Video: CBSE Class X • Pack 1 • 2018 • Question 24

CBSE Class X • Pack 1 • 2018 • Question 24

07:07

Video Transcript

The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 centimeters and 30 centimeters respectively. If its height is 24 centimeters, find the area of the metal sheet used to make the bucket. Use 𝜋 equals 3.14.

So here we have a bucket with the lower and upper diameters 10 centimeters and 30 centimeters, and a height of 24 centimeters. So our goal is to find the area of the metal sheet used to make this bucket. So essentially, we need to find the surface area of the bucket, which would include the bottom of the bucket and the sides of the bucket. But remember, there’s no lid on a bucket. So the bottom, we will find the area of a circle. In this side piece, we will have to find the lateral area. So we’ll just call it the lateral piece.

So how will we do this? A bucket isn’t exactly a shape. However, it’s in the form of a frustum of a cone. So it’s like a cone with the top cut off. So imagine this cone upside down. So we need the whole cone except for that bottom piece. So, so to find the surface area of the bucket, we could take the surface area of the larger cone and then subtract the surface area of the small cone. So before we begin doing that, let’s look at the formula needed for the surface area of a cone.

Technically, a cone is a pyramid. It’s a circular pyramid. And the formula for the surface area of a pyramid is equal to one-half times the perimeter times the slant height plus the area of the base. Well, the perimeter means the perimeter of the base. And the base is a circle. So the perimeter of a circle is the circumference. And we have a formula for that, two times 𝜋 times the radius. Slant height, we call 𝑙. And the area of the base will be the area of a circle, 𝜋 times radius squared. So let’s condense this.

So we have one-half times two times 𝜋 times 𝑟 times 𝑙 plus 𝜋𝑟 squared. The twos can cancel. So the surface area of a cone would be 𝜋𝑟𝑙 plus 𝜋𝑟 squared. So let’s plug this into our formula. So now that we have this, we have the formula for the surface area of the large cone and the formula for the surface area of the smaller cone. However, their radii are not the same and the slant heights will be different.

First, let’s begin looking at the radii. If the large diameter, 30, is the diameter, then half of that will be the radius. Half of 30 is 15 and half of 10 is five. So let’s go ahead and plug in these values. The radius of the large cone is 15. And the radius of the small cone is five. Now for the slant heights, here will be the slant height of the large cone. And here will be the slant height of the small cone, 𝑙 one and 𝑙 two. However, we don’t know either. But maybe this 𝑙 three could help. How could we find this 𝑙 three?

We can create a right triangle. Let’s move that 24-centimeter height over to the right. So we’ve moved it here, but it’s still equal to 24. This length would be 15. The top radius minus the bottom radius of five, so 15 minus five, giving us 10. And then this last side making a triangle would be the 𝑙 three. And the height would be perpendicular to the radius. So we have a right triangle. So here it is. And we can use the Pythagorean theorem to solve for 𝑙 three because it’s the hypotenuse.

The Pythagorean theorem states: The square of the hypotenuse, the side across from the 90-degree angle, is equal to the sum of the squares of the legs, the other two sides. So now, we can solve for 𝑙 three. 24 squared is 576. And then we add 100 to that and get 676. And then we take the square root and we find that 𝑙 three is equal to 26. So now that we have that piece, how do we find 𝑙 one or 𝑙 two?

We can use similar triangles. We just said with our triangle, 𝑙 three was equal to 26. So if we create another right triangle taking up the entire cone with that height, the radius of the top was 15, we don’t know 𝑙 one, and we actually don’t know the height of this triangle either. However, we’re just focused on 𝑙 one, that slant height. 10 can go with 15 and 26 can go with 𝑙 one. We can create proportion and solve. 10 is to 15 as 26 is to 𝑙 one. So we take 10 times 𝑙 one and set it equal to 15 times 26, which is 390. So to solve for 𝑙 one, we divide both sides of the equation by 10 and find that 𝑙 one is equal to 39.

So now that we have 𝑙 one, we can actually find 𝑙 two. Because 𝑙 three and 𝑙 two add to be equal to 𝑙 one. So again, we can find 𝑙 two by taking 𝑙 one, which is 39, and then subtracting 𝑙 three, 26. And 39 minus 26 is 13. So now that we have 𝑙 one and 𝑙 two, let’s plug them in to our formula. So here we go.

Before we start evaluating, we need to think about what we actually need. Because this is a bucket and the bucket doesn’t have a top on it. So we don’t need the circle at the top which would have been the area of the base of the larger cone, the 𝜋𝑟 squared, the area of the circle. We don’t want that. And when we go to take away the area of the smaller cone, we actually need the area of that circle because it’s the bottom of the bucket. So we want to keep this part and not subtract it. So let’s just leave that aside for a second.

So here we have 𝜋 times 15 times 39. That represents the surface area of the lateral part of the large cone. And then we’re subtracting the lateral part of the smaller cone. And that’s what gives us that green part. But we wanted to keep the area of the bottom of the bucket. So we want to add the 𝜋 times five squared. We wanna keep it.

Now, we will not multiply by 𝜋 till the very end. This will eliminate rounding errors. So we have 585𝜋 minus 65𝜋 plus 25𝜋. That gives us 545𝜋. So it tells us to use 𝜋 equals 3.14. And when we multiply, we get 1711.3 centimeters squared. So this is how much of a metal sheet we would need to use to be able to make this bucket.

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