{"id":25631,"date":"2021-06-23T16:54:42","date_gmt":"2021-06-23T11:24:42","guid":{"rendered":"https:\/\/mcqbooks.com\/?p=25631"},"modified":"2022-03-06T18:50:44","modified_gmt":"2022-03-06T13:20:44","slug":"ncert-solutions-for-class-9-maths-chapter-9-ex-9-4","status":"publish","type":"post","link":"https:\/\/mcqbooks.com\/ncert-solutions-for-class-9-maths-chapter-9-ex-9-4\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.4<\/h2>\n

Question 1.
\nParallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
\nSolution:
\nGiven: Parallelogram ABCD and rectangle ABEF with the same base AB and equal areas.
\n\"NCERT
\nTo prove that: Perimeter of ||gm ABCD > perimeter of rectangle ABEF.
\ni.e., AB + BC + CD + AD > AB + BE + EF + AF.
\nProof: Since opposite sides of a ||gm and a rectangle are equal.
\n\u2234 AB = CD (Opposite sides of || gm)
\nand AB = EF (Opposit sides of rectangle)
\n\u2234 CD = EF
\nor AB + CD = AB + EF ……(i)
\nNow, we know that all the segments tlat can be drawn to a given line from a point not lying on it, the perpendicular segment is the shortest.
\n\u2234 BE < BC and AF < AD
\n\u2234 BC + AD > BE + AF ……(ii)
\nAdding (i) and (ii) we get
\nAB + CD + BC + AD > AB + EF + BE + AF
\nor, AB + BC + CD + AD > AB + BE + EF + AF
\nTherefore, the perimeter of || gm ABCD > perimeter of rectangle ABEF.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nIn Fig. 9.30, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). Can you now answer the question that you have left in the ‘Introduction’, of this chapter, whether the Held of Budhia has been actually divided into three parts of equal area?
\n\"NCERT
\nSolution:
\nGiven: In \u0394ABC, D and E are two points on BC such that BD = DE = EC.
\nTo prove that: ar (\u0394ABD) = ar (\u0394ADE) = ar (\u0394AEC)
\nConstruction: Draw AL \u22a5 BC
\nProof: In \u0394ADE
\nar (\u0394ADE) = \\(\\frac {1}{2}\\) \u00d7 DE \u00d7 AL ……(i)
\nar (\u0394ABD) = \\(\\frac {1}{2}\\) \u00d7 BD \u00d7 AL
\nor, ar (\u0394ABD) = \\(\\frac {1}{2}\\) \u00d7 DE \u00d7 AL …(ii) (\u2235 BD = DE)
\nFrom equation (i) and (ii)
\nar (\u0394ADE) = ar (\u0394ABD) …..(iii)
\nAgain, ar (\u0394AEC) = \\(\\frac {1}{2}\\) \u00d7 EC \u00d7 AL
\nor, ar (\u0394AEC) = \\(\\frac {1}{2}\\) \u00d7 DE \u00d7 AL ……(iv) (\u2235 EC = DE given)
\nFrom equation (i) and (iv)
\nar (\u0394ADE) = ar (\u0394AEC) …..(v)
\nFrom equation (iii) and (v)
\nar (\u0394ABD) = ar (\u0394ADE) = ar (\u0394AEC)
\nYes, Budhiya has used the result of this question in dividing her land into three equal parts.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nIn Fig. 9.31, ABCD, DCFE, and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
\n\"NCERT
\nSolution:
\nIn \u0394ADE and \u0394BCF
\nAD = BC (Opposite sides of || gm ABCD)
\nDE = CF (Opposite sides of || gm DCFE)
\nand AE = BF (Opposite sides of || gm ABFE)
\nBy S-S-S congruency condition.
\n\u0394ADE \u2245 \u0394BCF
\nWe know that two congruent triangles have equal areas.
