{"id":4966,"date":"2023-05-18T12:08:22","date_gmt":"2023-05-18T06:38:22","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=4966"},"modified":"2023-05-19T17:27:44","modified_gmt":"2023-05-19T11:57:44","slug":"ap-board-9th-class-maths-solutions-chapter-11-ex-11-3","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-board-9th-class-maths-solutions-chapter-11-ex-11-3\/","title":{"rendered":"AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.3"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions<\/a> Chapter 11 Areas Ex 11.3 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 9th Class Maths Solutions 11th Lesson Areas Exercise 11.3<\/h2>\n

Question 1.
\nIn a triangle ABC, E is the midpoint of median AD. Show that
\n\"AP
\ni) ar \u0394ABE = ar \u0394ACE
\nSolution:
\nIn \u0394ABC; AD is a median.
\n\u2234 \u0394ABD = \u0394ACD …………….. (1)
\n(\u2235 Median divides a triangle in two equal triangles)
\nAlso in \u0394ABD; BE is a median.
\n\u2234 \u0394ABE = \u0394BED = \\(\\frac{1}{2}\\)\u0394ABD …………..(2)
\nAlso in \u0394ACD; CE is a median.
\n\u2234 \u0394ACE = \u0394CDE = \\(\\frac{1}{2}\\)\u0394ACD …………….(3)
\nFrom (1), (2) and (3);
\n\u0394ABE = \u0394ACE<\/p>\n

(OR)<\/p>\n

\u0394ABD = \u0394ACD [\u2235 AD is median in \u0394ABC]
\n\\(\\frac{1}{2}\\) \u0394ABD = \\(\\frac{1}{2}\\) \u0394ACD
\n[Dividing both sides by 2]
\n\u0394ABE = \u0394AEC
\n[\u2235 BE is median of \u0394ABD, CE is median of \u0394ACD]
\nHence proved.<\/p>\n

ii) ar\u0394ABE = \\(\\frac{1}{2}\\) ar(\u0394ABC)
\nSolution:
\n\u0394ABE = \\(\\frac{1}{2}\\) (\u0394ABD)
\n[From (i); BE is median of \u0394ABD]
\n\u0394ABE = \\(\\frac{1}{2}\\) [\\(\\frac{1}{2}\\) \u0394ABC]
\n[\u2235 AD is median of \u0394ABC]
\n= \\(\\frac{1}{4}\\) \u0394ABC
\nHence,proved.<\/p>\n

\"AP<\/p>\n

Question 2.
\nShow that the diagonals of a paral\u00aclelogram divide it into four triangles of equal area.
\nSolution:
\n\u25a1ABCD is a parallelogram.
\nThe diagonals AC and BD bisect each other at ‘O’.
\n\u0394ABC and \u25a1ABCD lie on the same base AB and between the same parallels AB\/\/CD.
\n\"AP
\n\u2234 \u0394ABC = \\(\\frac{1}{4}\\) \u25a1ABCD
\nNow in \u0394ABC; BO is a median
\n[\u2235 O is the midpoint of both diagonals AC and BD]
\n\u2234 \u0394AOB = \u0394BOC ………….(1)
\n[ \u2235 Median divides a triangle into two triangles of equal area]
\nSimilarly \u2235ABD and \u25a1ABCD lie on the same base AB and between the same parallels AB\/\/CD.]
\n\u2234 \u0394ABD = \\(\\frac{1}{2}\\)\u25a1 ABCD
\nAlso \u0394AOB = \u0394AOD …………..(2)
\n[ \u2235 AO is the median of AABD]
\nFrom (1) & (2)
\n\u0394AOB = \u0394BOC = \u0394AOD
\nSimilarly we can prove that
\n\u0394AOD = \u0394COD [ \u2235 OD is the median of \u0394ACD]
\n\u2234 \u0394AOB = \u0394BOC = \u0394COD = \u0394AOD
\nHence proved.<\/p>\n

