 90\u00b0
\n\u2234 \u2220AOB = \u2220BOC = \u2220COD = \u2220DOA = 90\u00b0 …………. (ii)
\nfrom (i) and (ii)
\n\u2234 Sides of a square are equal and bisect at right angles.<\/p>\n

Question 5.
\nShow that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
\n\"KSEEB
\nSolution:
\nData: ABCD is a quadrilateral and bisects each other at right angles, then it is a square.
\nAO = OC and AC = BD
\nBO = OD
\n\u2220AOB = \u2220BOC = \u2220COD + \u2220DOA = 90\u00b0.
\nTo Prove: ABCD is a square.
\nProof: In \u2206AOB and \u2206COD,
\nAO = OC
\nBO = OD (Data)
\n\u2220AOB = \u2220COD (Vertically opposite angles)
\n\u2234 \u2206AOB \u2245 \u2206COD (SAS Postulate)
\nAB = CD …………. (i)
\n\u2220ABO = \u2220CDO
\n\u2234 AB || CD ………… (ii)
\nFrom (i) and (ii) ABCD is a parallelogram.
\nNow, in \u2206AOD and \u2206COD,
\nAO = OC (Data)
\n\u2220AOD = \u2220COD = 90\u00b0 (Data)
\nOD is common
\n\u2234 \u2206AOD \u2245 \u2206COD (SAS Postulate)
\nAD = CD …………. (iii)
\nAD = BC ………….. (iv)
\nFrom (ii), (iii) and (iv)
\nAB = BC = CD = AD
\nFour sides of a quadrilateral are they are equal to each other and bisect each other at right angles, then it is a square.
\n\u2234 ABCD is a square.<\/p>\n

Question 6.
\nDiagonal AC of a parallelogram ABCD bisects \u2220A. Show that
\n\"KSEEB
\n(i) it bisects \u2220C also,
\n(ii) ABCD is a rhombus.
\nSolution:
\nData: Diagonal AC of a parallelogram ABCD bisects \u2220A.
\nTo Prove: (i) Diagonal AC also bisects \u2220C.
\n(ii) ABCD is a rhombus.
\nProof: i) Diagonal AC also bisects \u2220A.
\n\u2234 \u2220DAC = \u2220BAC …………… (i)
\nBut, \u2220DAC= \u2220BCA (alternate angles)(ii)
\n\u2220BAC=\u2220DCA (alternate angles)(iii)
\nFrom (i), (ii) and (iii),
\n\u2220BCA = \u2220DCA
\n\u2234 AC bisects \u2220C.<\/p>\n

(ii) \u2220DAC = \u2220DCA
\n\u2234 AD = DC
\nBut, AD = BC
\n\u2234 DC = AB
\n\u2234 AB = BC = CD = DA
\n\u2234 ABCD is a rhombus.<\/p>\n

Question 7.
\nABCD is a rhombus. Show that diagonal AC bisects \u2220A as well as \u2220C and diagonal BD bisects \u2220B as well as \u2220D.
\n\"KSEEB
\nSolution:
\nABCD is rhombus. Diagonal AC bisects \u2220A and \u2220C.
\nTo Prove: Diagonal BD bisects \u2220B and \u2220D.
\nProof: In a rhombus all sides are equal and opposite angles are equal to each other.
\nIn \u2206ABC,
\nAB = BC
\n\u2234 \u2220BAC = \u2220BCA
\n\u2220B AC = \u2220DCA (alternate angles)
\n\u2234 \u2220BCA = \u2220DCA …………… (i)
\n\u2234 AC bisects \u2220C.
\nNow, \u2220BCA = \u2220DAC (alternate angles)
\n\u2220BAC = \u2220DAC
\n\u2234 AC bisects \u2220A
\nIn \u2206ABD,
\nAD = DB
\n\u2234 \u2220ABD = \u2220ADB
\n\u2220ABD = \u2220CDB
\n\u2234 BD bisects \u2220D.
\n\u2220ADB = \u2220CBD
\n\u2220ABD = \u2220CBD
\n\u2234 BD bisects \u2220B.<\/p>\n