\nTherefore, ar (\u0394ADE) = ar (\u0394BCF)<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nIn Fig. 9.32, ABCD is a parallelogram, and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, show that ar (BPC) = ar (DPQ).
\n\"NCERT
\nSolution:
\nGiven: ABCD is a parallelogram in which side BC is produced to Q such that AD = CQ.
\nTo prove that: ar (\u0394BPC) = ar (\u0394DPQ)
\nConstruction: Join AC which intersects BP at O.
\nProof: \u0394BPC and \u0394APC lie on the same base PC and between the same parallel lines AB and CD.
\n\u2234 ar (\u0394BPC) = ar (\u0394APC) ……(i)
\n(\u0394s on the same base and between same parallels are equal in the area)
\nAgain, \u0394CQD and \u0394ACQ lie on the same base CQ and between the same parallel lines AD and BQ.
\n\u2234 (\u0394CQD) = ar (\u0394ACQ) …….(ii)
\n(As on the same base and between same parallels are equal in the area)
\nor, ar (\u0394CQD) – ar (\u0394CQP) = ar (\u0394ACQ) – ar (\u0394CQP)
\n(Subtract ar (\u0394CQP) both side)
\nar (\u0394DPQ) = ar (\u0394APC) ……(iii)
\nFrom equation (i) and (iii),
\nar (\u0394BPC) = ar (\u0394DPQ)<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nIn Fig. 9.33, ABC and BDE are two equilateral triangles such that D is the midpoint of BC. If AE intersects BC at F, show that
\n(i) ar (BDE) = \\(\\frac {1}{4}\\) ar (ABO)
\n(ii) ar (BDE) = \\(\\frac {1}{2}\\) ar (BAE)
\n(iii) ar (ABC) = 2ar (BEC)
\n(iv) ar (BFE) = ar (AFD)
\n(v) ar (BFE) = 2ar (FED)
\n(vi) ar (FED) = \\(\\frac {1}{8}\\) ar (AFC)
\n\"NCERT
\nSolution:
\nGiven: ABC and BED are two equilateral triangles. D is the midpoints of BC, and AE intersects BC at F.
\nTo prove that:
\n(i) ar (BDE) = \\(\\frac {1}{4}\\) ar (ABO)
\n(ii) ar (BDE) = \\(\\frac {1}{2}\\) ar (BAE)
\n(iii) ar (ABC) = 2ar (BEC)
\n(iv) ar (BFE) = ar (AFD)
\n(v) ar (BFE) = 2ar (FED)
\n(vi) ar (FED) = \\(\\frac {1}{8}\\) ar (AFC)
\n\"NCERT
\nConstruction: Join CE and AD and draw EG \u22a5 BC and join.
\nProof: \u0394ABC and \u0394DEF both equilateral \u0394s.
\n\u2234 \u2220ABC = \u2220BDE (each 60\u00b0)
\nBut it is the pair of alternate interior angle and we know that if pair of alternate interior angles are equal then the two lines are parallel.
\n\u2234 AB || DE
\nSimilarly, AC || BE and AD || EF
\n(i) ar (\u0394BDE) = \\(\\frac{\\sqrt{3}}{4}\\) (BD)2<\/sup>
\n(\u2235 Area of equilateral \u0394 = \\(\\frac{\\sqrt{3}}{4}\\) a2<\/sup>)
\n= \\(\\frac{\\sqrt{3}}{4}\\left(\\frac{1}{2} \\mathrm{BC}\\right)^{2}\\) (\u2235 D is mid point of BC)
\n\\(\\frac{\\sqrt{3}}{4} \\times \\frac{1}{4} \\mathrm{BC}^{2}\\)
\n= \\(\\frac {1}{4}\\) ar (\u0394ABC)
\n(\u2235 Area of equilateral \u0394 = \\(\\frac{\\sqrt{3}}{4}\\) a2<\/sup>)
\nSo, ar (\u0394BDE) = \\(\\frac {1}{4}\\) ar (\u0394ABC)<\/p>\n

(ii) ar(\u0394BDE)= \\(\\frac {1}{2}\\) ar(\u0394BEC)
\n(\u2235 ED is the median of \u0394EBC and median divide a \u0394 into two equal parts)
\n\u0394BEC and \u0394BAE lie on same base BE and between same parallels BE and AC.
\n\u2234 ar (\u0394BEC) = ar (\u0394BAE) ……(b)
\n(\u0394s on same base and between same parallels are equal in area)
\nFrom equation (a) and (b)
\nar (\u0394BDE) = \\(\\frac {1}{2}\\) ar (\u0394BAE)<\/p>\n

