# Video: CBSE Class X • Pack 3 • 2016 • Question 14

CBSE Class X • Pack 3 • 2016 • Question 14

06:40

### Video Transcript

The figure shows two concentric circles of radii seven centimeters and 14 centimeters. If the angle 𝐴𝑂𝐶 is 40 degrees, find the area of the shaded region. Use 𝜋 equals 22 over seven.

We can see the two concentric circles in the diagram. They have the same centre 𝑂. That’s what concentric means. And we’re told that their radii are seven centimeters and 14 centimeters. Clearly, the inner circle has radius seven centimeters and the larger outer circle has radius 14 centimeters. We’re also told that the angle 𝐴𝑂𝐶 is 40 degrees. And we can see this is already marked for us on the diagram.

Our task is to find the area of the shaded region. This is the region shaded blue in the figure, which is part of the ring or annulus bounded by the two circles. We can view our shaded region as a compound region. That’s a region which is built from simpler regions either by joining two together or removing one region from another.

We get the shaded region by taking the region inside the largest circle, removing a sector of this larger circle and also a sector of the smaller circle, whose central angle is a reflex angle. And hence, the area of the shaded region will be the area of the larger circle minus the area of those two sectors.

Let’s start by calculating the area of the circle. The area of a circle with radius 𝑟 is 𝜋𝑟 squared. And we know that the radius of the larger circle is 14 centimeters. So the area of the larger circle is 𝜋 times 14 squared centimetres squared.

How about the area of the sector of the larger circle? Well, the area of a sector whose central angle is 𝜃 degrees in a circle with radius 𝑟 is 𝜃 over 360 times 𝜋𝑟 squared. The central angle 𝜃 of our sector is 40 degrees and the radius of the larger circle we’ve seen is 14 centimeters. Substituting these values, we find that the area of the sector is 40 over 360 times 𝜋 times 14 squared centimeters squared.

Now, how about the area of the other sector? Its central angle is a reflex angle. This angle and the 40-degree angle 𝐴𝑂𝐶 together form an angle in a whole circle. The angle in a circle is 360 degrees. And so this angle has measure 360 minus 40 equals 320 degrees. So that’s the central angle 𝜃 of the sector.

How about the radius 𝑟? Well, we’re in the smallest circle where the radius is seven centimeters. So the area of this sector is 320 over 360 times 𝜋 times seven squared centimeters squared. We combined the units and we have an expression for the area of the shaded region. All we have to do now is simplify it.

Before we clear some room and start simplifying, let me first say that the way that we found the area of the shaded region was to build the shaded region out of regions, whose areas we could find easily. In this case, we built it out of a circle and two sectors. But other ways of building the shaded region are possible.

Okay, let’s simplify this expression then. We can take a common factor of 𝜋 times 14 squared out of the first two terms. And as one is 360 over 360, inside the brackets we have a 360 over 360 minus 40 over 360, which subtracting the numerators is 320 over 360. And so we get 320 over 360 times 𝜋 times 14 squared minus the other term.

Now, we notice that both terms have a factor of 320 over 360. If we take this factor out, you might recognize what remains in the brackets. 𝜋 times 14 squared minus 𝜋 times seven squared is the difference of the areas of the circles. And hence, it is the area of the ring or annulus which is bounded by the two circles. We can interpret 320 over 360 as the proportion of the ring that is shaded.

And if we look at the diagram, we can see that this is true. This gives us another way to think about the shaded region. 320 over 360 is not simplified. Both numerator and denominator are divisible by 40. 320 is eight times 40 and 360 is nine times 40. And so this simplifies to eight over nine.

We can of course also take the common factor of 𝜋 outside the brackets. And perhaps now, it’s a good time to use approximation. We’re told to use in the question 𝜋 equals 22 over seven. We also take this opportunity to write 14 squared minus seven squared as 147. The working for this is shown on the left below the diagram.

Do we get any further simplification? Yes, 147 divided by seven is 21. And we can cancel a common factor of three from 21 and nine. 21 is seven times three and nine is three times three. Tidying up then, we see that the area is eight times 22 times seven over three centimeters squared.

Now, let’s perform the multiplication in the numerator. Eight times 22 or 22 times eight is 176. And multiplying this by seven, we get 1232. Making this substitution then, we get our answer as an improper fraction. It’s 1232 over three centimeters squared. But we can choose to write this area as a decimal value if we like. Dividing 1232 by three, we get 410 remainder two.

So this fraction is 410 and two-thirds, which is 410.67 centimeters squared correct to two decimal places.