NCERT Solution Class 10th Maths Chapter – 11 Areas related to Circles Examples

NCERT Solution Class 10th Maths Chapter – 11 Areas related to Circles Examples
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NCERT Solution Class 10th Maths Chapter – 11 Areas related to Circles

TextbookNCERT
class10th
SubjectMathematics
Example11th
Chapter NameAreas Related to Circles
CategoryClass 10th Mathematics
Medium English
Sourcelast doubt

NCERT Solution Class 10th Maths Chapter – 11 Areas related to Circles Examples were prepared by Experienced Lastdoubt.com Teachers. Detailed all the Examples questions in Chapter – 11 Maths Class 10 Areas Related to Circles Examples provided in NCERT TextBook.

NCERT Solution Class 10th Maths Chapter – 11 Areas related to Circles

Chapter – 11

Areas Related to Circles

Examples

Example 1. Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Usе л = 3.14).
NCERT Solution Class 10th Maths Chapter – 11 Areas related to Circles Examples
Solution: Given sector is OAPB (see Fig. 11.5)
Area of the sector = θ/360 × лr2
= 30/360 × 3.14 × 4 cm2
= 12.56 × 3 cm2 = 4.19 cm2 (approx)

Area of the corresponding major sector 
= лr2 – area of sector OAPB

= (3.14 × 16 – 4.19) cm2
= 46.05 cm2 = 46.1 cm2 (approx)

Alternatively, area of the major sector = (360 – θ)/360 × лr2
= (360 – 30/360) × 3.14 × 16 cm2
= 330/360 × 3.14 × 16 cm2 46.05 cm2
= 46.1 cm2 (approx)

Example 2. Find the area of the segment AYB shown in Fig. 11.6, if radius of the circle is 21 cm and ∠AOB 120°. (Use л = 22/7)
NCERT Solution Class 10th Maths Chapter – 11 Areas related to Circles Examples
Solution: Area of the segment AYB
= Area of sector OAYB – Area of Δ OAB (1)
Now, area of the sector OAYB = 120/360 × 22/7 × 21 × 21 cm2 = 462 cm2 (2) For finding the area of Δ OAB, draw OM ⊥ AB as shown in Fig. 11.7.
Note that OA = OB. Therefore, by RHS congruence, Δ AMO = Δ BMO.
So, M is the mid-point of < ABM and < AOM = BOM = 1/2 × 120° = 60°.
NCERT Solution Class 10th Maths Chapter – 11 Areas related to Circles Examples
Let, OM = x cm
So, from A OMA, OM/OA = cos 60°
or, x/21 = 1/2 (cos 60° = 1/2)
or, x = 21/2
So, OM = 21/2 cm
Also, AM/OA = sin 60° = √3/2
So, AM = 21√3/2 cm
Therefore, AB = 2 AM = 2 × 21√3/2 cm = 21√3 cm
So, area of A OAB = 1/2 AB × OM = 1/2 × 21√3 × 21/2 cm² = 441/4 √3 cm² (3)
Therefore, area of the segment AYB = (462 – 441/4 √3) cm² [From (1), (2) and (3)] = 21/4 (88-21√3) cm²

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