Question

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.

Answer 1

Answer:

14 units

Step-by-step explanation:

We are given that the area between two concentric circles is $49\pi$

We have to find the length of chord in the larger circle that is tangent to the smaller circle.

Let $r_1,r_2$ be the radius of two circles.

$r_1$ be the radius of small circle and $r_2$ be the radius of large circle

We know that area pf circle=$\pi r^2$

Area of large circle =$\pi r^2_2$

Area of small circle =$\pi r^2_1$

Area between two circles =$49\pi$

Area of large circle -Area of small circle=$49\pi$

$\pi r^2_2-\pi r^2_1=49\pi$

$\pi(r^2_2-r^2_1)=49$

By pythagorus theorem

$AD^2=OA^2-OD^2$

$AD^2=r^2_2-r^2_1$

$AD=49$

$AD=\sqrt{49}=7$

Length of chord=$2\cdot AD$

Hence, the length of chord of the larger circle =$2\cdot7=14$ units

Answer 2

Answer :

Length of chord is 14 units long in the larger circle that is tangent to the smaller circle.

Explanation :

Given that,

Area between a pair of concentric circles = 49π  

We need to find the length of a chord in the larger circle that is tangent to the smaller circle.

Let R be the radius of larger circle.

Let r be the radius of smaller circle.

Let the length of chord be 2c

Area of space between concentric circle $=\pi(R^2-r^2)$

Further Explanation:

According to question, it becomes,

$\pi(R^2-r^2)=49\pi$

Therefore, $R^2-r^2=49$

In ΔOAB, ∠OAB = 90° (Please find attach figure)

Using the pythagorous theorem, we get that  

$c=\sqrt{R^2-r^2}$

$c=\sqrt{49}$               $\because R^2-r^2=49$

$c=7$

Length of chord = 2c

Length of chord = 2(7)

                          = 14 units

Learn more:

brainly.com/question/13034352  (Answered by wagonbelleville)

Keywords :

Length of chords, Pythagorous theorem, Concentric circles