{"id":595474,"date":"2022-04-29T06:02:27","date_gmt":"2022-04-29T06:02:27","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=595474"},"modified":"2022-05-07T08:42:37","modified_gmt":"2022-05-07T08:42:37","slug":"selina-class-9-icse-solutions-mathematics-chapter-12-mid-point-and-its-converse-including-intercept-theorem","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/selina-class-9-icse-solutions-mathematics-chapter-12-mid-point-and-its-converse-including-intercept-theorem\/","title":{"rendered":"Selina Class 9 ICSE Solutions Mathematics : Chapter 12-\u00a0Mid-point and Its Converse [ Including Intercept Theorem]"},"content":{"rendered":"\n

Class 9: Maths Chapter 12 solutions. Complete Class 9 Maths Chapter 12 Notes.<\/p>\n\n\n\n

Selina Class 9 ICSE Solutions Mathematics : Chapter 12- Mid-point and Its Converse [ Including Intercept Theorem]<\/h2>\n\n\n\n

Selina 9th Maths Chapter 12, Class 9 Maths Chapter 12 solutions<\/p>\n\n\n\n

Exercise 12(A)<\/strong><\/p>\n\n\n\n

Solution 1:<\/strong>
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Solution 2:<\/strong>
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Solution 3:<\/strong>
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Solution 7:<\/strong>
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Solution 8:<\/strong>
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Solution 9:<\/strong>
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Solution 11:<\/strong>
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Solution 12:<\/strong>
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Solution 16:<\/strong>
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Solution 17:<\/strong>
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Exercise 12(B)<\/strong><\/p>\n\n\n\n

Solution 1:<\/strong>
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Solution 2:<\/strong>
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Solution 3:<\/strong>
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Solution 4:<\/strong>
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Solution 5:<\/strong>
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Solution 6:<\/strong>
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Solution 7:<\/strong>
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Solution 8:<\/strong>
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Solution 9:<\/strong>
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Solution 10:<\/strong>
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Solution 11:<\/strong>
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Solution 12:<\/strong>
Given ABCD is parallelogram, so AD = BC, AB = CD.
Consider triangle APB, given EC is parallel to AP and E is midpoint of side AB. So by midpoint theorem, C has to be the midpoint of BP.
So BP = 2BC, but BC = AD as ABCD is a parallelogram.
Hence BP = 2AD
Consider triangle APB, AB || OC as ABCD is a parallelogram. So by midpoint theorem, O has to be the midpoint of AP.
Hence Proved<\/p>\n\n\n\n

Solution 13:<\/strong>
Consider trapezium ABCD.
Given E and F are midpoints on sides AD and BC, respectively.
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Consider LHS,
AB + CD = AB + CJ + JI + ID = AB + 2HF + AB + 2EG
So AB + CD = 2(AB + HF + EG) = 2(EG + GH + HF) = 2EF
AB + CD = 2EF
Hence Proved<\/p>\n\n\n\n

Solution 14:<\/strong>
Given \u0394 ABC
AD is the median. So D is the midpoint of side BC.
Given DE || AB. By the midpoint theorem, E has to be midpoint of AC.
So line joining the vertex and midpoint of the opposite side is always known as median. So BE is also median of \u0394 ABC.<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

Selina Class 9 ICSE Solutions Mathematics : Chapter 12- Mid-point and Its Converse [ Including Intercept Theorem]<\/p>\n\n\n\n

Download PDF<\/strong>: Selina Class 9 ICSE Solutions Mathematics : Chapter 12-\u00a0Mid-point and Its Converse [ Including Intercept Theorem] PDF<\/a><\/p>\n\n\n\n

Chapterwise Selina Publishers ICSE Solutions for Class 9 Mathematics :<\/strong><\/h2>\n\n\n\n