(Take \(\pi = \frac{22}{7}\))
Area Calculation for Concentric Circles
This problem asks us to find the area of the region between two concentric circles (circles with the same center). We are given the length of a chord of the outer circle that is tangent to the inner circle.
Understanding the Geometry
Let:
- \(R\) be the radius of the outer circle.
- \(r\) be the radius of the inner circle.
- The chord of the outer circle has length \(L = 14\) cm.
Since the chord of the outer circle touches the inner circle, it is tangent to the inner circle. Let the chord be \(AB\). Let \(O\) be the center of the circles. Let \(M\) be the point where the chord \(AB\) touches the inner circle. The line segment \(OM\) is the radius of the inner circle, \(r\), and it is perpendicular to the chord \(AB\) at point \(M\). Also, \(OA\) (or \(OB\)) is the radius of the outer circle, \(R\).
The line segment \(OM\) bisects the chord \(AB\) because a radius perpendicular to a chord bisects the chord.
Therefore, \(AM = MB = \frac{L}{2} = \frac{14 \text{ cm}}{2} = 7\) cm.
Now, consider the right-angled triangle \(\triangle OMA\). By the Pythagorean theorem:
\( OA^2 = OM^2 + AM^2 \)
Substituting the variables:
\( R^2 = r^2 + (7)^2 \)
\( R^2 = r^2 + 49 \)
Calculating the Area of the Annulus
The area of the region between the two concentric circles (also called the annulus) is the difference between the area of the outer circle and the area of the inner circle.
Area of outer circle = \(\pi R^2\)
Area of inner circle = \(\pi r^2\)
Area of the region = Area of outer circle - Area of inner circle
\( \text{Area} = \pi R^2 - \pi r^2 \)
We can factor out \(\pi\):
\( \text{Area} = \pi (R^2 - r^2) \)
From the Pythagorean theorem calculation above, we found that:
\( R^2 - r^2 = 49 \)
Now, substitute this value into the area formula:
\( \text{Area} = \pi \times 49 \)
Final Calculation
We are given that \(\pi = \frac{22}{7}\).
\( \text{Area} = \frac{22}{7} \times 49 \)
\( \text{Area} = 22 \times \frac{49}{7} \)
\( \text{Area} = 22 \times 7 \)
\( \text{Area} = 154 \text{ square cm} \)
Thus, the area of the region between the two concentric circles is 154 square cm.
Similar Questions
- The chord AB of a circle with centre at O is \(2\sqrt{3}\) times the height of the minor segment. If P is the area of the sector OAB and Q is the area of the minor segment of the circle, then what is the approximate value of \(\frac{P}{Q}\)?
(Take \(\sqrt{3} = 1.7\) and \(\pi = 3.14\)) What is the area of the shaded region?
What is the ratio of the area of the shaded region to the area of the non-shaded region?
What is the radius of the circle with centre at $O_1$?
What is the radius of the circle with centre at $O_2$?
What is the sum of the areas of the two circles?
What is the area of the shaded region?
What is the ratio of the area of the shaded region to that of the non-shaded region?
- In a quarter circle of radius R, a circle of radius \(r\) is inscribed. What is the ratio of R to \(r\)?
What is the perimeter of the shaded region?
Important Questions from Circles
The sum of the radius and diameter of a circle is 84 cm. What is the circumference of this circle?
The maximum area of a right-angled triangle inscribed in a circle of radius r is
The tangent at a point C of a circle and diameter AB when extended intersect at D, if ∠DCA = 110°, then ∠CBA is equal to
The equation of a circle with diameters are 2x - 3y + 12 = 0 and x + 4y - 5 = 0 and area of 154 sq. units is
The internal center of similitude of two circles $ (x - 1)^2 + (y - 3)^2 = 4 $ and $ (x + 5)^2 + (y - 9)^2 = 16 $ is
