{"id":25450,"date":"2021-06-23T10:09:27","date_gmt":"2021-06-23T04:39:27","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=25450"},"modified":"2021-06-23T10:12:02","modified_gmt":"2021-06-23T04:42:02","slug":"ncert-solutions-for-class-9-maths-chapter-8-ex-8-1","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 8 Quadrilaterals Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1<\/h2>\n

Question 1.
\nThe angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. the angles of the quadrilateral.
\nSolution:
\nLet the first angle of the quadrilateral be 3x.
\nSecond angle of quadrilateral is 5x.
\nThe third angle of the quadrilateral is 9x.
\nThe fourth angle of the quadrilateral is 13x.
\nTherefore,
\n3x + 5x + 9x + 13x = 360 (sum of all four angles of a quadrilateral is 360)
\n\u21d2 30x = 360
\n\u21d2 x = \\(\\frac{360}{30}\\)
\n\u21d2 x = 12
\nFirst angle of quadrilateral is 3x = 3 \u00d7 12 = 36\u00b0
\nSecond angle of quadrilateral is 5x = 5 \u00d7 12 = 60\u00b0
\nThird angle of quadrilateral is 9x = 9 \u00d7 12 = 108\u00b0
\nFourth angle of quadrilateral is 13x = 13 \u00d7 12 = 156\u00b0<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nIf the diagonals of a parallelogram are equal, then show that it is a rectangle.
\nSolution:
\nGiven: ABCD is a parallelogram, in which diagonal AC and BD are equal.
\nTo prove that: ABCD is a rectangle.
\n\"NCERT
\nProof: In \u2206ABC and \u2206DCB.
\nAC = BD (Given)
\nBC = GB (Common)
\nAB = DC (Opposite sides of a parallelogram)
\nTherefore, by S-S-S congruency condition
\n\u2206ABC \u2245 \u2206DCB
\nSo, \u2220ABC = \u2220BCD (By C.P.C.T)
\nBut, \u2220ABC + \u2220BCD = 180 (Sum of interior angles of the same side of transversal)
\n\u21d2 2\u2220ABC = 180 (\u2235 \u2220ABC = \u2220BCD)
\n\u21d2 \u2220ABC = 90\u00b0
\nTherefore, paraileiogram ABCD is a rectangle.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nShow that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
\nSolution:
\nGiven: ABCD is a quadrilateral, in which diagonal AC and BD intersect each other at a right angle.
\nTo prove that ABCD is a Rhombus.
\nProof: In \u2206AOD and \u2206AOB,
\nAO = OA (Common)
\nOD = OB (Given)
\n\u2220AOD = \u2220AOB (each 90\u00b0)
\nBy S-A-S congruency condition
\n\u2220AOU \u2245 \u2220AQB
\nSo, AD = AB …..(i) (By C.P.C.T)
\nSimilarly, AD = BC ……(ii)
\nand AB = CD ……(iii)
\nBy equation (i), (ii) and (iii)
\nAB = BC = CD = DA
\nTherefore, ABCD is a rhombus.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nShow that the diagonals of a square are equal and bisect each other at right angles.
\nSolution:
\nGiven: ABCD is a square, in which diagonal AC and BD intersect each other at Q.
\n\"NCERT
\nTo prove that:
\n(i) AC = BD
\n(ii) Diagonal bisects each other at right angle.
\nProof:
\n(i) In \u2206ABC and \u2206BAD
\nAB = BA (Common)
\nBC = AD (sides of square)
\n\u2220ABC = \u2220BAD (each 90\u00b0)
\nBy S-A-S congruency condition
\n\u2206ABC \u2245 \u2206BAD
\nAC = BD (By C.P.C.T.)<\/p>\n

