Video: CBSE Class X • Pack 5 • 2014 • Question 22

CBSE Class X • Pack 5 • 2014 • Question 22


Video Transcript

In the figure, 𝑃𝑆𝑅, 𝑅𝑇𝑄, and 𝑃𝐴𝑄 a three semicircles of diameters 10 centimeters, three centimeters, and seven centimeters, respectively. Find the perimeter of the shaded region. Take 𝜋 is equal to 3.14.

For this question, we’ve been given a shape and we’ve been asked to find its perimeter. The perimeter of the shape that we’ve been given is comprised of three semicircular arcs. If we define the perimeter as 𝑥, we can say that 𝑥 is the sum of these three arcs. The arcs are 𝑃𝑆𝑅, 𝑅𝑇𝑄, and 𝑃𝐴𝑄.

Now, the question tells us the diameter of the semicircles which form these arcs. This information can be used to answer the question in the following way. We first recall that the perimeter — also known as the circumference 𝑐 of a circle — is given by two 𝜋𝑟, where 𝑟 is the radius of the circle.

Since the semicircle is a circle that’s been cut in half, the arc length of a semicircle will be half that of the circumference of the original circle; that is, 𝑐 divided by two. Dividing both sides of our equation by two, we see that this is equal to two 𝜋𝑟 over two. Cancelling the two on the top and bottom half of our fraction, we see that this is equal to 𝜋 times 𝑟.

Now, for this question, we haven’t been given the radius of the circles, which have formed our semicircular arcs rather we’ve been given the diameters. We can now recall that the radius of a circle is half of its diameter. We can now put these two factors together by replacing the 𝑟 in our original formula by 𝑑 over two.

When we do so, we find that the length of a semicircular arc, here given by 𝑐 over two, is equal to 𝜋 times the diameter divided by two. This formula will allow us to move forward with our question. And we’ll put it to one side to make room for some working.

Looking back at the equation that we have for 𝑥, which we defined as the perimeter of the shaded region, we can now replace the length 𝑃𝑆𝑅 with 𝜋 times the diameter of the semicircle which formed 𝑃𝑆𝑅 divided by two. We can also do the same thing for the semicircles 𝑅𝑇𝑄 and 𝑃𝐴𝑄.

The question has given us that the diameter of 𝑃𝑆𝑅 is 10 centimeters. And so we can rewrite our first term as 𝜋 times 10 over two. The diameter of 𝑅𝑇𝑄 is three centimeters. And so our second term becomes 𝜋 times three over two. Finally, the diameter of 𝑃𝐴𝑄 is seven centimeters. And our last term becomes 𝜋 times seven over two.

We can now use the fact that all three of our terms have a factor of 𝜋. We can take out this factor to simplify our next line of working, which then becomes 𝜋 times 10 over two plus three over two plus seven over two. 10 plus three plus seven is equal to 20. And so our bracket simplify to 20 over two. 20 divided by two is 10. And so we can replace this in our brackets.

We’re now left with the fact that 𝑥, the perimeter of our shape, is equal to 𝜋 times 10. Here, we know that the question has given us an approximate value of 𝜋 to use, which is 3.14. We now substitute this value into our equation. 3.14 times 10 is equal to 31.4. And we also remember to add back in the units of length, which here are centimeters.

This is now our final answer for the question. And we have found that based on the semicircular diameters which we’ve been given the perimeter of the shaded region is 31.4 centimeters.

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