# Video: CBSE Class X • Pack 1 • 2018 • Question 21

CBSE Class X • Pack 1 • 2018 • Question 21

04:32

### Video Transcript

Find the area of the shaded region in the figure, where arcs drawn with centres 𝐴, 𝐵, 𝐶, and 𝐷 intersect in pairs at the midpoints 𝑃, 𝑄, 𝑅, and 𝑆 of the sides 𝐴𝐵, 𝐵𝐶, 𝐶𝐷, and 𝐷𝐴, respectively, of a square of side 12 centimetres. Use 𝜋 equals 3.14.

So let’s dissect this picture a little bit. We want to find the area of the shaded region, so what makes up the shaded region. Well, arcs are drawn, which are found here, here, here, and here, with their centres 𝐴, 𝐵, 𝐶, and 𝐷, and 𝑃, 𝑄, 𝑅, and 𝑆 are midpoints of the sides 𝐴𝐵, 𝐵𝐶, 𝐶𝐷, and 𝐷𝐴, respectively.

So 𝑃 is the midpoint of side 𝐴𝐵, 𝑄 is the midpoint of side 𝐵𝐶, 𝑅 is the midpoint of side 𝐶𝐷, and 𝑆 is the midpoint of side 𝐷𝐴. That’s why they use the word “respectively.” And it says that this square has a side of 12 centimetres. So if one side is 12 centimetres, they all must be 12 centimetres.

But we also know that 𝑃, 𝑄, 𝑅, and 𝑆 are midpoints essentially bisecting these sides. So on each side of the midpoints, they’re equal. So half of 12 is six. So all of these are six centimetres.

Okay, so now to find the area of the shaded region, well, the shaded region is on the inside of this square. So we would need to find the area of the square and then subtract those white spots.

But what are those white spots? They look like a fourth of a circle. And we know that it’s actually going to be a circle because it’s an arc where we have both of these sides being six, which will be the radius. So we have a fourth of a circle here, a fourth of a circle here, here, and here.

So we have one-fourth the area of a circle, but there are four of them, so we will need to multiply by four. But four and one-fourth cancel. So we will just have the area of a circle that needs to be subtracted. And that’s because if we would put all four of these one-fourth circles together, we would create one whole circle. And they all have the same radius, so together they make one whole circle.

So we need to take the area of the square and then subtract the area of a circle. The area of a square is simply length times width. But if the length and the width are the same, because it’s a square, we could just say side times side, since the sides are the same, but a side times a side could also be written as side squared.

The area of a circle is 𝜋 times the radius squared. So we know the side lengths are 12 centimetres. So we have 12 squared or 12 times 12 minus 𝜋 times six squared. So 12 times 12 is 144, and six squared is 36. So we have 36 times 𝜋. We’re told to use 3.14 for 𝜋, so we need to take 36 times 3.14, and we get 113.04. So 144 minus 113.04 gives us 30.96.

And how did we get that? Well, we need to add point zero, zero to 144. And now all the way to the right, we have zero minus four, but we can’t do that. We can’t borrow from the zero on the left, but we can borrow from the four and make it a three, so 10 minus four. 10 minus four is six.

Now this zero becomes a nine, and nine minus zero is nine. Bring down our decimal point. Three minus three is zero, four minus one is three, and then the one minus one is zero, so they cancel. So again, we get 30.96. And since this is an area, we should have centimetres squared. So the area of this shaded region in the figure would be 30.96 square centimetres.