(ii) In \u2206AOB and \u2206COD,
\nAB = CD (Sides of square)
\n\u2220AOB = \u2220COD (Vertically opposite angles)
\n\u2220OBA = \u2220OCD (Alternate interior angle)
\nTherefore, by A-S-A congruency condition
\n\u2206AOB \u2245 \u2206COD
\nSo, OA = OC (By C.P.C.T.)
\nand OB = OD (By C.P.C.T)
\nNow, In \u2206AOD and \u2206COD
\nAD = CD (Sides of square)
\nOA = OC (Prove above)
\nOD = OD (Common)
\nBy S-S-S ccngruency condition
\n\u2206AOD = \u2206COD
\nSo, \u2220AOD = \u2220COD (By C.P.C.T)
\nBut, \u2220AOD + \u2220COD = 180 (Linear pair)
\n\u21d2 \u2220AOD + \u2220AOD = 180 (\u2235 \u2220AOD = \u2220COD)
\n\u21d2 2\u2220AOD = 180
\n\u21d2 \u2220AOD = 90\u00b0 …..(ii)
\nTherefore, from equations (i) and (ii) it is dear that diagonal of a square bisect each other at a right angle.<\/p>\n

\"NCERT<\/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.
\nSolution:
\nGiven: ABCD is a quadrilateral in which diagonal AC and BD bisect each other at right angles.
\n\"NCERT
\nTo prove that: ABCD is a square.
\nProof: In \u2206AOB and \u2206COD
\nAO = CO (Given)
\n\u2220AOB = \u2220COD (Vertically opposite angles)
\nOB = OD (Given)
\nBy S-A-S Congruency Condition.
\n\u2206AOB \u2245 \u2206COD
\nSo, AB = CD …..(i) (By C.P.C.T)
\nand \u2220OAB = \u2220OCD (By C.P.C.T)
\nBut it is the pair of alternate interior angles and we know that if pair of alternate interior angles are equal then the two lines are parallel.
\n\u2234 AB || CD …..(ii)
\nFrom (i) and (ii)
\nABCD is a parallelogram
\nNow, in \u2206AOD and \u2206COD
\nAO = OC (Given)
\n\u2220AOD = \u2220COD (Each 90\u00b0)
\nOD = OD (Common)
\nBy S-A-S congruency condition
\n\u2206AOD \u2245 \u2206COD
\nAD = CD …..(iii) (ByC.P.C.T.)
\nAgain, AD = BC ……(iv)
\n(Opposite sides of parallel gram ABCD)
\nFrom equation (i), (iii) and (iv)
\nTherefore ABCD is a square.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nDiagonal AC of a parallelogram ABCD bisects \u2220A [see Fig. 8.19]. Show that
\n(i) it bisects \u2220C also.
\n(ii) ABCD is a rhombus.
\n\"NCERT
\nSolution:
\n(i) In \u2206DAC and \u2206BCA
\nDA = BC (Opposite sides of parallelogram)
\nDC = BA (Opposite sides of parallelogram)
\nAC = CA (Common)
\nBy S-S-S congruency condition
\n\u2206DAC = \u2206BCA
\nSo, \u2220DAC = \u2220BCA …..(i) (By C.P.C.T)
\nand \u2220ACD = \u2220BAC ……(ii) (By C.P.C.T)
\nAdd equation (i) and (ii)
\n\u2220DAC + \u2220ACD = \u2220BCA + \u2220BAC
\n\u2220DAC + \u2220ACD = \u2220BCA + \u2220DAC (\u2235 \u2220DAC = \u2220BAC given)
\nTherefore, \u2220ACD = \u2220BCA
\nor, AC bisect \u2220C.<\/p>\n

(ii) We have,
\n\u2220ACD = \u2220BAC …..(iii) (From equation (ii))
\nBut \u2220BAC = \u2220CAD …..(iv)
\nFrom (iii) and (iv)
\n\u2220ACD = \u2220CAD
\n\u2234 DA = DC
\n(Side opposite to equal angles are equal)
\nBut DA = BC (Opposite side of || gm)
\n\u2234 AB = BC = CD = DA
\nTherefore, ABCD is a rhombus.<\/p>\n

