{"id":15595,"date":"2024-02-29T06:00:58","date_gmt":"2024-02-29T00:30:58","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=15595"},"modified":"2024-02-29T15:19:31","modified_gmt":"2024-02-29T09:49:31","slug":"selina-icse-solutions-class-9-maths-mid-point-converse-including-intercept-theorem","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/selina-icse-solutions-class-9-maths-mid-point-converse-including-intercept-theorem\/","title":{"rendered":"Selina Concise Mathematics Class 9 ICSE Solutions Mid-point and Its Converse [ Including Intercept Theorem]"},"content":{"rendered":"

Selina Concise Mathematics Class 9 ICSE Solutions Mid-point and Its Converse [ Including Intercept Theorem]<\/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 12 Mid-point and Its Converse [ Including Intercept Theorem]. 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

Download Formulae Handbook For ICSE Class 9 and 10<\/a><\/p>\n

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

Exercise 12(A)<\/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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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>
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Solution 17:<\/strong><\/span>
\n\"Selina<\/p>\n

Exercise 12(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>
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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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Solution 10:<\/strong><\/span>
\n\"Selina<\/p>\n

Solution 11:<\/strong><\/span>
\n\"Selina<\/p>\n

Solution 12:<\/strong><\/span>
\nGiven ABCD is parallelogram, so AD = BC, AB = CD.
\nConsider 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.
\nSo BP = 2BC, but BC = AD as ABCD is a parallelogram.
\nHence BP = 2AD
\nConsider triangle APB, AB || OC as ABCD is a parallelogram. So by midpoint theorem, O has to be the midpoint of AP.
\nHence Proved<\/p>\n

Solution 13:<\/strong><\/span>
\nConsider trapezium ABCD.
\nGiven E and F are midpoints on sides AD and BC, respectively.
\n\"Selina
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\nConsider LHS,
\nAB + CD = AB + CJ + JI + ID = AB + 2HF + AB + 2EG
\nSo AB + CD = 2(AB + HF + EG) = 2(EG + GH + HF) = 2EF
\nAB + CD = 2EF
\nHence Proved<\/p>\n

Solution 14:<\/strong><\/span>
\nGiven \u0394 ABC
\nAD is the median. So D is the midpoint of side BC.
\nGiven DE || AB. By the midpoint theorem, E has to be midpoint of AC.
\nSo 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

More Resources for Selina Concise Class 9 ICSE Solutions<\/strong><\/p>\n