{"id":25539,"date":"2021-06-23T12:11:25","date_gmt":"2021-06-23T06:41:25","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=25539"},"modified":"2022-03-02T10:36:19","modified_gmt":"2022-03-02T05:06:19","slug":"ncert-solutions-for-class-9-maths-chapter-9-ex-9-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-9-ex-9-2\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 9 Areas of Parallelograms and Triangles Ex 9.2 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 9.2<\/h2>\n

Question 1.
\nIn fig. 9.15, ABCD is a parallelogram and AE \u22a5 DC and CF \u22a5 AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm. Find AD.
\n\"NCERT
\nSolution:
\nIn this fig. we have given AB = 16 cm, AE = 8 cm, CF = 10 cm and ABCD is a parallelogram.
\n\u2234 AB = CD (opp side of parallelogram)
\n\u2234 CD = 16 cm
\nAgain we know that,
\nArea of parallelogram = Base \u00d7 Corresponding altitude
\n\u2234 Area of parallelogram ABCD = CD \u00d7 AE
\n= 16 \u00d7 8
\n= 128 cm2<\/sup> …..(i)
\nAgain Area of parallelogram = AD \u00d7 CF
\n128 = AD \u00d7 10 (from equ. (i))
\nAD = \\(\\frac{128}{10}\\) = 12.8 cm
\nAD = 12.8 cm<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nIf E, F, G and H are respectively the mid points of the sides of a parallelogram ABCD. show that ar(EFGH) = \\(\\frac {1}{2}\\) ar(ABCD).
\nSolution:
\nGiven: ABCD is a parallelogram, in which E, F, G, and H are the midpoints of sides AB, BC, CD, and AD respectively.
\n\"NCERT
\nTo prove that
\nar(||gm EFGH) = \\(\\frac {1}{2}\\) ar (||gm ABCD)
\nConstruction: Join HF
\nProof: ABCD is a ||gm
\n\u2234 AD || BC and AD = BC
\n\u2234 AH = BF and AH || BF
\nTherefore, ABFH is a ||gm.
\nSince, \u2206HEF and || gm HABF are on the same base HF and between the same parallel lines, AB and HF
\n\u2234 ar(\u2206HEF) = \\(\\frac {1}{2}\\) ar(||gm HABF) …….(i)
\nSimilarly, \u2206HGF and ||gm HDCF are on same base HF and between the same parallel lines, DC and HF.
\n\u2234 ar (\u2206HGF) = \\(\\frac {1}{2}\\) ar (|| gm HDCF) …….(ii)
\nAdding (i) and (ii) we get
\nar(\u2206HEF) + ar(\u2206HGF)
\n= \\(\\frac {1}{2}\\) ar(|| gm HABF) + \\(\\frac {1}{2}\\) ar(||gm HDCF) + ar (|| gm FFGH)
\n= \\(\\frac {1}{2}\\) [ar(||gm HABF) + ar (||gm HDCF)]
\nor, ar (||gm EFGH) = \\(\\frac {1}{2}\\) ar (|| gm ABCD)<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nP and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD shows that ar(\u2206APB) = ar(\u2206BQC).
\nSolution:
\nGiven: ABCD is a || gm in which P and Q are the points on side DC and AD respectively.
\nTo prove that: ar(\u2206APB) = ar(\u2206BQC)
\nProof: ABQC and || gm ABCD lie on the same base BC and between the same parallel lines.
\n\"NCERT
\n\u2234 ar (\u2206BQC) = \\(\\frac {1}{2}\\) ar(||gm ABCD) ……(i)
\nSimilarly, \u2206APB and ||gm ABCD lie on the same base AB and between same parallel lines.
\n\u2234 ar(\u2206APB) = \\(\\frac {1}{2}\\) ar (||gm ABCD) ……(ii)
\nFrom equation (i) and (ii), we can say
\nar(\u2206BQC) = ar(\u2206APB)<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nIn fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
\n(i) ar(\u2206APB) + ar(\u2206PCD) = \\(\\frac {1}{2}\\) ar (||gm ABCD)
\n(ii) ar(\u2206APD) + ar(\u2206PBC) = ar(\u2206APB) + ar(\u2206PCD)
\n\"NCERT
\nSolution:
\nGiven: ABCD is a || gm in which P is any point interior to it.
\nTo prove that:
\n(i) ar(\u2206APB) + ar(\u2206PCD) = \\(\\frac {1}{2}\\) ar (||gm ABCD)
\n(ii) ar (\u2206APD) + ar(\u2206PBC) = ar(\u2206APB) + ar(\u2206PCD)
\n\"NCERT
\nConstruction: Through P draw a line parallel to AB which interacts with AD at M and BC at N.
\nProof: In parallelogram ABCD,
\nAB || CD (opp sides of a ||gm)
\nBut, AB || MN (by construction)
\n\u2234 MM || CD
\nIn ABNM
\nAB || MN and AM || BN (opp sides of || gm)
\n\u2234 ABNM is a || gm
\nAgain, \u2206APB and || ABNM lie on the same base AB and between the same parallel lines AB and MN.
\n\u2234 ar (\u2206APB) = \\(\\frac {1}{2}\\) ar (|| gm ABNM) ……(i)
\nSimilarly, MNCD is a parallelogram.
\nSince \u2206PCD and || gm MNCD lie on same base CD and between same parallel lines CD and MN.
\nTherefore, ar (\u2206PCD) = \\(\\frac {1}{2}\\) ar (|| gm MNCD) ……(ii)
\nAdding (i) and (ii) we get,
\nar(\u2206APB) + ar(\u2206PCD) = \\(\\frac {1}{2}\\) [ar (|| gm ABNM) + ar (|| gm MNCD)]
\n\u2234 ar(\u2206APB) + ar(\u2206PCD) = \\(\\frac {1}{2}\\) [ar (|| gm ABCD) ……(iii)<\/p>\n