Question 8.
\nABCD is a rectangle in which diagonal
\n\"KSEEB
\n(i) ABCD is a square,
\n(ii) Diagonal BD bisects \u2220B as well as \u2220D.
\nSolution:
\nData: ABCD is a rectangle in which diagonal AC bisects \u2220A as well as \u2220C.
\nTo Prove:
\n(i) ABCD is a square.
\n(ii) Diagonal BD bisects \u2220B as well as \u2220D.
\nProof: (i) ABCD is a rectangle. AC and BD diagonals intersect at O’.
\nAB = DC (Opposite sides of rectangle)
\nAD = BC
\n\u2220BAD = \u2220BCD (Opposite angles)
\n\u2234 \\(\\frac{1}{2}\\) \u2220BAD = \\(\\frac{1}{2}\\) \u2220BCD
\n\u2234 \u2220DAC = \u2220DCA (AC bisects \u2220A)
\n\u2234 DC = DA
\nBut, AB = CD and DA = BC (Data)
\n\u2234 AB = BC = CD = AD
\n\u2234 ABCD is a square.<\/p>\n

(ii) In \u2206ABD.
\nAB = AD
\n\u2220ABD = \u2220ADB
\n\u2220ABD = \u2220CDB (alternate angles)
\n\u2220ADB = \u2220CDB
\n\u2234 BD bisects \u2220D.
\n\u2220ADB = \u2220CDB (alternate angles)
\n\u2220ABD = \u2220CBD (alternate angles)
\n\u2234 BD bisects \u2220B.<\/p>\n

Question 9.
\nIn parallelogram ABCD. two points P and Q are taken on diagonal BD such that DP = BQ. Show that:
\n\"KSEEB
\n(i) \u2206APD \u2245 \u2206CQB
\n(ii) AP = CQ
\n(iii) \u2206AQB \u2245 \u2206CPD
\n(iv) AQ = CP
\n(v) APCQ is a parallelogram.
\nSolution:
\nData: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ.
\nTo Prove:
\n(i) \u2206AAPD \u2245 \u2206CQB
\n(ii) AP = CQ
\n(iii) \u2206AQB \u2245 \u2206CPD
\n(iv) AQ = CP
\n(v) APCQ is a parallelogram.
\nProof: ABCD is a parallelogram.
\n(i) In \u2206APD and \u2206CQB,
\nAD = BC (opposite sides)
\n\u2220ADP = \u2220CBQ (alternate angles)
\nDP = BQ (Data)
\n\u2234 \u2206APD \u2245 \u2206CQB (ASA Postulate)<\/p>\n

(ii) AP = CQ<\/p>\n

(iii) In \u2206AQB and \u2206CPD,
\nAB = CD (opposite sides)
\n\u2220ABQ = \u2220CDP (alternate angles)
\nBQ = DP (Data)
\n\u2234 \u2206AQB \u2245 \u2206CPD (ASA Postulate)<\/p>\n

(iv) \u2234 AQ = CP<\/p>\n

(v) In \u2206AQPand \u2206CPQ,
\nAP = CQ (proved)
\nAQ = PC PQ is common.
\n\u2234 \u2206AQP \u2245 \u2206CPQ
\n\u2234 \u2220APQ = \u2220CQP
\nThese are a pair of alternate angles.
\n\u2234AP || PC
\nSimilarly, AQ || PC
\n\u2234 APCQ is a parallelogram.<\/p>\n

Question 10.
\nABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that
\n\"KSEEB
\n(i) \u2206APB \u2245 \u2206CQD
\n(ii) AP = CQ.
\nSolution:
\nData: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.
\nTo Prove:
\n(i) \u2206APB \u2245 \u2206CQD
\n(ii) AP = CQ.
\nProof: ABCD is a parallelogram. BD is diagonal.
\nAP \u22a5 BD and CQ \u22a5 BD.<\/p>\n

(i) In \u2206APB and \u2206CQD,
\n\u2220APB = \u2220CQD = 90\u00b0
\nAB = CD (opposite sides)
\n\u2220ABP = \u2220CQD (alternate angles)
\n\u2234 \u2206APB \u2245 \u2206CQD (AAS postulate)<\/p>\n

(ii) \u2234 AP = CQ.<\/p>\n

Question 11.
\nIn \u2206ABC and \u2206DEF, 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
\n\"KSEEB
\n(i) Quadrilateral ABED is a parallelogram.
\n(ii) Quadrilateral BEFC is a parallelogram.
\n(iii) AD || CF and AD = CF.
\n(iv) quadrilateral ACFD is a parallelogram.
\n(v) AC = DF
\n(vi) \u2206ABC \u2245 \u2206DEF.
\nSolution:
\nData: In \u2206ABC and \u2206DEF, AB = DE, AB || DE, BC = EF and BC || EF.
\nVertices A, B, and C are joined to vertices D, E, and F respectively.
\nTo Prove:
\n(i) Quadrilateral ABED is a parallelogram.
\n(ii) Quadrilateral BEFC is a parallelogram.
\n(iii) AD || CF and AD = CF.
\n(iv) quadrilateral ACFD is a parallelogram.
\n(v) AC = DF
\n(vi) \u2206ABC \u2245 \u2206DEF.
\nProof: (i) AB = DE and AB||DE (Data)
\n\u2234 BE = AD and BE || AD
\n\u2234 ABCD is a parallelogram.<\/p>\n