\"NCERT<\/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.
\nSolution:
\nIn this Fig. ABCD is a rhombus.
\n\"NCERT
\nTherefore, AB = BC = CD = AD
\nIn \u2206ABC,
\nAB = BC (Because ABCD is a rhombus)
\n\u2234 \u22201 = \u22202 (Angle opposite to equal sides are equal)
\nBut, \u22201 = \u22203 …….(ii) (Alternate interior angles)
\nFrom (i) and (ii)
\n\u22201 = \u22203
\n\u2234 AC bisects \u2220C.
\nSimilarly, we can prove AC bisects \u2220A, and BD bisects both \u2220B as well as \u2220D.<\/p>\n

Question 8.
\nABCD is a rectangle in which diagonal AC bisects. \u2220A as well as \u2220C. Show that
\n(i) ABCD is a square.
\n(ii) diagonal BD bisects \u2220B as well as \u2220D.
\n\"NCERT
\nSolution:
\n(i) We have given ABCD is a rectangle.
\n\u2220A = \u2220C (each 90\u00b0)
\nor, \\(\\frac {1}{2}\\) \u2220A = \\(\\frac {1}{2}\\) \u2220C
\n\u2220DAC = \u2220DCA (AC bisects \u2220A as well as \u2220C)
\nSo, AD = CD ……(i) (Sides opposite to equal angles are equal)
\nBut, AD = BC ……(ii) (Opposite sides of rectangle)
\nFrom equation (i) and (ii)
\nAB = BC = CD = AD
\n\u2234 ABCD is rhombus.
\nBut, each angle of rhombus ABCD is 90\u00b0.
\nTherefore, ABCD is a square.<\/p>\n

(ii) We know that diagonals of squares bisect opposite angles.
\nTherefore, diagonal BD bisects both \u2220B and \u2220D.<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nIn parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig 8.20) Show that:
\n(i) \u2206APD \u2245 \u2206CQB
\n(ii) AP = CQ
\n(iii) \u2206AQB = \u2206CPD
\n(iv) AQ = CP
\n(v) APCQ is a parallelogram.
\n\"NCERT
\nSolution:
\nGiven: ABCD is a parallelogram, and two points P and Q are taken on diagonal BD such that DP = BQ.
\nTo prove that:
\n(i) \u2206APD \u2245 \u2206CQB
\n(ii) AP = CQ
\n(iii) \u2206AQB = \u2206CPD
\n(iv) AQ = CP
\n(v) APCQ is a parallelogram.
\n\"NCERT
\nConstruction: Draw diagonal AC of parallelogram ABCD intersecting BD at O.
\nProof:
\n(i) In \u2206APD and \u2206CQB
\n\u2220ADP = \u2220CBQ (Pair of alternate interior angles)
\nAD = BC (Opposite sides of || gm ABCD)
\nPD = QB (Given)
\nBy S-A-S congruency condition.
\n\u2220APD \u2245 \u2220CQB<\/p>\n

(ii) We have
\n\u2220APD \u2245 \u2220CQB (Prove above)
\n\u2234 AP = CQ (By C.P.C.T.)<\/p>\n

(iii) In \u2206AQB and \u2206CPD
\n\u2220LBQ = \u2220CDP (Pair of alternate interior angles)
\nAB = CD (Opposite sides of parallelogram ABCD)
\nBQ = DP (Given)
\nBy S-A-S congruency condition
\n\u2206AQB \u2245 \u2206CPD<\/p>\n

(iv) We have
\n\u2206AQB \u2245 \u2206CPD (Prove above)
\n\u2234 AQ = CP (By C.P.C.T)<\/p>\n

(v) In \u2206QCP,
\nAQ = CP (Prove above from (iv))
\nand AP = CQ (Prove above from (ii))
\nWe know that if both opposite pairs are equal then the parallelogram.
\nHence, APCQ is a parallelogram.<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nABCD is a parallelogram and AP and CQ are perpendiculars from vertices A arid C on diagonal BD respectively (see fig 8.21). Show that
\n(i) \u2206APB \u2245 \u2206CQD
\n(ii) AP = CQ
\n\"NCERT
\nSolution:
\n(i) In \u2206APB and \u2206CQD
\n\u2220APB = \u2220CQD (Each 90\u00b0)
\nAB = CD (Opposite sides of || gm ABCD)
\nand, \u2220ABP = \u2220CDQ (Alternate interior angles)
\nBy A-S-A Congruency condition
\n\u2206APB = \u2206CQD<\/p>\n