\"NCERT<\/p>\n

(iii) We have
\nar (\u0394BEC) = ar (\u0394BAE) ……(i)
\n(Prove above equ. (b))
\nNow, \u0394BAE and \u0394BAD lie on same base BA and between same parallels AB and DE.
\n\u2234 ar (\u0394BAE) = ar (\u0394BAD) ……(ii)
\nBut ar (\u0394BAD) = \\(\\frac {1}{2}\\) ar (\u0394ABC) ……(iii)
\n(\u2235 median divides a triangle into two equal parts)
\nFrom equation (i), (ii) and (iii)
\nar (\u0394BEC) = \\(\\frac {1}{2}\\) ar (\u0394ABC)
\nor, ar (\u0394ABC) = 2 ar (\u0394BEC)<\/p>\n

(iv) \u0394BDE and \u0394ADE lie on same base DE and between same parallels DE and AB.
\n\u2234 ar (\u0394BDE) = ar (\u0394ADE)
\nor, ar (\u0394BDE) – ar (\u0394DEF) = ar (\u0394ADE) – ar (\u0394DEF)
\n[Subtract ar (\u0394DEF) both side]
\nor, ar (\u0394BFE) = ar (\u0394AFD)<\/p>\n

(v) In \u0394ABC and \u0394DEB
\n\"NCERT
\nAgain, \u0394DEB and \u0394DEA lie on same base DE and between same parallels DE and AB.
\n\u2234 ar (\u0394DEB) = ar (\u0394DEA)
\nor, ar (\u0394DEB) – ar (\u0394DEA) = ar (\u0394DEA) – ar (\u0394DEF)
\n(Subtract ar (\u0394DEF) both side)
\nor, ar (\u0394BEF) = ar (\u0394ADF) ……(b)
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

(vi) We have given that
\nBC = 2BD = 2 (BF + FD) (\u2235 BD = BF + FD)
\n= 2 (2FD + FD) [From equation (c), BF = 2DF]
\nor, BC = 6FD ……(d)
\n\"NCERT<\/p>\n

Question 6.
\nDiagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar (APB) \u00d7 ar (CPD) = ar (APD) \u00d7 ar (BPC)
\nSolution:
\nGiven: ABCD is a quadrilateral in which diagonal, AC, and BD intersect each other at P.
\nTo prove that: ar (APB) \u00d7 ar (CPD) = ar (APD) \u00d7 ar(BPC)
\n\"NCERT
\nConstruction: Draw, AM \u22a5 BD, and CN \u22a5 BD.
\nProof: ar (\u0394APB) \u00d7 ar (\u0394CPD) = \\(\\frac {1}{2}\\) \u00d7 PB \u00d7 AM \u00d7 \\(\\frac {1}{2}\\) \u00d7 PD \u00d7 CN
\n= \\(\\frac {1}{2}\\) \u00d7 PB \u00d7 CN \u00d7 \\(\\frac {1}{2}\\) \u00d7 PD \u00d7 AM
\n= ar (\u0394BPC) \u00d7 ar (\u0394APD)
\n\u2234 ar (\u0394APB) \u00d7 ar (\u0394CPD) = ar (\u0394APD) \u00d7 ar (\u0394BPC)<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nP and Q are respectively the midpoints of sides AB and BC of a triangle ABC and R is the midpoint of AP. Show that
\n(i) ar (PRQ) = \\(\\frac {1}{2}\\) ar (ARC)
\n(ii) ar (RQC) = \\(\\frac {3}{8}\\) ar(ABC)
\n(iii) ar (PBQ) = ar (ARC)
\n\"NCERT
\nSolution:
\nGiven: P and Q are the midpoint of the side AB and BC respectively and R is the midpoint of AP.
\nTo prove that:
\n(i) ar (PRQ) = \\(\\frac {1}{2}\\) ar (ARC)
\n(ii) ar (RQC) = \\(\\frac {3}{8}\\) ar(ABC)
\n(iii) ar (PBQ) = ar (ARC)
\nConstruction: Join AQ, CP, CR and PQ.
\nProof: (i) In \u0394APC,
\nR is the mid point of AP.
\nSo, ar (\u0394ARC) = ar (\u0394PRC)
\n(Median divide a triangle into two equal parts)
\nor, ar (\u0394ARC) = \\(\\frac {1}{2}\\) ar (\u0394APC) ……(i)
\nNow, ar (\u0394APC) = \\(\\frac {1}{2}\\) ar (\u0394ABC) ……(ii)
\n(PC is the median of \u0394ABC)
\nand ar (\u0394AQB) = \\(\\frac {1}{2}\\) ar (\u0394ABC) ……(iii)
\n(\u2235 AQ is the median of \u0394ABC)
\nFrom equation (i) and (ii) we get
\nar (\u0394APC) = ar (\u0394AQB) ……(A)
\nFrom equation (ii) and (iii) we get
\nar (\u0394APC) = ar (\u0394AQB) …..(iv)
\nAgain, ar (\u0394APQ) = \\(\\frac {1}{2}\\) ar (\u0394ABQ) ……(v)
\n(\u2235 PQ is the median of \u0394AQB)
\nar (\u0394PRQ) = \\(\\frac {1}{2}\\) ar (\u0394APQ)
\n(\u2235 QR is the median \u0394AQP)
\nor, ar (\u0394PRQ) = \\(\\frac {1}{2}\\) \u00d7 \\(\\frac {1}{2}\\) ar (\u0394ABQ) (\u2235 from equation (v))
\n= \\(\\frac {1}{4}\\) ar (\u0394AQB) (\u2235 from equation (A))
\nor, ar (\u0394PRQ) = \\(\\frac {1}{4}\\) . 2 ar (\u0394ARC)
\n(\u2235 ar \u0394ARC = \\(\\frac {1}{2}\\) ar (\u0394APC))
\nor, ar (\u0394PRQ) = \\(\\frac {1}{2}\\) ar (\u0394ARC)<\/p>\n

