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Frank Modern Certificate Solution for Class 9 Mathematics Chapter 15 - Mid-point and Intercept Theorems

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Frank Textbook Solutions Chapter 15 - Mid-point and Intercept Theorems

Frank Textbook Solutions are considered extremely helpful for solving difficult questions in the ICSE Class 9 Mathematics exam. TopperLearning Textbook Solutions are compiled by our subject experts. Herein, you can find all the answers to the questions of   Chapter 15 - Mid-point and Intercept Theorems for the Frank textbook.

Frank Textbook Solutions for class 9  are in accordance with the latest ICSE syllabus, and they are amended from time to time to be most relevant. Our free Frank Textbook Solutions for ICSE Class 9 Mathematics will give you deeper insight on the chapters and will help you to score more marks in the final examination. ICSE Class 9 students can refer to our solutions while doing their homework and while preparing for the exam.

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Exercise/Page

Frank Modern Certificate Solution for Class 9 Mathematics Chapter 15 - Mid-point and Intercept Theorems Page/Excercise 15.1

Solution 1

Solution 2

Solution 3(a)

 

Solution 3(b)

 

Solution 3(c)

Solution 4

Solution 5

 

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

 

Solution 12

Solution 13

Solution 14

Solution 15(a)

Let us draw a diagonal AC which meets PQ at O as shown below:

 

 

 

 

Solution 16

Solution 17

Solution 18

Construction: Draw DS BF, meeting AC at S.

 

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

 

Frank Modern Certificate Solution for Class 9 Mathematics Chapter 15 - Mid-point and Intercept Theorems Page/Excercise 15.2

Solution 1

 

 

 

Solution 2

Solution 3

Solution 4

Solution 5

The figure is as below:

Let ABCD be a quadrilateral where P, Q, R, S are the midpoints of sides AB, BC, CD, DA respectively.

Diagonals AC and BD intersect at right angles at point O. We need to show that PQRS is a rectangle

Solution 6

Note: This question is incomplete.

According to the information given in the question,

F could be any point on BC as shown below:

So, this makes it impossible to prove that DP = DE, since P too would shift as F shift because P too would be any point on DE as F is.

Note: If we are given F to be the mid-point of BC, the result can be proved.

Solution 7

 

From the figure EF AB and E is the midpoint of BC. 

Therefore, F is the midpoint of AC. 

Here EF BD, EF = BD as D is the midpoint of AB. 

BE DF, BE = DF as E is the midpoint of BC. 

Therefore BEFD is a parallelogram.

Remark: Figure modified

Solution 8

Solution 9

TopperLearning provides step-by-step solutions for each question in each chapter in the Frank textbook recommended by ICSE schools. Access Chapter 15 - Mid-point and Intercept Theorems here. Our Frank Textbook Solutions for ICSE Class 9 Mathematics are designed by our subject matter experts. These solutions will help you to revise the whole chapter, so you can clear your fundamentals before the examination.

Text Book Solutions

ICSE IX - Mathematics

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