{"id":2251,"date":"2020-09-08T06:34:50","date_gmt":"2020-09-08T01:04:50","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=2251"},"modified":"2021-07-02T15:03:44","modified_gmt":"2021-07-02T09:33:44","slug":"kseeb-solutions-class-9-maths-chapter-11-ex-11-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-11-ex-11-3\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 11 Areas of Parallelograms and Triangles Ex 11.3"},"content":{"rendered":"

Karnataka Board Class 9 Maths Chapter 11 Areas of Parallelograms and Triangles Ex 11.3<\/h2>\n

Question 1.
\nIn Fig., E is any point on median AD of a \u2206ABC. Show that ar. (ABE) = ar.(ACE).
\n\"KSEEB
\nSolution:
\nData: E is any point on Median AD of an \u2206ABC.
\nTo Prove: ar.(\u2206ABE) = ar. (\u2206ACE)
\nProof: In \u2206ABC, AD is the median.
\n\u2234 \u2206ABD = \u2206ACD ……….. (i)
\nIn \u2206EBC, DE is the median.
\n\u2234 \u2206EBC = \u2206ECD …………. (ii)
\nSubtracting (ii) from (i),
\n\u2206ABD – \u2206EBD = \u2206ACD – \u2206ECD
\n\u2234 ar. (\u2206ABE) = ar. (\u2206ACE).<\/p>\n

Question 2.
\nIn a triangle ABC, E is the mid-point of median AD. Show that ar.(\u2206BED) = \\(\\frac{1}{4}\\) ar.(\u2206ABC).
\n\"KSEEB
\nSolution:
\nData: In a triangle ABC, E is the midpoint of median AD.
\nTo Prove: ar.(\u2206BED) = \\(\\frac{1}{4}\\) ar.(\u2206ABC)
\nProof: D is the midpoint of BD.
\n\u2234 BD = \\(\\frac{1}{2}\\) BC ……… (i)
\nE is the midpoint of AD.
\n\u2234 DE = \\(\\frac{1}{2}\\) AD …………(ii)
\n\"KSEEB<\/p>\n

Question 3.
\nShow that the diagonals of a parallelogram divide it into four triangles of equal area.
\n\"KSEEB
\nSolution:
\nData: Diagonals of ABCD, AC and BD intersect at ‘O’.
\nTo Prove: ar.(\u2206AOD) = ar.(\u2206DOC) = ar.(\u2206COB) = ar.(\u2206AOB)
\nProof: ‘O’ is the mid-point of diagonals AC and BD.
\nDO is the median of \u2206ADC.
\n\u2234 ar.(\u2206AQD) = ar.(\u2206DOC) ……….. (i)
\nCO is the median of \u2206DCB.
\n\u2234 ar. (\u2206DOC) = ar.(\u2206BOC) ……….. (ii)
\nBO is the median of \u2206ABC.
\n\u2234 ar.(\u2206BOC) = ar.(\u2206AOB) …………. (iii)
\nAO is the median of \u2206ADB.
\n\u2234 ar.(\u2206AOB) = ar.(\u2206AOD) …………. (iv)
\nFrom (i), (ii), (iii) and (iv),
\nar. (\u2206AOD) = ar. (\u2206DOC) = ar. (\u2206COB) = ar. (\u2206AOB)
\n\u2234 The diagonals of a parallelogram divide it into four triangles of equal area.<\/p>\n

Question 4.
\nIn Fig., ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that ar.(\u2206ABC) = ar.(\u2206ABD).
\n\"KSEEB
\nSolution:
\nData \u2206ABC and \u2206ABD are two triangles on the same base AB.
\nLine segment CD bisected by AB at O’.
\nTo Prove: ar.(\u2206ABC) = ar.(\u2206ABD)
\nProve: AB bisects line segment CD at O’.
\n\u2234 OC = OD
\nIn \u2206ADC, AO is the median on CD.
\n\u2234 ar.(\u2206AOD) = ar.(\u2206AOC) …………… (i)
\nIn \u2206BDC, BO is the median on CD.
\n\u2234 ar.(\u2206BOD) = ar.(\u2206BOC) ……………. (ii)
\nAdding (i) and (ii),
\nar.(\u2206AOD) + ar.(\u2206BOD) = ar.(\u2206AOC)+ ar.(\u2206BOC)
\n\u2234 ar.(\u2206ABD) = ar.(\u2206ABC)<\/p>\n