\"NCERT<\/p>\n

(ii) In \u0394RBC, RQ is the median.
\n\u2234 ar (\u0394RBQ) = ar (\u0394RCQ) …….(I)
\n(median divide a \u0394 into tqo equal parts)
\nBut, ar (\u0394RBQ) = ar (\u0394AQB) – ar (\u0394AQR)
\n= \\(\\frac {1}{2}\\) ar (\u0394ABC) – \\(\\frac {1}{8}\\) ar (\u0394ABC)
\n(\u2235 ar (\u0394AQB) = \\(\\frac {1}{2}\\) ar (\u0394ABC) and ar(\u0394AQR) = \\(\\frac {1}{8}\\) ar (\u0394ABC))
\nar (\u0394RBQ) = \\(\\frac {3}{8}\\) ar (\u0394ABC) ……..(II)
\nFrom equation I and II we can say.
\nar (\u0394RQC) = \\(\\frac {3}{8}\\) ar (\u0394ABC)<\/p>\n

(iii) In \u0394ABQ, QP is the median
\nar (\u0394PBQ) = \\(\\frac {1}{2}\\) ar (\u0394ABQ)
\n(\u2235 QP is the median)
\n= \\(\\frac {1}{2}\\) \u00d7 \\(\\frac {1}{2}\\) ar (\u0394ABC)
\n(\u2235 ar (\u0394ABQ) = \\(\\frac {1}{2}\\) ar (\u0394ABC))
\nor, ar(\u0394PBQ) = \\(\\frac {1}{4}\\) ar(\u0394ABC) ……(a)
\nAgain, ar (\u0394ARC) = \\(\\frac {1}{4}\\) ar (\u0394ACP)
\n(\u2235 CR is the median of \u0394ACP)
\n= \\(\\frac {1}{2}\\) \u00d7 \\(\\frac {1}{2}\\) ar (\u0394ABC)
\n(\u2235 ar (\u0394ACP) = \\(\\frac {1}{2}\\) ar (\u0394ABC))
\nor, ar (\u0394ARC) = \\(\\frac {1}{4}\\) ar (\u0394ABC) ……(b)
\nFrom equation (a) and (b) we get
\nar (\u0394PBQ) = ar (\u0394ARC).<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nIn Fig. 9.34, ABC is a right triangle right angled at A. BCFD, ACFG and ABMN are squares on the sides BC, CA, and And AB respectively. Line AX \u22a5 DE meets BC at Y. Show that:
\n(i) \u0394MBC \u2245 \u0394ABD
\n(ii) ar (BYXD) = 2 ar (MBC)
\n(iii) ar (BYXD) = ar (ABMN)
\n(iv) \u0394FCB \u2245 \u0394ACE
\n(v) ar (CYXE) = 2 ar (FCB)
\n(vi) ar (CYXE) = ar (ACFG)
\n(vii) ar (BCED) = ar (ABMN) + ar (ACFG)
\n\"NCERT
\nSolution:
\n(i) In \u0394MBC and \u0394ABD
\nMB = AB (Sides of square ABMN)
\nBC = BD (Sides of square BCED)
\n\u2220MBA = \u2220CBD (each 90\u00b0)
\nor, \u2220MBA + \u2220ABC = \u2220CBD + \u2220ABC
\nor, \u2220MBC = \u2220ABD
\nSo, By S-A-S congruency condition
\n\u0394MBC \u2245 \u0394ABD …….(i)<\/p>\n

