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Given:
As shown above in the figure, two concentric circles with centre At point O. Radius of the bigger circle = R cm, of the smaller circle = r cm. Area of between the two concentric circles = 286 cm2.Difference between the radii of the two circles = 7 cm = R – r ____ (1)
Formula Used:
Area of a circle = πR2 (Where R = Radius of the circle)
Calculation:
⇒ πR2 – πr2 = 286, 22/7 × (R2 – r2) = 286
⇒ (R – r)(R + r) = 13× 7, (R + r)× 7 = 13× 7, (R + r) = 13 cm ____(2)
From (1) & (2), we get
∴ R = 10 cm, r = 3 cm
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Here we will discuss about the area of a circular ring along with some example problems.
The area of a circular ring bounded by two concentric circle of radii R and r (R > r)
= area of the bigger circle – area of the smaller circle
= πR\(^{2}\) - πr\(^{2}\)
= π(R\(^{2}\) - r\(^{2}\))
= π(R + r) (R - r)
Therefore, the area of a circular ring = π(R + r) (R - r), where R and r are the radii of the outer circle and the inner circle respectively.
Solved example problems on finding the area of a circular ring:
1. The outer diameter and the inner diameter of a circular path are 728 m and 700 m respectively. Find the breadth and the area of the circular path. (Use π = \(\frac{22}{7}\)).
Solution:
The outer radius of a circular path R = \(\frac{728 m}{2}\) = 364 m.
The inner radius of a circular path r = \(\frac{700 m}{2}\) = 350 m.
Therefore, breadth of the circular path = R - r = 364 m -
350 m = 14 m.
Area of the circular path = π(R + r)(R - r)
= \(\frac{22}{7}\)(364 + 350) (364 - 350) m\(^{2}\)
= \(\frac{22}{7}\) × 714 × 14 m\(^{2}\)
= 22 × 714 × 2 m\(^{2}\)
= 31,416 m\(^{2}\)
Therefore, the area of the circular path = 31416 m\(^{2}\)
2. The inner diameter and the outer diameter of a circular path are 630 m and 658 m respectively. Find the area of the circular path. (Use π = \(\frac{22}{7}\)).
Solution:
The inner radius of a circular path r = \(\frac{630 m}{2}\) = 315 m.
The outer radius of a circular path R = \(\frac{658 m}{2}\) = 329 m.
Area of the circular path = π(R + r)(R - r)
= \(\frac{22}{7}\) (329 + 315)(329 - 315) m\(^{2}\)
= \(\frac{22}{7}\) × 644 × 14 m\(^{2}\)
= 22 × 644 × 2 m\(^{2}\)
= 28,336 m\(^{2}\)
Therefore, the area of the circular path = 28,336 m\(^{2}\)
Here we will solve different types of problems on finding the area and perimeter of combined figures. 1. Find the area of the shaded region in which PQR is an equilateral triangle of side 7√3 cm. O is the centre of the circle. (Use π = \(\frac{22}{7}\) and √3 = 1.732.)
Here we will discuss about the area and perimeter of a semicircle with some example problems. Area of a semicircle = \(\frac{1}{2}\) πr\(^{2}\) Perimeter of a semicircle = (π + 2)r. Solved example problems on finding the area and perimeter of a semicircle
Here we will discuss about the area and circumference (Perimeter) of a circle and some solved example problems. The area (A) of a circle or circular region is given by A = πr^2, where r is the radius and, by definition, π = circumference/diameter = 22/7 (approximately).
Here we will discuss about the perimeter and area of a Regular hexagon and some example problems. Perimeter (P) = 6 × side = 6a Area (A) = 6 × (area of the equilateral ∆OPQ)
Here we will get the ideas how to solve the problems on finding the perimeter and area of irregular figures. The figure PQRSTU is a hexagon. PS is a diagonal and QY, RO, TX and UZ are the respective distances of the points Q, R, T and U from PS. If PS = 600 cm, QY = 140 cm
9th Grade Math
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