{"id":27673,"date":"2020-07-11T23:09:29","date_gmt":"2020-07-11T17:39:29","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=27673"},"modified":"2021-07-02T15:07:05","modified_gmt":"2021-07-02T09:37:05","slug":"kseeb-solutions-for-class-8-maths-chapter-15-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-8-maths-chapter-15-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 15 Quadrilaterals Additional Questions"},"content":{"rendered":"

Students can Download Maths Chapter 15 Quadrilaterals Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 8 Maths Chapter 15 Quadrilaterals Additional Questions<\/h2>\n

Question 1.
\nComplete the following
\na. A quadrilateral has four<\/span> sides.
\nb. A quadrilateral has two<\/span> diagonals
\nc. A quadrilateral in which one pair of sides are parallel to each other is trapezium<\/span>.
\nd. In an isosceles trapezium the base angles are equal<\/span>.
\ne. In a rhombus, the diagonals bisect each other in right<\/span> angles.
\nf. In a square, all the sides are equal<\/span>.<\/p>\n

Question 2.
\nLet ABCD be a parallelogram, what special name will you give if.
\na. AB = BC
\nSolution:
\nRhombus<\/p>\n

b. \u2220BAD – 90\u00b0
\nSolution:
\nRectangle<\/p>\n

c. AB = AD and \u2220BAD = 90\u00b0
\nSolution:
\nSquare<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nA quadrilateral has three acute angles each measuring 70\u00b0. What is the measure of the fourth angle?
\nSolution:
\nMeasure of the fourth angle
\n= 360 – (70 + 70 + 70) = 360 – 210
\n= 140\u00b0<\/p>\n

Question 4.
\nThe difference between two adjacent angles of a parallelogram is 20\u00b0 . Find the measures of all the angles \u2018 of the parallelogram.
\nSolution:
\nLet the adjacent angles be \u2018x\u2019 and \u2018y\u2019
\nx + y = 180\u00b0 [Adjacent angles are supplementary]<\/p>\n

\"KSEEB<\/p>\n

x = \\(\\frac{200}{2}\\) ; x = 100\u00b0
\nx + y = 180
\n100 + y = 180\u00b0
\ny = 180 – 100
\ny = 180\u00b0
\n\u2234The angles are 100\u00b0, 80\u00b0, 100\u00b0 and 80\u00b0<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nThe angles of a quadrilateral are in the ratio 1 : 2 : 3 : 4 find all the angles.
\nSolution:
\nRatio of the angles 1 : 2 : 3 : 4
\nLet the angles be x, 2x, 3x and 4x
\nx + 2x + 3x + 4x = 360\u00b0
\n[Angle sum property of a quadrilateral]
\n10x = 360\u00b0
\nx = \\(\\frac{360}{10}\\) ; x = 36\u00b0
\n2x = 2 x 36 = 72\u00b0, 3x = 3 x 36 = 108\u00b0
\n4x = 4 x 36 = 144\u00b0
\n\u2234 The angles are 36\u00b0, 72\u00b0, 108\u00b0 and 144\u00b0<\/p>\n

Question 6.
\nLet PQRS be a parallelogram with PQ = 10 cm and QR = 6 cm. Calculate the measures of the other two sides and measures of the other two sides and perimter of PQRS.
\nSolution:
\nPQ = RS = 10 cm
\n[opposite sides of a parallelogram]
\nQR = SP = 6 cm
\nPerimeter = PQ + QR + RS + SP
\n= 10 + 6 + 10 + 6
\n= 32 cm<\/p>\n

Question 7.
\nThe perimeter of a square is 60 cm. Find the length of its side.
\nSolution:
\nPerimeter of a square = 4a
\n60 = 4a
\n\u2234 a = \\(\\frac{60}{4}\\) a = 15 cm
\nThe length of the side is 15 cm<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nLet ABCD be a square and let AC = BD = 10 cm let AC and BD intersect at O. Find OC and OD.
\nSolution:
\n\"KSEEB
\nThe diagonals of a square bisect each other
\n\"KSEEB<\/p>\n

