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secondary 3 | A Maths
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Date Posted: 3 years ago
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Hope this helps eh, thanks.
Date Posted:
3 years ago
Thank you!!!
Let the radii of the bigger and smaller circle be R cm and r cm respectively.
Then their diameters would be 2R cm and 2r cm respectively since diameter = 2 × radius
Sum of diameters = 2R + 2r = 20
R + r = 10 ①
Next, the sum of their areas is 52π cm²
πR² + πr² = 52π
R² + r² = 52 ②
①² or ① x ① :
(R + r)² = 10²
R² + 2Rr + r² = 100
Since from ② we know that R² + r² = 52,
Then 2Rr + 52 = 100
2Rr = 48
Rr = 24
r = 24/R ③
Sub ③ into ①, R + 24/r + R = 10
Multiply both sides by R, 24 + R² = 10R
R² - 10R + 24 = 0
(R - 6)(R - 4) = 0
R = 6 or R = 4
So r = 24/6 = 4 or r = 24/4 = 6
Since R > r , R = 6 and not 4.
Circumference of bigger circle = 2πR cm = 2π(6) cm = 12π cm
Then their diameters would be 2R cm and 2r cm respectively since diameter = 2 × radius
Sum of diameters = 2R + 2r = 20
R + r = 10 ①
Next, the sum of their areas is 52π cm²
πR² + πr² = 52π
R² + r² = 52 ②
①² or ① x ① :
(R + r)² = 10²
R² + 2Rr + r² = 100
Since from ② we know that R² + r² = 52,
Then 2Rr + 52 = 100
2Rr = 48
Rr = 24
r = 24/R ③
Sub ③ into ①, R + 24/r + R = 10
Multiply both sides by R, 24 + R² = 10R
R² - 10R + 24 = 0
(R - 6)(R - 4) = 0
R = 6 or R = 4
So r = 24/6 = 4 or r = 24/4 = 6
Since R > r , R = 6 and not 4.
Circumference of bigger circle = 2πR cm = 2π(6) cm = 12π cm
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