{"id":2231,"date":"2020-09-08T05:00:20","date_gmt":"2020-09-07T23:30:20","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=2231"},"modified":"2021-07-02T15:03:45","modified_gmt":"2021-07-02T09:33:45","slug":"kseeb-solutions-class-9-maths-chapter-11-ex-11-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-11-ex-11-2\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 11 Areas of Parallelograms and Triangles Ex 11.2"},"content":{"rendered":"

Karnataka Board Class 9 Maths Chapter 11 Areas of Parallelograms and Triangles Ex 11.2<\/h2>\n

Question 1.
\nIn Fig., ABCD is a parallelogram, AE \u22a5 DC and CF \u22a5 AD. AB = 16 cm., AE = 8 cm, CF = 10cm, find AD.
\n\"KSEEB
\nSolution:
\nABCD is a parallelogram.
\nAE \u22a5 DC and CF \u22a5 AD.
\nAB = 16 Cm, AE = 8 cm, CF = 10 cm, then AD = ?
\n(i) Area of ABCD = base \u00d7 altitude
\n= DC \u00d7 AE ( \u2235 AB=DC)
\n= 16 \u00d7 8
\n= 128 sq. cm.
\nWhether ABCD is quadrilateral, ADCB is a quadrilateral.
\n\u2234 Area of ADCB = 128 sq. cm.
\nBase = AD = ?
\nAltitude, CF= 100 cm.
\nBase \u00d7 Altitude = Area
\nAD \u00d7 CF = 128
\nAD \u00d7 10 = 128
\n\u2234 AD = \\(\\frac{128}{10}\\)
\n\u2234 AD = 12.8 cm.<\/p>\n

Question 2.
\nIf E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that
\n\"KSEEB
\nar(EFGH) = \\(\\frac{1}{2}\\) ar(ABCD)
\nSolution:
\nData: E, F, G and H are mid-points of the sides of a parallelogram ABCD,
\nTo Prove: area (EFGH) = \\(\\frac{1}{2}\\) area (ABCD)
\nConstruction: HF is joined.
\nProof: Now, AD = BC and AD || BC
\n\u2234 2AH = 2BF
\n\u2234 AH = BF and AH || BF
\n\u2234 AHFB is a parallelogram.
\nSimilarly, HDCF is a parallelogram.
\nNow, DEHF and Quadrilateral AHFD are on base HF and in between HF || AB.
\n\"KSEEB<\/p>\n

Question 3.
\nP and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (PAB) = ar (BQC).
\n\"KSEEB
\nSolution:
\nData: P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.
\nTo Prove: ar(\u2206APB) = ar(\u2206BQC)
\nProof: ABCD is a parallelogram.
\n\u2234 AB || DC AB = DC
\nAD || BC AD = BC
\nNow \u2206APB and ABCD are on same base AB and in between AB || DC
\n\u2234 Area(\u2206APB) = \\(\\frac{1}{2}\\) \u00a0Area\u00a0ABCD) \u2026\u2026\u2026 (i)
\nSimilarly, \u2206BQC and BADC are on the same base BC and in between BC || AD.
\n\u2234 Area(\u2206BQC) = \\(\\frac{1}{2}\\) \u00a0Area(ABCD) \u2026\u2026\u2026. (ii)
\nFrom (i) and (ii),
\n\u2234 Area(\u2206APB) = Area (\u2206BQC).<\/p>\n

Question 4.
\nIn Fig., P is a point in the interior of a parallelogram ABCD, Show that
\n\"KSEEB
\n(i) ar.(ABP)= ar(PCD) = \\(\\frac{1}{2}\\) ar (\"KSEEBABCD)
\n(ii) ar(APD)+ ar(PBC) = ar(APB) + ar(PCD)
\n(Hint: Through P, draw a line parallel to AB).
\nSolution:
\nData : P is any point in the interior of a parallelogram ABCD. PA, PB, PC and PD are joined.
\nTo Prove: (i) ar(ABP= ar(PCD)= \\(\\frac{1}{2}\\) ar (\"KSEEBABCD)
\n(ii) ar(APD) + ar(PBC)= ar(APB) + ar(PCD)
\n\"KSEEB
\nConstruction: AB || XY is drawn through P.
\nProof: (i) XY || AB || DC
\n\u2234 ABYX and XDCY are parallelograms.
\n\u2206APB and \"KSEEBABYX are on base AB and in between AB||XY.
\n\"KSEEB
\nConstruction: AD || MN is drawn through P.
\nProof: \u2206ADP and \"KSEEBADMN lie on the base AD and in between AD || MN
\n\"KSEEB<\/p>\n

Question 5.
\nIn Fig., PQRS and ABRS are parallelograms and X is any point on side BR. Show that
\n\"KSEEB
\n(i) ar.(PQRS) = ar.(\"KSEEBABRS)
\n(ii) ar.(AXS) + = \\(\\frac{1}{2}\\) ar. (\"KSEEBPQRS).
\nSolution:
\nData: PQRS and ABRS are parallelograms and X is any point on side BR.
\n\"KSEEB<\/p>\n

Question 6.
\nA farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
\n\"KSEEB
\nSolution:
\nIf AP and AQ are joined \"KSEEBPQRS is divided into three triangles, \u2206SAP, \u2206APQ and \u2206ARQ.
\n\u2206APQ and \"KSEEBPQRS are on base PQ and in between PQ || SR.
\n\"KSEEB
\nFrom (i) and (ii).
\n\u2234 ar.(\u2206APQ) = ar.(\u2206SAP) + ar.(\u2206ARQ)
\n\u2234 Farmer can use the part of the field to sow wheat,
\ni.e. Ar. (\u2206SAP + \u2206ARQ) and in the same area, he can use the Area of \u2206APQ to sow pulses.<\/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 11.2 Question 1. In Fig., ABCD is a parallelogram, AE \u22a5 DC and CF \u22a5 AD. AB = 16 cm., AE = 8 cm, CF = 10cm, find AD. Solution: ABCD is a parallelogram. AE \u22a5 DC and CF \u22a5 AD. AB …<\/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\/2231"}],"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=2231"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2231\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2231"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2231"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2231"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}