Question 9.
\nLet PQRS be a rhombus with PR = 15 cm and QS = 8 cm. Find the area of the rhombus.
\nSolution:
\nArea of rhombus PQRS = Area of \u2206 PQS + Area of \u2206 RQS
\n\"KSEEB
\n\"KSEEB
\nArea of rhombus = 60 cm2<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nLet ABCD be a parallelogram and suppose the bisectors of \u2220A and \u2220B meet at P prove that \u2220APB = 90\u00b0
\nSolution:
\n\"KSEEB
\n\u2220A + \u2220B = 180\u00b0
\n[Adjacent angles of a prallelogram]
\n\\(\\frac{1}{2}\\)\u2220A + \\(\\frac{1}{2}\\) \u2220B = \u2220B . 180\u00b0 [Multiplying by \\(\\frac{1}{2}\\) ]
\n\u2220PAB + \u2220PBA = 90\u00b0 [AP and BP are angular bisectors]
\n\u2220PAB + \u2220PBA +\u2220APB = 180\u00b0 [Angle sum property of a triangle]
\n90 +\u2220APB = 180\u00b0
\n\u2220APB = 180 – 90
\n\u2220APB = 90\u00b0<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nLet ABCD be a sqaure. Locate points P, Q, R, S on the sides AB, BC, CD, DA respectively such that AP = BQ = CR = DS. Prove that PQRS is a square.
\nSolution:
\nLet AP = BQ = CR = DS = x
\n\"KSEEB
\nAP + PB = AB
\nx + PB = AB
\nPB = AB – x
\nAB = BC
\n\u2234 QC = AB – x
\nIn \u2206 PBQ and \u2206 QCR
\nBQ = CR[= x]
\nBP = QC [AB – x]
\n\u2220B = \u2220C [= 90\u00b0]
\n\u2234 \u2206 PBQ = A QCR [SAS postulate]
\n\u2234 PQ = QR
\n[corresponding sides] similarly we can show that QR = SR and SR = SP
\ni.e., PQ = QR = RS = SP …… (i)
\n\u2220BQP = \u2220CRQ [corresponding angles]
\n\u2220BQP + \u2220PQR + \u2220RQC = 180\u00b0 [Angle sum property of a triangle]
\n\u2220BQP + \u2220PQR + 90 – \u2220CRQ = 180\u00b0
\n[\u2234 \u2220CQR = 90 – \u2220CRQ]
\n\u2220BQP + \u2220PQR – \u2220BQP = 180 – 90
\n[\u2235 \u2220DQP = \u2220CRQ]
\n\u2220PQR = 90\u00b0
\n\u2234 PQRS is a square.<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nLet ABCD be a rectangle and let P, Q, R, S be the midpoints of AB, BC, CD, DA respectively prove that PQRS is a Rhombus.
\nSolution:
\nIn \u2206 APS and \u2206 CQR
\n\"KSEEB
\nAP = CR [P and R are the midpoints of AB and CD)
\nAS = CQ [S and Q are the midpoints of AD and BC]
\n\u2220SAP = \u2220RCQ [=90\u00b0 angles of a rectangle]
\n\u2234 \u2206 APS \u2245 \u2206 CQR [SAS postulate]
\n\u2234 SP = QR [corresponding sides] similarly we can show PQ = QR = RS
\ni.e., PQ = QR = RS = SP
\n\u2234 PQRS is a rhombus<\/p>\n

Question 13.
\nLet ABCD be a quadrilateral in which diagonals intersect at \u2018O\u2019 perpendicularly. Prove that AB + BC + CD + DA > (AC + BD)
\nSolution:
\nIn A AOD, AD > AO + OD … (i)
\nIn A AOB, AB > AO + OB … (ii)
\nIn A BOC, BC > BO + OC … (iii)
\nIn A COD, CD > DO + OC … (iv)
\n\"KSEEB
\nBy adding (i), (ii), (iii) and (iv)
\nAD + AB + BC + CD > AO + OD + AO + OB + BO + OC + DO + CO > 2 AO + 2 OC + 2 OB + 2 OD > 2 (AO + OC + OB + OD)
\nAD + AB + BC + CD > 2 (AC + BD) [\u2235 AO + OC = AC OB + OD = BD]
\n\u2234 AD + AB + BC + CD > AC + BD<\/p>\n

