2<\/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
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(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 \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 <\/p>\n
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(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 <\/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 \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
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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 \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
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(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
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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 \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
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(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
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(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 …<\/p>\n
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n