(ii) Similarly, BC = EF and BC || EF.
\n\u2234 BE = CF
\nBE || CF
\n\u2234 BEFC is a parallelogram.<\/p>\n

(iii) ABED is a parallelogram.
\n\u2234 AD = BE
\nAD || BE ………. (i)
\nSimilarly, BEFC is a parallelogram.
\n\u2234 CF = BE
\nCF || BE ……………. (ii)
\nComparing (i) and (ii),
\nAD = CF and AD || CF,<\/p>\n

(iv) In a quadrilateral ACFD,
\nAD = CF
\nAD || CF (proved)
\n\u2234 AC = DF
\nAC || DF
\n\u2234 ACFD is a parallelogram.<\/p>\n

(v) ACFD is a parallelogram.
\nAC = DF (opposite sides).<\/p>\n

(vi) In \u2206ABC and \u2206DEF,
\nAB = DE (Data)
\nBC = EF (opposite sides of a
\nparallelogram)
\nAC = DF (Opposite sides of a
\nparallelogram)
\n\u2234 \u2206ABC \u2245 \u2206DEF (SSS postulate).<\/p>\n

Question 12.
\nABCD is a trapezium in which AB||CD and AD = BC . Show that
\n\"KSEEB
\n(i) \u2220A = \u2220B
\n(ii) \u2220C = \u2220D
\n(iii) \u2206ABC \u2245 \u2206BAD
\n(iv) diagonal AC = diagonal BD.
\n(Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
\nSolution:
\nData: ABCD is a trapezium in which AB || CD and AD = BC.
\nTo Prove:
\n(i) \u2220A = \u2220B
\n(ii) \u2220C = \u2220D
\n(iii) \u2206ABC \u2245 \u2206BAD
\n(iv) diagonal AC = diagonal BD.
\nConstruction: Produce straight line AB.
\nDraw a parallel line DA, parallel to AB such that the produced line meet this at E.
\nProof: (i) ABCD is a trapezium.
\nAB || CD and AD = BC (Data)
\nNow, AD || CD and AB || DC.
\n\u2234 ADCE is a parallelogram.
\n\u2234 AD = CE (Opposite sides)
\nAD = BC (Data)
\nCE = BC
\n\u2234 \u2220CBE = \u2220CEB
\nNow, \u2220DAB + \u2220CEB = 180\u00b0 (Sum of consecutive angles.)
\n\u2234 \u2220DAB + \u2220CEB = 180\u00b0
\n( \u2235 \u2220CEB = \u2220CBE) ……………… (i)
\nNow, \u2220ABC + \u2220CDE = 180\u00b0
\n(linear pair) ……………….. (ii)
\nComparing (i) and (ii),
\n\u2220DAB + \u2220CBE = \u2220ABC + \u2220CBE
\n\u2234 \u2220DAB = \u2220ABC Or \u2220A = \u2220B.<\/p>\n

(ii) AB || CD
\n\u2220DAB + \u2220ADC =180\u00b0 ……….. (iii)
\n\u2220ABC + \u2220BCD = 180\u00b0 ………… (iv)
\nCompairing (iii) and (iv),
\n\u2220DAB + \u2220ADC = \u2220ABC + \u2220BCD
\n\u2234 \u2220ADC = \u2220BCD (\u2234\u2220A = \u2220B)
\nOr \u2220D = \u2220C.<\/p>\n

(iii)
\n\"KSEEB
\nDraw diagonals AC and BD.
\nIn \u2206ABC and \u2206ABD,
\nBC = AD (Data)
\n\u2220ABC = \u2220BAD (proved)
\nAB is common.
\n\u2234 \u2206ABC \u2245 \u2206ABD (SAS Postulate).<\/p>\n

(iv) \u2206ABC \u2245 \u2206BAD (Proved)
\n\u2234 Diagonal AC = Diagonal BD.<\/p>\n

We hope the KSEEB Solutions for Class 9 Maths Chapter 7 Quadrilaterals Ex 7.1 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 7 Quadrilaterals Exercise 7.1, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 7 Quadrilaterals Ex 7.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 7 Quadrilaterals Exercise 7.1. Karnataka Board Class 9 Maths Chapter 7 Quadrilaterals Ex 7.1 Question 1. The angles of a quadrilateral are in the ratio …<\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2015"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=2015"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2015\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2015"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2015"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2015"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}