OP Malhotra Class 9 ICSE Maths Solutions Chapter 9

Warning: Undefined array key "https://nctbsolution.com/op-malhotra-class-9-icse-maths-solutions/" in /home/862143.cloudwaysapps.com/hpawmczmfj/public_html/wp-content/plugins/wpa-seo-auto-linker/wpa-seo-auto-linker.php on line 192

OP Malhotra Class 9 ICSE Maths Solutions Chapter 9 Mid-Point and Intercept Theorems

Welcome to NCTB Solutions. Here with this post we are going to help 9th class students for the Solutions of OP Malhotra Class 9 ICSE Math Book, Chapter 9, Mid-Point and Intercept Theorems. Here students can easily find step by step solutions of all the problems for Mid-Point and Intercept Theorems, Exercise 9a Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 9 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.

Mid-Point and Intercept Theorems Exercise 9(a) Solution :

Question no – (1) 

Solution : 

(i) The line joining the mid-points of two sides of a triangle is Parallel to the third side.

(ii) The line drawn through the mid-point of one side of a triangle parallel to another side bisects the Third side.

(iii) In Fig. 9.16, S and 7 are the mid-points of PQ and PR respectively. If ST = 3 cm, then QR = 6 cm.

(iv) In Fig 9.17, D and E are the mid-points of AB and AC. If DE-7.5 cm, then BC = 15 cm

Here, DE = 7.5cm

BC = (2 × 7.5) cm

= 15 cm

Question no – (2) 

Solution : 

In △ABC, AB = 6cm, AC = 3cm M is the midpoint of AB

∴ MN = 1/2 × AC

= 1/2 × 3

= 1.5 cm

Question no – (3) 

Solution : 

From the figure we know,

△ABC, D is the midpoint of AB.

E is the midpoint of AC

(i) If BC = 6 cm, then DE = 1/2 × 6

= 3 cm

(ii) If ∠DBC = 140° and DE || BC

∠ADE  = ∠DBC = ∠ABC

∠ADE = 140°

Question no – (4) 

Solution : 

D, E F are the mid point of side BC, AB and AC.

AD and EF are joined.

Here, AD bisects EF

In △ABC,

E and F are the mid point of CA and AB.

So, EF || BC and, EF = 1/2 BC

ED || BA ad, = 1/2 AB

FD || AC and = to 1/2 AC

Thus, AEDF is a parallelogram if diagonals bisect each other, AD Bisects EF …(Proved)

Next Chapter Solution : 

👉 Chapter 11 👈

Updated: June 20, 2023 — 5:03 am

Leave a Reply

Your email address will not be published. Required fields are marked *