\"KSEEB<\/p>\n

Question 14.
\nLet ABCD be a quadrilateral with diagonals AC and BD. Prove the follwoing statements (compare these with the previous problem)
\na. AB + BC + CD > AD
\nb. AB + BC + CD + DA > 2AC
\nc. AB + BC + CD + DA > 2BD
\nd. AB + BC + CD + DA > AC + BD
\nSolution:
\na. AB + BC + CD + DA > AC + BD [From previous problem]
\nAB + BC + CD > AC + BD – AD
\n\u2234 AB + BC + CD > – AD<\/p>\n

b. AB + BC + CD + DA > 2 (AC + BD) [From the previous problem]
\nAB + BC + CD + DA > 2AC + 2BD
\n\u2234 AB + BC + CD + DA > 2AC<\/p>\n

c. AB + BC + CD + DA > 2AC + 2BD [From the previous problem]
\n\u2234 AB + BC + CD + DA > 2BD<\/p>\n

d. AB + BC + CD + DA > AC + BD [From the previous problem]<\/p>\n

Question 15.
\nLet PQRS be a kite such that PQ > PS. Prove that \u2220PSR > \u2220PQR (Hint : Join QS)
\nSolution:
\n\"KSEEB
\nJoin QS
\nIn \u2206 PQS, PQ > PS [data]
\n\u2234 \u2220PSQ > \u2220PQS …(i)
\nIn \u2206 SRQ, RQ > SR
\n[\u2235 PQ = QR and SP = SR]
\n\u2234 \u2220QSR. > \u2220SQR … (ii)
\nBy adding (i) and (ii)
\n\u2220PSQ + \u2220QSR > \u2220PQS + \u2220SQR
\n\u2220PSR >\u2220PQR<\/p>\n

\"KSEEB<\/p>\n

Question 16.
\nLet ABCD be a quadrilateral in which AB is the smallest side and CD is a
\nlargest side \u2220A > \u2220C and \u2220B > \u2220D ( Hint: join AC and BD)
\nSolution:
\n\"KSEEB
\nJoin AC and BD In \u2206 ABD, AB is the smallest side
\nAD > AB
\n\u2220ABD > \u2220APB
\n[Angle opposite to larger side] …(i)
\nIn \u2206 CBD, CD is the largest side
\n\u2234 CD > CB
\n\u2234 [CBD > \u2220CDB [Angle opposite to larger side] …. (ii)
\nBy adding (i) and (ii) .
\n\u2220ABD + \u2220CBD >\u2220ADB + \u2220CDB
\n\u2220ABC > \u2220ADC
\n\u2234 \u2220B >\u2220D
\nIn \u2206 ABC, AB is the smallest side] ……. (iv)
\n\u2234 BC > AB \u2220BAC > \u2220ACB [Angle opposite to larger side] … (iii)
\nIn \u2206 ADC, CD is the largest side
\n\u2234 CD > AD
\n\u2220CAD > \u2220ACD [Angle opposite to larger side] …(iv)
\nBy adding (iii) and (iv)
\n\u2220BAC + \u2220CAD > \u2220ACB + \u2220ACD
\n\u2220BAD > \u2220ACD
\n\u2220A > \u2220C<\/p>\n

\"KSEEB<\/p>\n

Question 17.
\nIn a triangle ABC, let D be the midpoint of BC. Prove that AB + AC > 2 AD (what property of quadrilateral is needed here)
\nSolution:
\nIn \u2206 ABD, AB + BD > AD … (i)
\nIn \u2206 ACD, AC + DC > AD …(ii)
\nBy adding (i) and (ii)
\nAB + BD + AC + DC > AD + AD
\nAB + AC + BD + DC > 2AD
\nAB + AC + BC > 2AD
\n\"KSEEB
\nor
\nAB + AC > 2AD<\/p>\n