Question 5.
\nD, E, and F are respectively the midpoints of the sides BC, CA, and AB of an \u2206ABC. Show that
\n\"KSEEB
\n(i) BDEF is a parallelogram.
\n(ii) ar.(\u2206DEF) = \\(\\frac{1}{4}\\) ar. (\u2206ABC)
\n(iii) ar.(BDEF) = \\(\\frac{1}{2}\\) ar.(\u2206ABC).
\nSolution:
\nData : D, E, and F are respectively the mid-points of the sides BC, CA, and AB of an \u2206ABC.
\nTo Prove:
\n(i) BDEF is a parallelogram.
\n(ii) ar.(ADEF) = \\(\\frac{1}{4}\\) ar.(\u2206ABC)
\n(iii) ar.(BDEF) = \\(\\frac{1}{2}\\) ar.(\u2206ABC).
\nProof: (i) In \u2206ABC, D, E and F are respectively mid-points of the sides BC, CA and AB.
\n\u2234 BD = \\(\\frac{1}{2}\\) BC
\nand EF = \\(\\frac{1}{2}\\) BC (mi-point theorem) z .
\nEF || BC
\n\u2234 BD = EF
\nBD || EF
\n\u2234 BDEF is a parallelogram.<\/p>\n

(ii) Diagonals of BDEF divides into two congruent triangles.
\n\u2234 \u2206BDF \u2245 \u2206DEF ………… (i)
\nSimilarly, in DCEF,
\n\u2206DCE \u2245 \u2206DEF …………. (ii)
\nSimilarly, in AEDF,
\n\u2234 \u2206AEF \u2245 \u2206DEF ……….. (iii)
\nFrom (i), (ii) and (iii),
\nar.(\u2206BDF) = ar.(\u2206DEF) = ar.(\u2206DCE) = ar.(\u2206AEF)
\n\u2234 4 \u00d7 ar.(\u2206DEF) = ar.(\u2206ABC)
\n\u2234 ar.(\u2206DEF) = \\(\\frac{1}{4} \\)ar.(\u2206ABC)<\/p>\n

(iii) ar.(\u2206BED) + ar.(\u2206DCE) + ar.(\u2206AEF) + ar.(\u2206DEF) = ar.(\u2206ABC)
\n\u2206BDF + \u2206BDF + \u2206DEF + \u2206DEF = ar.(\u2206ABC)
\n2 \u00d7 \u2206BDE + 2 \u00d7 \u2206DEF = ar.(\u2206ABC)
\n2(\u2206BDF + \u2206DEF) = ar.(\u2206ABC)
\n2 \u00d7 ar. (\"KSEEBBDEF) = ar.(\u2206ABC)
\nar.(\"KSEEBBDEF) = \\(\\frac{1}{2}\\) \u00d7 ar.(\u2206ABC)<\/p>\n