(ii) We have
\n\u2206APB = \u2206CQD (Prove above)
\n\u2234 AP = CQ (By C.P.C.T)<\/p>\n

Question 11.
\nIn \u2206ABC and \u2206DEF, AB = DE and AB || DE, BC = EF and BC || EF. Vertices A, B, and C are joined to vertices D, E and F respectively (see fig 8.22). Show that
\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
\n\"NCERT
\nSolution:
\nGiven: AB = DE and AB || DE,
\nBC = EF and BC || EF
\nTo prove that:
\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:
\n(i) We have
\nAB = DE and AB || DE
\nWe know that if one opposite pair of a quadrilateral are parallel and equal, then the quadrilateral is a parallelogram.
\n\u2234 Quadrilateral ABDE is a parallelogram.<\/p>\n

(ii) We have
\nBC = EF and BC || EF
\nTherefore, quadrilateral BCEF is a parallelogram.<\/p>\n

(iii) From (i) we have
\nABED is a parallelogram.
\nAD = BE and AD || BE …….(A)
\n(Opposite sides of parallelogram ABED)
\nAgain, From (ii)
\nBEFC is a parallelogram,
\nBE = CF and BE || CF ……..(B)
\n(Opposite sides of parallelogram BEFC)
\nFrom equation (A) and (B)
\nAD || CF and AD = CF<\/p>\n

\"NCERT<\/p>\n

(iv) We have
\nAD || CF and AD = CF (Prove above from part (iii))
\nWe know that, if one opposite pair of a quadrilateral are parallel and equal, then the quadrilateral is a parallelogram.
\n\u2234 ACFD is a parallelogram.<\/p>\n

(v) We have,
\nACFD is a parallelogram
\n(Prove above from part IV)
\n\u2234 AC || DF
\n(Opposite sides of parallelogram ACFD)
\nand AC = DF<\/p>\n

(vi) In \u2206ABC and \u2206DEF
\nAB = DE (Given)
\nBC = EF (Given)
\nAC = DF (Prove above)
\nBy S-S-S Congruency condition,
\n\u2206ABC \u2245 \u2206DEF.<\/p>\n

\"NCERT<\/p>\n

Question 12.
\nABCD is, a trapezium in which AB || CD and AD = BC (see fig 8.23). Show that
\n(i) \u2220A = \u2220B
\n(ii) \u2220C = \u2220D
\n(iii) \u2206ABC \u2245 \u2206BAD
\n(iv) Diagonal AC = Diagonal BD.
\n\"NCERT
\nSolution:
\nGiven: ABCD is a trapezium in which AB || CD and AD = BC.
\nTo prove that:
\n(i) \u2220A = \u2220B
\n(ii) \u2220C = \u2220D
\n(iii) \u2206ABC \u2245 \u2206BAD
\n(iv) Diagonal AC = diagonal BD.
\nConstruction: Through C draw CE || AD which intersect AB produced at E, and join ACBD.
\n\"NCERT
\nProof: (i) We have,
\nAB || CD (Given)
\nAD || CE (By construction)
\nTherefore, AECD is a parallelogram.
\nSo, AD = CE …..(i)
\n(Opposite sides of || gm)
\nBut AD = BC ……(ii) (Given)
\nFrom (i) and (ii)
\nBC = CE
\n\u2234 \u2220CBE = \u2220CEB
\n(Angle opposite to equal sides are equal)
\nNow, \u2220A + \u2220CEB = 180\u00b0
\n(Sum of interior angles of the same side of transversal)
\nor, \u2220A + \u2220CBE = 180\u00b0 ……(iii) (\u2235 \u2220CEB = \u2220CBE)
\nBut, \u2220B + \u2220CBE = 180\u00b0 ……(iv) (Linear pair)
\nFrom equation (iii) and (iv)
\nor, \u2220A = \u2220B<\/p>\n