\"KSEEB<\/p>\n

Question 18.
\nLet ABCD be a quadrilateral and let P, Q, R, S be the midpoints of AB, BC, CD, DA respectively prove that PQRS is a parallelogram [what extra result you need to prove this] It can be proved if ABCD is a rectangle
\nSolution:
\nIn \u2206 APS and \u2206 CQR.
\n\"KSEEB
\nAP = CR [P and R are the midpoints of AB and CD]
\nAS = CQ [\u2235 S and Q are the midpoints of AD and BC]
\n\u2220SAP = \u2220RCQ [= 90\u00b0 angles of a rectangle]
\n\u2234 \u2206 APS \u2245 \u2206 CQR [SAS posltulate]
\n\u2234 SP = QR [corresponding sides]
\nsimilarly we can show PQ = QR = RS
\ni.e., PQ = QR = RS = SP
\n\u2234 PQRS is a rhombus
\n\u2234 PQRS is a parallelogram.<\/p>\n

\"KSEEB<\/p>\n

Question 19.
\nProve that the base angles of an isosceles trapezium are equal.
\nSolution:
\nIn trap ABCD, Draw \u22a5rs<\/sup> to base
\n\\(\\overline{\\mathrm{BE}}\\) and \\(\\overline{\\mathrm{CF}}\\)
\n\"KSEEB
\n\\(\\overline{\\mathrm{BE}}\\) \u2245 \\(\\overline{\\mathrm{CF}}\\) (\u2235 Altitudes of trapezium)
\n\\(\\overline{\\mathrm{AB}}\\) = \\(\\overline{\\mathrm{CD}}\\) (Given)
\n\\(\\overline{\\mathrm{AE}}\\) = \\(\\overline{\\mathrm{FD}}\\) (Sides of Right angle)
\nFrom SSS congruence
\n\u2220A \u2245 \u2220D [CPCT]<\/p>\n

Question 20.
\nSuppose in a quadrilateral ABCD, AC = BD and AD = BC. Prove that ABCD is a trapezium.
\nSolution:
\nData: ABCD is a quadrilateral in which
\nAD = BC and AC = BD
\n\"KSEEB
\nTo Prove: ABCD is a trapezium
\nProof: In \u2206 ADB and \u2206 ACB AB = AB (Common side)
\nAD = BC (data)
\nBD = AC (data)
\n\u2234 \u2206 ADB \u2245 \u2206 ACB (SSS postulate)
\n\u2234 \u2220DAB = \u2220ABC (CPCT) …(i)
\nIn \u2206 ADC and \u2206 BDC
\nDC = DC (common side)
\nAD = BC (data)
\nAC = BD (data)
\n\u2234 \u2206 ADC \u2245 \u2206 BDC (SSS postulate)
\n\u2234 \u2220ADC = \u2220BCD (IPCT) …(ii)
\nIn Quadrilateral ABCD
\n\u2220BAD + \u2220ABC + \u2220DCD + \u2220ADC = 360\u00b0
\n\u2220ABC + \u2220ABC + \u2220BCD + \u2220BCD = 360\u00b0
\n[\u2220ABC = \u2220DAB, \u2220ADC = \u2220BDC] 2\u2220ABC + 2\u2220DCD = 360\u00b0
\n2 (\u2220ABC + \u2220BCD = 360\u00b0)
\n\u2220ABC +\u2220BCD = \\(\\frac{360^{\\circ}}{2}\\) = 180\u00b0
\nBut theSe are co-interior angles AB || DC
\n\u2234 ABCD is a trapezium.<\/p>\n

\"KSEEB<\/p>\n

Question 21.
\nProve logically the diagonals of a parallelogram bisect each other. Show conversely that a quadrilateral in which diagonals bisect each other is a parallelogram.
\nSolution:
\nGiven: A parallelogram ABCD with diagonal AC and BD intersecting at O.
\nTo prove: AO = OC and BO = OD
\n\"KSEEB
\nProof: Consider \u2206 AOD and \u2206 COB.
\nWe observe
\n\u2220ADO = \u2220CBO (alternate angles)
\n\u2220DAO = \u2220BCO (alternate angles)
\nAD = BC (opposite sides of a parallelogram)
\nBy ASA, \u2206 AOD \u2245 \u2206 COB.
\nHence AO = OC and OD = OB
\nGiven: a quadrilateral ABCd such that AC and BD bisect each other.
\nTo prove: ABCD is a parallelogram.
\nProof: Let AC and BD intersect at K.
\nIn triangles AKD and CKB, we observe
\nAK = KC (given)
\nBK = KD (given)
\n\u2220AKD = CKB (vertically opposite angles)
\n\"KSEEB
\nHence by SAS congruency,
\n\u2206 AKD \u2245 \u2206 CKB. lt follows that
\n\u2220DAK = \u2220BCK and AD = CB.
\nThus AD || BC. Similarly, we can prove AB || CD and AB = CD.
\nTherefore ABCD is a parallelogram.<\/p>\n