(ii) BYXD and \u0394ABD lie on the same base BD and between the same parallels BD and AX.
\n\u2234 ar (BYXD) = 2 ar (\u0394ABD)
\n[If a rectangle and a triangle lie on the same base and between same || is then the area of \u0394 is half of the rectangle]
\nor, a (DBYXD) = 2 ar (\u0394MBC) [from (i) congruent \u0394 is have equal area]<\/p>\n

\"NCERT<\/p>\n

(iii) We have
\nar (BYXD) = 2 ar (\u0394MBC) ……(ii)
\n(Prove above)
\nand ar (ABMN) = 2 ar (OMBC) …….(iii)
\n(Because they lie on same base BM and between same || lines MB and NC]
\nFrom equation (ii) and (iii)
\nar (BYXD) = ar (ABMN)<\/p>\n

(iv) In \u0394FCB and \u0394ACE
\nFC = AC (Sides of square)
\nCB = CE (Sides of square)
\n\u2220FCA = \u2220BCE (each 90\u00b0)
\n\u2220FCA + \u2220ACB = \u2220BCE + \u2220ACB (Add \u2220ACB both side)
\nor, \u2220FCB = \u2220ACE
\nSo, by S-A-S Congruency Condition
\n\u0394FCB = \u0394ACE …….(iv)<\/p>\n

(v) CYXE and \u0394ACE lie on samp base CE and between the same parallels CE and AX.
\n(\u2235 ar (CYXE) = 2ar (\u0394ACE))
\n(If a rectangle and a \u0394 lie on the same base and between same || lines then ar if the rectangle is double the area of \u0394)
\nor, ar (CYXE) = 2 ar (\u2220FCB)
\n(From equ. (iv) congruent \u0394 have equal area)<\/p>\n

(vi) ACFG and \u0394FCB lie on same base CF and between same parallels CF and BG.
\n\u2234 ar (ACFG) = 2ar (\u0394FCB) …….(vi)
\nFrom equation (v) and (vi)
\nar (CYXE) = ar (ACFG)<\/p>\n

\"NCERT<\/p>\n

(vii) \u0394ABC, is a right angle triangle right angled at A.
\n\u2234 By pythagorus theorem,
\nBC2<\/sup> = AB2<\/sup> + AC2<\/sup>
\nor, ar (Square BCED) = ar (Square ABMN) + ar (Square ACFG) [\u2235 Area of square = (side)2<\/sup>]
\nar(BCED) = ar(ABMN) + ar(ACFG).<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.4 Question 1. Parallelogram ABCD and rectangle ABEF are on the same base AB … Read more<\/a><\/p>\n","protected":false},"author":10,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/posts\/25631"}],"collection":[{"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/users\/10"}],"replies":[{"embeddable":true,"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/comments?post=25631"}],"version-history":[{"count":1,"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/posts\/25631\/revisions"}],"predecessor-version":[{"id":37457,"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/posts\/25631\/revisions\/37457"}],"wp:attachment":[{"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/media?parent=25631"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/categories?post=25631"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/tags?post=25631"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}