(ii) We have AB || CD
\n\u2234 \u2220A + \u2220D = 180\u00b0 …..(v)
\n(Sum of an interior angle of the same side of transversal)
\nand, \u2220C + \u2220B = 180\u00b0 ……(vi)
\n(Sum of interior angle of the same side of transversal)
\nFrom equation (v) and (vi)
\n\u2220A + \u2220D = \u2220C + \u2220B
\nBut, \u2220A = \u2220B (Prove above)
\n\u2234 \u2220C = \u2220D<\/p>\n

\"NCERT<\/p>\n

(iii) In \u2206ABC and \u2206BAD
\nAB = BA (Common)
\nBC = AD (Given)
\n\u2220B = \u2220A (Prove above)
\nBy S-A-S congruency condition
\n\u2206ABC \u2245 \u2206BAD<\/p>\n

(iv) We have
\n\u2206ABC \u2245 \u2206BAD (Prove above)
\n\u2234 AC = BD (By CPCT)
\nTherefore, diagonal AC = diagonal BD.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1 Question 1. The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. the angles of …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1 Question 1. The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. the angles of … NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 Read More »\" \/>\n<meta property=\"og:url\" content=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/\" \/>\n<meta property=\"og:site_name\" content=\"NCERT Solutions\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/\" \/>\n<meta property=\"article:published_time\" content=\"2021-06-23T04:39:27+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2021-06-23T04:42:02+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@ncertsolguru\" \/>\n<meta name=\"twitter:site\" content=\"@ncertsolguru\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Prasanna\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"14 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Organization\",\"@id\":\"https:\/\/ncertsolutions.guru\/#organization\",\"name\":\"NCERT Solutions\",\"url\":\"https:\/\/ncertsolutions.guru\/\",\"sameAs\":[\"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/\",\"https:\/\/www.instagram.com\/ncertsolutionsguru\/\",\"https:\/\/www.linkedin.com\/in\/ncertsolutionsguru\/\",\"https:\/\/in.pinterest.com\/ncertsolutionsguru\/\",\"https:\/\/en.wikipedia.org\/wiki\/National_Council_of_Educational_Research_and_Training\",\"https:\/\/twitter.com\/ncertsolguru\"],\"logo\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/ncertsolutions.guru\/#logo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1\",\"contentUrl\":\"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1\",\"width\":460,\"height\":45,\"caption\":\"NCERT Solutions\"},\"image\":{\"@id\":\"https:\/\/ncertsolutions.guru\/#logo\"}},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/ncertsolutions.guru\/#website\",\"url\":\"https:\/\/ncertsolutions.guru\/\",\"name\":\"NCERT Solutions\",\"description\":\"NCERT Solutions for Class 1 to 12\",\"publisher\":{\"@id\":\"https:\/\/ncertsolutions.guru\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/ncertsolutions.guru\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#primaryimage\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1\",\"contentUrl\":\"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1\",\"width\":170,\"height\":17,\"caption\":\"NCERT Solutions\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#webpage\",\"url\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/\",\"name\":\"NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 - NCERT Solutions\",\"isPartOf\":{\"@id\":\"https:\/\/ncertsolutions.guru\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#primaryimage\"},\"datePublished\":\"2021-06-23T04:39:27+00:00\",\"dateModified\":\"2021-06-23T04:42:02+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/ncertsolutions.guru\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1\"}]},{\"@type\":\"Article\",\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#webpage\"},\"author\":{\"@id\":\"https:\/\/ncertsolutions.guru\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3\"},\"headline\":\"NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1\",\"datePublished\":\"2021-06-23T04:39:27+00:00\",\"dateModified\":\"2021-06-23T04:42:02+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#webpage\"},\"wordCount\":1817,\"publisher\":{\"@id\":\"https:\/\/ncertsolutions.guru\/#organization\"},\"image\":{\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png\",\"articleSection\":[\"CBSE Class 9\"],\"inLanguage\":\"en-US\"},{\"@type\":\"Person\",\"@id\":\"https:\/\/ncertsolutions.guru\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3\",\"name\":\"Prasanna\",\"image\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/ncertsolutions.guru\/#personlogo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g\",\"caption\":\"Prasanna\"},\"url\":\"https:\/\/ncertsolutions.guru\/author\/prasanna\/\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 - NCERT Solutions","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/","og_locale":"en_US","og_type":"article","og_title":"NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 - NCERT Solutions","og_description":"These NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1 Question 1. The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. the angles of … NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 Read More »","og_url":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/","og_site_name":"NCERT Solutions","article_publisher":"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","article_published_time":"2021-06-23T04:39:27+00:00","article_modified_time":"2021-06-23T04:42:02+00:00","og_image":[{"url":"https:\/\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png"}],"twitter_card":"summary_large_image","twitter_creator":"@ncertsolguru","twitter_site":"@ncertsolguru","twitter_misc":{"Written by":"Prasanna","Est. reading time":"14 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Organization","@id":"https:\/\/ncertsolutions.guru\/#organization","name":"NCERT Solutions","url":"https:\/\/ncertsolutions.guru\/","sameAs":["https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","https:\/\/www.instagram.com\/ncertsolutionsguru\/","https:\/\/www.linkedin.com\/in\/ncertsolutionsguru\/","https:\/\/in.pinterest.com\/ncertsolutionsguru\/","https:\/\/en.wikipedia.org\/wiki\/National_Council_of_Educational_Research_and_Training","https:\/\/twitter.com\/ncertsolguru"],"logo":{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/#logo","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","width":460,"height":45,"caption":"NCERT Solutions"},"image":{"@id":"https:\/\/ncertsolutions.guru\/#logo"}},{"@type":"WebSite","@id":"https:\/\/ncertsolutions.guru\/#website","url":"https:\/\/ncertsolutions.guru\/","name":"NCERT Solutions","description":"NCERT Solutions for Class 1 to 12","publisher":{"@id":"https:\/\/ncertsolutions.guru\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/ncertsolutions.guru\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#primaryimage","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1","width":170,"height":17,"caption":"NCERT Solutions"},{"@type":"WebPage","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#webpage","url":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/","name":"NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 - NCERT Solutions","isPartOf":{"@id":"https:\/\/ncertsolutions.guru\/#website"},"primaryImageOfPage":{"@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#primaryimage"},"datePublished":"2021-06-23T04:39:27+00:00","dateModified":"2021-06-23T04:42:02+00:00","breadcrumb":{"@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/ncertsolutions.guru\/"},{"@type":"ListItem","position":2,"name":"NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1"}]},{"@type":"Article","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#article","isPartOf":{"@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#webpage"},"author":{"@id":"https:\/\/ncertsolutions.guru\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3"},"headline":"NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1","datePublished":"2021-06-23T04:39:27+00:00","dateModified":"2021-06-23T04:42:02+00:00","mainEntityOfPage":{"@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#webpage"},"wordCount":1817,"publisher":{"@id":"https:\/\/ncertsolutions.guru\/#organization"},"image":{"@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-8-ex-8-1\/#primaryimage"},"thumbnailUrl":"https:\/\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png","articleSection":["CBSE Class 9"],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/ncertsolutions.guru\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3","name":"Prasanna","image":{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/#personlogo","inLanguage":"en-US","url":"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g","caption":"Prasanna"},"url":"https:\/\/ncertsolutions.guru\/author\/prasanna\/"}]}},"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/posts\/25450"}],"collection":[{"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/users\/9"}],"replies":[{"embeddable":true,"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/comments?post=25450"}],"version-history":[{"count":0,"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/posts\/25450\/revisions"}],"wp:attachment":[{"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/media?parent=25450"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/categories?post=25450"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/tags?post=25450"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}