\"AP<\/p>\n

Question 3.
\nIn the figure, \u0394ABC and \u0394ABD are two triangles on the same base AB. If line segment CD is bisected by \\(\\overline{\\mathbf{A B}}\\) at O, show that ar (\u0394ABC) = ar (\u0394ABD).
\n\"AP
\nSolution:
\nFrom the figure, in \u0394AOC; \u0394BOD
\nOA = OB [ \u2235 given]
\n\u2220AOC = \u2220BOD [Vertically opp. angles]
\n\u2234 \u0394AOC \u2245 \u0394BOD (SAS congruence)
\nThus AC = BD (CPCT)
\n\u2220OAC = \u2220OBD (CPCT) .
\nBut these are alternate interior angles for the lines AC, BD.
\n\u2234 AC \/\/ BD
\nAs AC = BD and AC \/\/ BD;
\n\u25a1ABCD is a parallelogram.
\nAB is a diagonal of oABCD
\n\u21d2 \u0394ABC \u2245 \u0394ABD
\n(\u2235 diagonal divides a parallelogram into two congruent triangles)
\n\u2234 ar(\u0394ABC) = ar (\u0394ABD)<\/p>\n

\"AP<\/p>\n

Question 4.
\nIn the figure \u0394ABC; D, E and F are the midpoints of sides BC, CA and AB respectively. Show that
\ni)BDEF is a parallelogram
\nii) ar (\u0394DEF) = \\(\\frac { 1 }{ 4 }\\) ar(\u0394ABC)
\niii) ar (BDEF) = \\(\\frac { 1 }{ 2 }\\) ar(\u0394ABC)
\n\"AP
\nSolution:
\ni) In \u0394ABC; D, E and F are the mid\u00acpoints of the sides.
\n\u2234 EF\/\/BC FD\/\/AC ED\/\/AB
\nEF = \\(\\frac { 1 }{ 2 }\\)BC FD = \\(\\frac { 1 }{ 2 }\\)AC ED = \\(\\frac { 1 }{ 2 }\\)AB
\n[ \u2235 line joining the mid points of any two sides of a triangle is parallel to third side and equal to half of it]
\n\u2234 In \u25a1BDEF
\nBD = EF [ \u2235 D is mid point of BC and \\(\\frac { 1 }{ 2 }\\) BC = EF]
\nDE = BF
\n\u2234 \u25a1BDEF is a parallelogram.<\/p>\n

ii) \u25a1BDEF is a parallelogram (from (i))
\nThus \u0394BDF = \u0394DEF
\nSimilarly \u25a1CDFE; \u25a1AEDF are also parallelograms.
\n\u2234 \u0394DEF = \u0394CDE =\u0394AEF
\n\u2234 \u0394ABC = \u0394AEF+ \u0394BDF + \u0394CDF + \u0394DEF
\n= 4\u0394DEF
\n\u21d2 \u0394DEF = \\(\\frac { 1 }{ 4 }\\)\u0394ABC<\/p>\n

iii) \u25a1BDEF = 2 \u0394DEF …………..(1)
\n(from (ii))
\n\u0394ABC = 4 \u0394DEF (2)
\n(from (ii))
\nFrom (1) & (2);
\n\u0394ABC = 2 (2\u0394DEF) = 2 \u25a1BDEF
\nHence proved.<\/p>\n

\"AP<\/p>\n

Question 5.
\nIn the figure D, E are points on the sides AB and AC respectively of \u0394ABC such that ar (\u0394DBC) = ar (\u0394EBC). Prove that DE \/\/ BC.
\n\"AP
\nSolution:
\n\u0394DBC = \u0394EBC
\nThe two triangles are on the same base BC and between the same pair of lines BC and DE.
\nAs they are equal in area.
\n\u2234 BC \/\/ DE.<\/p>\n

Question 6.
\nIn the figure, XY is a line parallel to BC is drawn through A. If BE \/\/ CA and CF \/\/ BA are drawn to meet XY at E and F respectively. Show that ar (\u0394ABE) = ar (\u0394ACF).
\n\"AP
\nSolution:
\nGiven that XY\/\/BC; BE\/\/CA; CF\/\/BA
\nIn quad ABCF; AB\/\/CF and BC\/\/AF
\nHence \u25a1ABCF is a parallelogram.
\nAlso in \u25a1ACBE ; BC\/\/AE and AC\/\/BE
\nHence \u25a1ACBE is a parallelogram.
\nNow in \u25a1ABCF and \u25a1ACBE
\n\u0394ABC = \u0394ACF …………..(1);
\n\u0394ABC = \u0394ABE …………..(2)
\n[\u2235 Diagonal divides a parallelogram into two congruent triangles]
\n\u2234 \u0394ACF = \u0394ABE [from (1) & (2)]
\nHence proved.<\/p>\n