\"KSEEB<\/p>\n

Question 22.
\nProve logically that the diagonal of a rectangle are equal.
\nSolution:
\nIn rectangle ABCD, AC and BD are diagonal
\n\"KSEEB
\nIn \u2206 AOD and \u2206 BOC
\n\u2220AOD = \u2220BOC (V.O.A)
\n\u2220ADO = \u2220OBC (Alternate angle]
\n\u2220DAO = \u2220BCQ (Alternate angle]
\n\u2234 \u2206 AOD \u2245 \u2206 BOC OA = OC, OB = OD
\nIn \u2206 AOB and \u2206 COD
\n\u2220AOB = \u2220COD (V.O.A)
\n\u2220OAB = \u2220PCD (Alternate Angle)
\n\u2220QBA = \u2220ODC (Alternate Angle)
\n\u2206 AOB \u2245 \u2206 COD
\n\u2234 OA = OC, OB = OD
\n\u2234 Diagonal AC = Diagonal BD.<\/p>\n

\"KSEEB<\/p>\n

Question 23.
\nIn a rhombus PQRS, ZSQR = 40\u00b0 and PQ = 3 cm. Find ZSPQ, ZQSR and the perimeter of the rhombus.
\nSolution:
\nGiven PQRS is a rhombus
\n\"KSEEB
\n\u2220SQK = 40\u00b0, PQ = 3cm
\nPerimeter of rhombus PQRS
\n= PQ + QR + SR + SP 3 + 3 + 3 + 3 = 12 cm<\/p>\n

Question 24.
\nLet ABCD be a rhombus and \u2220ABC = 124\u00b0. Calculate. \u2220A, \u2220D and \u2220C.
\nSolution:
\n\"KSEEB
\nIn a rhombus ABCD,
\nIf \u2220ABC = 124\u00b0 then
\n\u2220ADC = 124\u00b0 (Opposite angles are equal)
\n\u2220DAD = 180\u00b0 – 124\u00b0
\n\u2220DAD = 56\u00b0
\n\u2220DCD = \u2220DAD = 56\u00b0
\n\u2234 \u2220A = 56\u00b0, \u2220D = 124\u00b0, \u2220C = 56\u00b0<\/p>\n

\"KSEEB<\/p>\n

Question 25.
\nFour congruent rectangles are placed as shown in the figure. Area of the outer square is 4 times that of the inner square is 4 times that of the inner square. What is the ratio of length to breadth of the congruent rectangles?
\nSolution:
\n\"KSEEB
\nOuter square
\nABCD = 4 \u00d7 Inner square PQRS<\/p>\n

Question 26.
\nLet ABCD be a quadrilateral in which \u2220A = \u2220C, and \u2220B = \u2220D. Prove that ABCD is a parallelogram.
\nSolution:
\nABCD is a Quadrilateral in which AC and BD are the diagonals intersecting Each other at O.
\nOA = OC, and OB = OD.
\n\u2235 O is the midpoint of diagonals
\n\u2234 Each pair of opposite angles are equal. It is a parallelogram.<\/p>\n

\"KSEEB<\/p>\n

Question 27.
\nIn a quadrilateral ABCD, suppose AB = CD and AD = BC. Prove that ABCD is a parallelogram.
\nSolution:
\n\"KSEEB
\nGiven: Quadrilateral ABCD in which AB = DC and AD = BC
\nTo Prove: ABCD is a parallelogram.
\nConstruction: Draw diagonal AC.
\nProof:
\n\"KSEEB
\n\u2234 ABCD is a parallelogram.<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 15 Quadrilaterals Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 8 Maths Chapter 15 Quadrilaterals Additional Questions Question 1. Complete the following a. A quadrilateral has …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27673"}],"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\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=27673"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27673\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=27673"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=27673"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=27673"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}