Math, asked by sneha395378, 1 year ago

two circles each of radius 7cm intersect each other the distance between their centres is 7 under root 2 centimetre find the area common to both the circles​

Answers

Answered by wajahatkincsem
6

Answer:

The answer is 49/2 (π-2)cm²

Step-by-step explanation:

step 1 - ∆ ABC is an O scales and 45° - 45° - 90°

Triangle drop a perpendicular OA the the side BS is ∆ ABC

In triangle AOB ,

< AOB = 90° , < ABO = 45° , AB = 7cm

Sin 45° = AO/AB

1/√2 = AO/7 ( AO = 7/√2 cm)

Step 2- Length OA is an bisector in triangle ABC therefore,

BC = 7√2 cm

BO = 7√2/2 = 7\√2 cm

Area of triangle AOB= 1/2× 7/√2 × 7/×2

= 49/4 cm^2

= Area of sector ABO = 1/2 ×r^2× θ = 1/2 × 7² × π/4  (44° = π/4)

= 49π/8 cm²

Step 3- Area of Triangle AOB and AOC combined =49/4×2 = 49/2cm²

= Area of sector of both sides combined 49π\8× 2 = 49π/4cm²

= Area of overlap region = 2×(49π/4 -49/2 )cm²

Thus the area common to both the circles is 49/2 (π-2)cm²

Answered by chandgautam64
0

Answer:

Yes here is ur answer that I have explained.

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