Question 6.
\nIn Fig., diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
\n\"KSEEB
\n(i) ar.(\u2206DOC) = ar.(\u2206AOB)
\n(ii) ar.(\u2206DCB) = ar.(\u2206ACB)
\n(iii) DA || CB or ABCD is a parallelogram.
\n[Hint: From D and B, draw perpendiculars to AC.]
\nSolution:
\nData: Diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD,
\nTo Prove:
\n(i) ar.(\u2206DOC) = ar.(\u2206AOB)
\n(ii) ar.(\u2206DCB) = ar.(\u2206ACB)
\n(iii) DA || CB or ABCD is a parallelogram
\n\"KSEEB
\nConstruction: Draw DM \u22a5 AC and BN \u22a5 AC.
\nProof: (i) DM \u22a5 CO, BN \u22a5 AO.
\nIn \u2206DOM and \u2206BON,
\nDO = BO (data)
\n\u2220DMO = \u2220BNO = 90\u00b0 (Construction)
\n\u2220DOM = \u2220BON (Vertically opposite angles)
\n\u2234 \u2206DOM \u2245 \u2206BON (AAS Postulate)
\n\u2234 DM = BN
\n\u2234 ar.(\u2206DOM) = ar.(\u2206BON) ………… (i)
\nIn \u2206DMC, and \u2206BNA,
\n\u2220DMC = \u2220BNA = 90\u00b0
\nDC = AB (Data)
\nDM = BN (Proved)
\n\u2234 \u22a5\u2206DMC = \u22a5\u2206BNA
\n\u2234 ar. (\u2206DMC) = ar, (\u2206BNA) ………… (ii)
\nAdding (i) and (ii),
\nar.(\u2206DOM)+ ar.(\u2206DMC) = ar.(\u2206BON) + ar.(\u2206BNA)
\n\u2234 (\u2206DOC) = (\u2206AOB) …………….. (iii)<\/p>\n

(ii) (\u2206DOC) = (\u2206AOB) (Proved)
\nAdding \u2206BOC to both sides,
\n\u2206DOC + \u2206BOC = \u2206AOB + \u2206BOC
\n\u2234 \u2206DCB = \u2206ACB.<\/p>\n

(iii) From equation (ii), \u2206DMC = \u2220BNA
\n\u2234 \u2220DCM = \u2220BAN.
\nThese are a pair of alternate angles.
\n\u2234 AB || DC
\nand also AB = DC (Data)
\n\u2234 ABCD is a parallelogram OR DA || CB.<\/p>\n

Question 7.
\nD and E are points on sides AB and AC respectively of \u2206ABC such that ar.(\u2206DBC) = ar.(\u2206EBC). Prove that, DE || BC.
\n\"KSEEB
\nSolution:
\nData : D and E are mints on sides AB and AC respectively of \u2206ABC such that ar.(\u2206DBC) = ar.(\u2206EBC).
\nTo Prove DE || BC.
\nConstruction: BE and DC joined.
\nProof: \u2206DBC and \u2206EBC are on base BC.
\nar.(\u2206DBC) = ar. (\u2206EBC) (Data)
\nTwo triangles on the same base (or equal bases) and between the same parallel are equal in area.
\n\u2234 BC || DE.<\/p>\n

Question 8.
\nXY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar.(\u2206ABE) = ar.(\u2206ACF).
\n\"KSEEB
\nSolution:
\nData: XY is a line parallel to side BC of a triangle ABC.
\nIf BE || AC and CF || AB meet XY at E and F respectively.
\nTo Prove: ar.(\u2206ABE) = ar.(\u2206ACF).
\nProof: XY || BC
\n\u2234 EY || BC
\nand BE || AC
\nBE || YC
\n\u2234 BEYC is a parallelogram.
\nSimilarly, BCFX is a parallelogram.
\nBEYC and BCFX are on base BC and between BC || EF.
\n\u2234 ar.(BEYC) = ar.(BCFX) ………….. (i)
\nNow, \u2206ABE and BEYC are on base BE and in between BE || AC.
\n\u2234 ar.(\u2206ABE) = \\(\\frac{1}{2}\\) ar.(BCYX) …………… (ii)
\nSimilarly, \u2206ACF and BCFX are on base CF and in between CF || AB.
\n\u2234 ar.(\u2206ACF) = \\(\\frac{1}{2}\\) ar.(imggrectBCFX) …………….. (iii)
\nComparing (i), (ii) and (iii),
\n\u2234 ar. (\u2206ABE) = ar. (\u2206ACF)<\/p>\n

