2 circles touch each other externally and the sum of their areas is 52 cm2.if the distance between the centres is 10 cm,find their radii.
Answers
Let their radius be R and r
Area of 1st circle = πR²
Area of 2nd circle = πr²
Given the sum of their areas = 52 cm²
=> πR² + πr² = 52
=> R² + r² = 52/π = 16.55
Also given,
R + r = 10
=> R = 10 - r
Hence ,
(10-r)² + r² = 16.55
=> 100 + r² - 20r + r² = 16.55
=> 2r² - 20r + 100 = 16.55
=> r² - 10r + 50 = 8.28
=> r² - 10r + 41.72 = 0
=> r = [10 ± √(100 - 166.88)]/2
=> r = [10 ± √- 66.88)]/2
which gives r an imaginary value, so the given data in question are wrong. Please check your question.
Answer:
6 & 4 cm
Step-by-step explanation:
2 circles touch each other externally and the sum of their areas is 52 cm2.if the distance between the centres is 10 cm,find their radii.
Correction : Area = 52π cm²
Let the radii of each circle = r1 & r2
Then r1 + r2 = 10 cm (distance between the centres is 10 cm)
π (r1)² + π (r2)² = 52π cm² (sum of their areas is 52 cm²)
π ((r1)² + (r2)²) = 52π
(r1)² + (r2)² = 52π/π
(r1)² + (r2)² = 52
r1 + r2 = 10 cm
Squaring both sides
(r1)² + (r2)² + 2(r1((r2 ) = 100
52 + 2(r1((r2 ) = 100
(r1((r2 ) = 24
r1 + r2 = 10
r1 & r2 are roots of equation
r² - 10r + 24
=> r² - 10r + 24 = 0
=> r² - 6r -4r + 24 = 0
=> r(r-6)-4(r-6) = 0
=> (r-4)(r-6) = 0
=> r = 4 & 6
r = 6 & 4 cm