\"AP<\/p>\n

Question 7.
\nIn the figure, diagonals AC and BD of a trapezium ABCD with AB\/\/DC inter\u00acsect each other at ‘O’. Prove that ar (\u0394AOD) = ar (\u0394BOC)
\n\"AP
\nSolution:
\nGiven that AB \/\/ CD
\nNow \u0394ADC and \u0394BCD are on the same base and between the same parallels AB \/\/ CD.
\n\u2234 \u0394ADC = \u0394BCD
\n\u21d2 \u0394ADC – \u0394COD = \u0394BCD – \u0394COD
\n\u21d2 \u0394AOD = \u0394BOC [from the figure]<\/p>\n

Question 8.
\nIn the figure ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
\n(i) ar (\u0394ACB) = ar (\u0394ACF)
\n(ii) ar (AEDF) = ar (ABCDE)
\n\"AP
\nSolution:
\nABCDE is a pentagon.
\nAC\/\/BF
\ni) \u0394ACB and \u0394ACF are on the same base AC and between the same parallels AC\/\/BF.
\n\u2234 \u0394ACB = \u0394ACF<\/p>\n

ii) \u25a1AEDF = \u25a1AEDC + \u0394ACF
\n= \u25a1AEDC + \u0394ABC
\n[ \u2235 \u0394ACF = \u0394ACB]
\n= area (ABCDE)
\nHence proved.<\/p>\n

\"AP<\/p>\n

Question 9.
\nIn the figure, if ar (\u0394RAS) = ar (\u0394RBS) and ar (\u0394QRB) = ar (\u0394PAS) then show that both the quadrilaterals PQSR and RSBA are trapeziums.
\n\"AP
\n\u0394RAS = \u0394RBS ………….. (1)
\nBoth the triangles are on the same base RS and between the same pair of lines RS and AB.
\nAs their areas are equal RS must be parallel to AB.
\n\u21d2 RS\/\/AB
\n\u2234 \u25a1ABRS is a quadrilateral in which AB\/\/RS.
\n\u2234 \u25a1ABRS (or) \u25a1RSBA is a trapezium.
\nNow AQRB = APAS (given)
\n\u21d2 \u0394QRB – \u0394RBS = \u0394PAS – \u0394RAS
\n[from (1) \u0394RBS = \u0394RAS]
\n\u21d2 \u0394QRS = \u0394PRS
\nThese two triangles are on the same base RS and between the same pair of lines RS and PQ.
\nAs these two triangles have same area RS must be parallel to PQ.
\n\u21d2 RS \/\/ PQ
\nNow in quad PQRS; PQ\/\/RS.
\nHence \u25a1PQRS is a trapezium.<\/p>\n

\"AP<\/p>\n

Question 10.
\nA villager Ramayya has a plot of land in the shape of a quadrilateral. The grampanchayat of the village decided to take over some portion of his plot from one of the comers to construct a school. Ramayya agrees to the above proposal with the condition that he should be given equal amount of land in exchange of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will implemented. (Draw a rough sketch of the plot.)
\n\"AP
\nSolution:
\nLet \u25a1ABCD is the plot of Ramayya.
\nSchool be constructed in the region \u0394MCD where M is a point on BC such that \u25a1ABCD \u2245 \u0394ADE
\nDraw the diagonal BD.
\nDraw a line parallel to BD through C which meets AB produced at E.
\nJoin D, E
\n\u0394ADE is the required triangle.
\n\"AP
\nAnalysis:
\n\u0394CED and \u0394CEB are on the same base CE and between the same parallels CE and DB.
\n\u2234 \u0394CED = \u0394CEB [also from the figure]
\n\u0394CEM + \u0394CMD = \u0394CEM + \u0394BME
\n\u2234 \u0394CMD = \u0394BME
\n\u2234 \u0394ADE = \u25a1ABCD<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.3 Textbook Questions and Answers. AP State Syllabus 9th Class Maths Solutions 11th Lesson Areas Exercise 11.3 Question 1. In a triangle ABC, E is the midpoint of median AD. Show that i) ar \u0394ABE = ar \u0394ACE Solution: In \u0394ABC; AD is … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/4966"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=4966"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/4966\/revisions"}],"predecessor-version":[{"id":40569,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/4966\/revisions\/40569"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=4966"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=4966"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=4966"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}