Question 9.
\nThe side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed. Show that ar.(ABCD) = ar(PBQR).
\n[Hint: Join AC and PQ. Now compare ar.(\u2206ACQ) and ar.(\u2206APQ)]
\nSolution:
\nData: The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed.
\n\"KSEEB
\nTo Prove: ar.(ABCD) = ar.(PBQR)
\nConstruction: AC and CQ are joined.
\n\"KSEEB
\nProof: \u2206ACQ and \u2206APQ are on base AQ and in between AQ || CP.
\n\u2234 (\u2206ACQ) = (\u2206APQ)
\nNow, \u2206ACQ – \u2206ABQ = \u2206APQ – \u2206ABQ
\n\u2234 \u2206ABC = \u2206QBP ………….. (i)
\nDiagonals divides quadrilateral into two congruent triangles.
\n\"KSEEB<\/p>\n

Question 10.
\nDiagonals AC and BD of a trapezium ABCD with AB || DC, intersect each other at ‘O’. Prove that ar.(\u2206AOD) = ar.(\u2206BOC).
\n\"KSEEB
\nSolution:
\nData: Diagonals AC and BD of a trapezium ABCD with AB || DC, intersect each other at ‘O’.
\nTo Prove: ar.(\u2206AOD) = ar.(\u2206BOC)
\nProof: \u2206ABD and \u2206ABC are on base AB and between AB || DC.
\n\u2234 ar. (\u2206ABD) = ar. (\u2206ABC)
\nSubtracting ar.(\u2206ABO) on both sides,
\nar.(\u2206ABD) – ar.(\u2206ABO) = ar.(\u2206ABC) – ar.(\u2206ABO)
\n\u2234 ar.(\u2206AOD) = ar.(\u2206BOC).<\/p>\n

Question 11.
\nIn fig., ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
\n\"KSEEB
\n(i) ar.(\u2206ACB) = ar.(\u2206ACF)
\n(ii) ar.(\u2206EDF) = ar.(ABCDE).
\nSolution:
\nData: ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.
\nTo Prove: (i) ar.(\u2206ACB) = ar.(\u2206ACF)
\n(ii) ar.(AEDF) = ar.(ABCDE)
\nProof: (i) \u2206ACB and \u2206HCF are on base AC and between AC || BF.
\n\u2234 ar.(\u2206ACB) = ar.(\u2206ACF)
\n(ii) Now, W(\u2206ACB) = W(\u2206ACF) proved,
\nadding ACDE on both sides,
\nar.(\u2206ACB) + ar.(ACDE) = ar.(\u2206ACF) + ar.(ACDE)
\n\u2234 ar.(ABCDE) = ar.(AEDF)<\/p>\n

Question 12.
\nA villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given an equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
\n\"KSEEB
\nSolution:
\nLet the shape of plot be quadrilateral ABCD. Itwaari agrees to give land as follows:
\nDiagonal AC of a quadrilateral is joined. BA side of land ABCD is produced to E.
\nCE and DE || AC are drawn.
\nNow, \u2206ACD and \u2206ACE are on same base aC and in between AC || DE.
\n\u2234 ar.(\u2206ACD) = ar.(\u2206ACE)
\nSubtracting ar.(\u2206AOC) on both sides
\nar (\u2206ACD) – ar(\u2206AOC) = ar(\u2206ACE) – ar(\u2206AOC)
\nar.(\u2206DOC) = ar.(\u2206AOC).
\nNow, Itwaari agrees to give ADOC to construct a Health Centre. Area of \u2206AOE and \u2206CEB is remains with Itrwaari.<\/p>\n

Question 13.
\nABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that area(ADx) = area(ACy). [Hint: Join Cx.)
\n\"KSEEB
\nSolution:
\nData : ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at x and BC at y.
\nTo Prove: ar.(ADx) = ar.(ACy)
\nConstruction: Join CX.
\nProof: ABCD is a trapezium.
\n\u2234 AB || DC
\n\u2206ADx and \u2206ACx are on same base Ax and in between AB || DC.
\n\u2234 (\u2206ADx) = (\u2206ACx) ………… (i)
\nBut, AC || xy (Data)
\n\u2206ACx and \u2206ACy are on same base AC and in between AC || xy.
\n\u2234 (\u2206ACx) = (\u2206ACy) …………. (ii)
\nFrom comparing (i) and (ii),
\n\u2234 (\u2206ADx) = (\u2206ACy).<\/p>\n

