Request a call back

Join NOW to get access to exclusive study material for best results

CBSE Class 10 Answered

area of shaded region?
Asked by chiku999 | 03 Feb, 2010, 07:42: PM
Expert Answer

DB = 6 cm

DE = EB = 6/2 = 3 cm

DG = radius = 6 cm.

Hence GE = 33 cm, using Pythagorus's theorem in DEG,

sin EDG = 3/2

 EDG = 60

 FDG = 2x60 = 120

Area of segment GBF = Area of sector DFBG - Area of triangle DFG

= πR2/3 - (1/2)(R/2)(3R)

= 22.09 sq. cm.

Area of region GEB = 22.09/2 = 11.05 sq cm.

Area of region DEBG = 2xArea of region GEB = 22.09 sq. cm.

Shaded are of one part of the total black shaded area = (1/4) area of circle with radius R - Area of region DEBG

= (1/4)(3.14)(36) - 22.09 = 6.17 sq cm.

REQUIRED SHADED AREA = 2X6.17 = 12.34 sq cm.

Regards,

Team,

TopperLearning.

Answered by | 05 Feb, 2010, 03:35: PM
CBSE 10 - Maths
Asked by tejasdd | 28 May, 2010, 08:18: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by ketanvai | 23 Apr, 2010, 06:35: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by srprusty | 10 Apr, 2010, 05:43: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by s.sreeram | 11 Mar, 2010, 08:14: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by | 10 Mar, 2010, 07:26: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by rinki94 | 10 Mar, 2010, 03:59: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by BhavSimran | 10 Mar, 2010, 12:24: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by BhavSimran | 10 Mar, 2010, 12:08: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by BhavSimran | 10 Mar, 2010, 12:05: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by nishac | 10 Mar, 2010, 09:12: AM
ANSWERED BY EXPERT

Start your hassle free Education Franchisee and grow your business!

Know more
×