Suppose that the area between a pair of concentric circles is 49pi. Find the length of a chord in the larger circle that is tangent to the smaller circle.

Suppose that the area between a pair of concentric circles is 49pi.

Of we call the two circles r and R, and the center O, draw the tangent and let P be where it touches circle r, and Q be where it intersects R.

Then the triangle OPQ is a right triangle with one leg r and the hypotenuse R. The other leg is thus sqrt(R^2-r^2)

But, pi*R^2 - pi*r^2 = 49pi.
So, R^2-r^2 = 49, and the length of the tangent chord is 14.

14 Please do not post HW questions here.

To find the length of the chord, we can start by visualizing the problem. We have two concentric circles, where one is larger than the other. The chord we are looking for is a line segment that touches the smaller circle at one point, while also lying entirely within the larger circle.

Let's call the radius of the smaller circle r and the radius of the larger circle R. We know that the area between the two circles is 49π, which means the area of the larger circle minus the area of the smaller circle equals 49π.

The area of a circle can be calculated using the formula A = πr^2. So, the area between the two circles can be expressed as (πR^2) - (πr^2) = 49π.

Simplifying this equation gives us π(R^2 - r^2) = 49π.

We can cancel out the π terms on both sides of the equation, which leaves us with R^2 - r^2 = 49.

Now, we need to find the length of the chord. To do that, we can use the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse (the longest side) equals the sum of the squares of the other two sides.

In our case, the chord forms a right-angled triangle with the radius of the larger circle (R), the radius of the smaller circle (r), and half the length of the chord (d/2).

Using the Pythagorean theorem, we have (d/2)^2 = R^2 - r^2.

We already know that R^2 - r^2 = 49, so substituting this into the equation gives (d/2)^2 = 49.

To find the length of the chord (d), we can multiply both sides of the equation by 4, which gives d^2 = 4 * 49.

Taking the square root of both sides, we get d = 2 * √(49), which simplifies to d = 2 * 7 = 14.

Therefore, the length of the chord in the larger circle that is tangent to the smaller circle is 14 units.

Suppose that the area between a pair of concentric circles is 49pi. Find the length of a chord in the larger circle that is tangent to the smaller circle.

Of we call the two circles r and R, and the center O, draw the tangent and let P be where it touches circle r, and Q be where it intersects R.

14

Please do not post HW questions here.

To find the length of the chord, we can start by visualizing the problem. We have two concentric circles, where one is larger than the other. The chord we are looking for is a line segment that touches the smaller circle at one point, while also lying entirely within the larger circle.

Aops hacker

nah nobody care what u saybud

reheheheh

L^2=49.

That's true

Ditto and COPYCAT!