{"id":15625,"date":"2022-06-02T06:00:57","date_gmt":"2022-06-02T00:30:57","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=15625"},"modified":"2023-01-25T11:15:45","modified_gmt":"2023-01-25T05:45:45","slug":"selina-icse-solutions-class-9-maths-area-theorems-proof-use","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/selina-icse-solutions-class-9-maths-area-theorems-proof-use\/","title":{"rendered":"Selina Concise Mathematics Class 9 ICSE Solutions Area Theorems [Proof and Use]"},"content":{"rendered":"

Selina Concise Mathematics Class 9 ICSE Solutions Area Theorems [Proof and Use]<\/span><\/h2>\n

ICSE Solutions<\/a>Selina ICSE Solutions<\/a><\/p>\n

APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 16 Area Theorems [Proof and Use]. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.<\/p>\n

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Selina ICSE Solutions for Class 9 Maths Chapter 16 Area Theorems [Proof and Use]<\/strong><\/p>\n

Exercise 16(A)<\/strong><\/span><\/p>\n

Solution 1:<\/strong><\/span>
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Solution 2:<\/strong><\/span>
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\nSince from the figure, we get CD\/\/FE therefore FC must parallel to DE. Therefore it is proved that the quadrilateral CDEF is a parallelogram.
\nArea of parallelogram on same base and between same parallel lines is always equal and area of parallelogram is equal to the area of rectangle on the same base and of the same altitude i.e, between same parallel lines.
\nSo Area of CDEF= Area of ABDC + Area of ABEF
\nHence Proved<\/p>\n

Solution 3:<\/strong><\/span>
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Solution 4:<\/strong><\/span>
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\nGiven ABCD is a parallelogram. P and Q are any points on the sides AB and BC respectively, join diagonals AC and BD.
\nproof:
\nsince triangles with same base and between same set of parallel lines have equal areas
\narea (CPD)=area(BCD)\u2026\u2026 (1)
\nagain, diagonals of the parallelogram bisects area in two equal parts
\narea (BCD)=(1\/2) area of parallelogram ABCD\u2026\u2026 (2)
\nfrom (1) and (2)
\narea(CPD)=1\/2 area(ABCD)\u2026\u2026 (3)
\nsimilarly area (AQD)=area(ABD)=1\/2 area(ABCD)\u2026\u2026 (4)
\nfrom (3) and (4)
\narea(CPD)=area(AQD),
\nhence proved.
\n(ii)
\nWe know that area of triangles on the same base and between same parallel lines are equal
\nSo Area of AQD= Area of ACD= Area of PDC = Area of BDC = Area of ABC=Area of APD + Area of BPC
\nHence Proved<\/p>\n

Solution 5:<\/strong><\/span>
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Solution 6:<\/strong><\/span>
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Solution 7:<\/strong><\/span>
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Solution 8:<\/strong><\/span>
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Solution 9:<\/strong><\/span>
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Solution 10:<\/strong><\/span>
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Solution 11:<\/strong><\/span>
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Solution 12:<\/strong><\/span>
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Solution 13:<\/strong><\/span>
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Solution 14:<\/strong><\/span>
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Solution 15:<\/strong><\/span>
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Solution 16:<\/strong><\/span>
\nWe know that area of triangles on the same base and between same parallel lines are equal.
\nConsider ABED quadrilateral; AD||BE
\nWith common base, BE and between AD and BE parallel lines, we have
\nArea of \u0394ABE = Area of \u0394BDE
\nSimilarly, in BEFC quadrilateral, BE||CF
\nWith common base BC and between BE and CF parallel lines, we have
\nArea of \u0394BEC = Area of \u0394BEF
\nAdding both equations, we have
\nArea of \u0394ABE + Area of \u0394BEC = Area of \u0394BEF + Area of \u0394BDE
\n=> Area of AEC = Area of DBF
\nHence Proved<\/p>\n

Solution 17:<\/strong><\/span>
\nGiven: ABCD is a parallelogram.
\nWe know that
\nArea of \u0394ABC = Area of \u0394ACD
\nConsider \u0394ABX,
\nArea of \u0394ABX = Area of \u0394ABC + Area of \u0394ACX
\nWe also know that area of triangles on the same base and between same parallel lines are equal.
\nArea of \u0394ACX = Area of \u0394CXD
\nFrom above equations, we can conclude that
\nArea of \u0394ABX = Area of \u0394ABC + Area of \u0394ACX = Area of \u0394ACD+ Area of \u0394CXD = Area of ACXD Quadrilateral
\nHence Proved<\/p>\n

Solution 18:<\/strong><\/span>
\nJoin B and R and P and R.
\nWe know that the area of the parallelogram is equal to twice the area of the triangle, if the triangle and the parallelogram are on the same base and between the parallels
\nConsider ABCD parallelogram:
\nSince the parallelogram ABCD and the triangle ABR lie on AB and between the parallels AB and DC, we have
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Exercise 16(B)<\/strong><\/span><\/p>\n

Solution 1:<\/strong><\/span>
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Solution 2:<\/strong><\/span>
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Solution 3:<\/strong><\/span>
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Solution 4:<\/strong><\/span>
\nWe have to join PD and BD.
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Solution 5:<\/strong><\/span>
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Solution 6:<\/strong><\/span>
\nRatio of area of triangles with same vertex and bases along the same line is equal to ratio of their respective bases. So, we have
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Solution 7:<\/strong><\/span>
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Solution 8:<\/strong><\/span>
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Exercise 16(C)<\/strong><\/span><\/p>\n

Solution 1:<\/strong><\/span>
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Solution 2:<\/strong><\/span>
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Solution 3:<\/strong><\/span>
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Solution 4:<\/strong><\/span>
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Solution 5:<\/strong><\/span>
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Solution 6:<\/strong><\/span>
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Solution 7:<\/strong><\/span>
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Solution 8:<\/strong><\/span>
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Solution 9:<\/strong><\/span>
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More Resources for Selina Concise Class 9 ICSE Solutions<\/strong><\/p>\n