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
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
Answer:
Yes here is ur answer that I have explained.