RS Aggarwal Chapter 11 Class 9 Maths Exercises 11.2 (ex 11b) Solutions 2024

RS Aggarwal Chapter 11 Class 9 Maths Exercise 11.2 Solutions: You have studied many properties of a triangle in Chapters 6 and 7 and you know that on joining three non-collinear points in pairs, the figure so obtained is a triangle. Now, let us mark four points and see what we obtain by joining them in pairs in some order. Although most of the objects we see around are of the shape of a special quadrilateral called a rectangle, we shall study more about quadrilaterals and especially parallelograms because a rectangle is also a parallelogram and all properties of a parallelogram are true for a rectangle as well. Know more about the Chapter here.

Download RS Aggarwal Chapter 11 Class 9 Maths Exercise 11.2 Solutions

EXERCISE 10B

Access The RS Aggarwal Chapter 11 Class 9 Maths Exercise 11.2 Solutions

Question 1.
Solution:
Let AB be a chord of a circle with centre O. OC⊥AB and OA be the radius of the circle, then
AB = 16cm, OA = 10cm .

OC ⊥ AB.
OC bisects AB at C
AC = 12 AB = 12 x 16 = 8cm

Question 2.
Solution:
Let AB be the chord of the circle with centre O and OC ⊥ AB, OA be the radius of the circle,
then OC = 3cm, OA = 5cm

Now in right ∆ OAC,
OA² = AC² = OC² (Pythagoras Theorem)

Question 3.
Solution:
Let AB be the chord, OA be the radius of
the circle OC ⊥ AB
then AB = 30 cm, OC = 8cm

Question 4.
Solution:
AB and CD are parallel chords of a circle with centre O.

Question 5.
Solution:
Let AB and CD be two chords of a circle with centre O.
OA and OC are the radii of the circle. OL⊥AB and OM⊥CD.

Question 6.
Solution:
In the figure, a circle with centre O, CD is its diameter AB is a chord such that CD⊥AB.
AB = 12cm, CE = 3cm.
Join OA.
∵ COD⊥AB which intersects AB at E

Question 7.
Solution:
A circle with centre O, AB is diameter which bisects chord CD at E
i.e. CE = ED = 8cm and EB = 4cm
Join OC.
Let radius of the circle = r

Question 8.
Solution:
Given : O is the centre of a circle AB is a chord and BOC is the diameter. OD⊥AB
To prove : AC || OD and AC = 20D
Proof : OD⊥AB
∵ D is midpoint of AB

Question 9.
Solution:
Given : O is the centre of the circle two
chords AB and CD intersect each other at P inside the circle. PO bisects ∠BPD.
To prove : AB = CD.

Question 10.
Solution:
Given : PQ is the diameter of the circle with centre O which is perpendicular to one chord AB and chord AB || CD.
PQ intersects AB and CD at E and F respectively
To prove : PQ⊥CD and PQ bisects CD.

Important Definition for RS Aggarwal Chapter 11 Class 9 Maths Ex 11b Solutions

A parallelogram is a type of quadrilateral which contains parallel opposite sides.

1. Area of parallelogram = Base × Height
2. Area of Triangle = (1/2)× Base × Height

Know more at the official website.

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