Question 14.
\nIn Fig., AP || BQ || CR.
\nProve that ar.(AQC) = ar.(PBR).
\n\"KSEEB
\nSolution:
\nData: AP || BQ || CR.
\nTo Prove: ar.(\u2206AQC) = ar.(\u2206PBR)
\nProof: AP || BQ. CR is a line segment.
\n\u2206ABQ and \u2206PBQ are on base BQ and in between AP || BQ
\nar.(\u2206ABQ) = ar.(\u2206PBQ) ………….. (i)
\n\u2206BQC and \u2206BQR are on base BQ and in between BQ || CR.
\n\u2234 ar.(\u2206BQC) = ar.(\u2206BQR) ………….. (ii)
\nAdding (i) and (ii),
\nar.(\u2206ABQ) + ar.(\u2206BQC) = ar.(\u2206PBQ) + ar.(\u2206BQR)
\n\u2234 ar.(\u2206AQC) = ar.(\u2206PBR).<\/p>\n

Question 15.
\nDiagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar.(AOD) = ar.(BOC). Prove that ABCD is a trapezium.
\n\"KSEEB
\nSolution:
\nData: Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar.(AOD) = ar.(BOC).
\nTo Prove: ABCD is a trapezium.
\nProof: ar.(\u2206AOD) = ar.(\u2206BOC) (Data)
\nAdding ar.(\u2206ODC) on both sides,
\nar.(\u2206AOD) + ar.(\u2206ODC) = ar.(\u2206BOC) + ar.(\u2206ODC)
\n\u2234 ar.(\u2206ADC) = ar.(\u2206BDC)
\nArea of these triangles are equal, they are on base DC and in between DC, AB straight lines.
\n\u2234 DC || AB.
\nNow, one pair of opposite sides of a quadrilateral are parallel, hence ABCD is a trapezium.<\/p>\n

Question 16.
\nIn Fig., ar.(DRC) = ar.(DPC) and ar.(BDP) = ar.(ARC). Show that both the quadrilaterals ABCD and CPR are trapeziums.
\n\"KSEEB
\nSolution:
\nData: in fig. ar.(DRC) = ar.(DPC) and
\nar.(BDP) = ar.(ARC).
\nTo Prove: (i) ABCD is a trapezium
\n(ii) DCPR is a trapezium.
\nProof: (i) ar.(\u2206BDP) = ar.(\u2206ARC) (Data)
\nar.(\u2206DPC) + ar.(\u2206DCB) = ar.(\u2206DCR) + ar.(\u2206DCA)
\nBut, \u2206DRC = \u2206DPC. (Data)
\n\u2234 ar.(\u2206DCB) = ar.(\u2206DCA)
\nThese are on same base DC and in between straight lines DC and AB.
\n\u2234 DC || AB.
\nNow, one pair of opposite sides of quadrilateral ABCD are parallel, hence ABCD is a trapezium.<\/p>\n

(ii) \u2206DRC and \u2206DPC are on the same base DC and in between DC and RP and they are equal in area.
\n\u2234 DC || RP.
\n\u2234 Hence ‘Quadrilateral’ DCPR is a trapezium.<\/p>\n

KSEEB Solutions for Class 9 Maths<\/a><\/h5>\n","protected":false},"excerpt":{"rendered":"

Karnataka Board Class 9 Maths Chapter 11 Areas of Parallelograms and Triangles Ex 11.3 Question 1. In Fig., E is any point on median AD of a \u2206ABC. Show that ar. (ABE) = ar.(ACE). Solution: Data: E is any point on Median AD of an \u2206ABC. To Prove: ar.(\u2206ABE) = ar. (\u2206ACE) Proof: In \u2206ABC, …<\/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\/2251"}],"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=2251"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2251\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2251"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2251"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2251"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}