(ii) Similarly, itive draw a line through P and parallel to D.
\nThen, ar(\u2206APD) + ar(\u2206PBC) = \\(\\frac {1}{2}\\) ar (|| gm ABCD) ……(iv)
\nFrom equation (iii) and (iv) we can say
\nar(\u2206APP) + ar(\u2206PBC) = ar(\u2206APB) + ar(\u2206PCD)<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nIn fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
\n(i) ar (|| gm PQRS) = ar (|| gm ABRS)
\n(ii) ar (\u2206AXS) = \\(\\frac {1}{2}\\) ar (|| gm PQRS)
\n\"NCERT
\nSolution:
\nParallelograms PQRS and ABRS are lying on the same base SR and in between the same parallel lines SR and PB.
\n\"NCERT
\nWe know that parallelograms lie on the same base and between the same parallel lines are equal in area,
\n\u2234 ar(|| gm PQRS) = ar(|| gm ABRS) ……(i)<\/p>\n

(ii) Again, \u2206AXS and || gm ABRS lie on the same base AS and between the same parallel lines AS and BR.
\n\u2234 ar (\u2206AXS) = \\(\\frac {1}{2}\\) ar (|| gm ABRS)
\nor, ar (\u2206AXS) = \\(\\frac {1}{2}\\) ar (|| gm PQRS) (from equ. (i))<\/p>\n

\"NCERT<\/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 point P and Q. In how many parts the field is divided? What is the shape of these parts? The fanner wants to saw wheat and pulses in equal portions of the field separately. How should she do it?
\nSolution:
\nAccording to the question, fields is divided into three-part. The shape of the three parts is triangular as \u2206PSA, \u2206PAQ, and \u2206QRA.
\n\"NCERT
\nAgain, according to the question, the farmer wants to sow wheat and pulses in an equal portion of the field separately.
\nWe know that if a parallelogram and triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
\n\u2234 ar(\u2206PAQ) = \\(\\frac {1}{2}\\) ar (|| gm PQRS)
\nor, ar(\u2206PAQ) = ar(\u2206PSA) + ar(\u2206QRA)
\nSo, farmers must sow wheat in triangular parts PAQ and pluses in other two triangular parts PSA and QRA, or, pulses in triangular parts PAQ and wheat in other two triangular parts PSA and QRA.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2 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 9.2 Question 1. In fig. 9.15, ABCD is a parallelogram and AE \u22a5 DC …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2<\/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 9.2 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-9-ex-9-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2 Questions and Answers are prepared by our highly skilled subject experts. 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