Kaliyuga Ekalavya

NCERT Class X Chapter 12: Surface Areas And Volumes Example 4 Question: Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. (Use \( \pi = \frac{22}{7} \)) Given: Height of the cylinder, \( h = 1.45\,\text{m} = 145\,\text{cm} \) Radius of the cylinder, \( r = 30\,\text{cm} \) \( \pi = \frac{22}{7} \) To Find: Total surface area of the bird-bath. Formula: Total surface area = Curved surface area of cylinder + Area of base + Curved surface area of hemisphere Curved surface area of cylinder: \( 2\pi r h \) Area of base: \( \pi r^2 \) Curved surface area of hemisphere: \( 2\pi r^2 \) Solution: Step 1: Write the expression for total surface area. $$ \text{Total surface area} = 2\pi r h + \pi r^2 + 2\pi r^2 = 2\pi r h + 3\pi r^2 $$ Step 2: Substitute the given val...

NCERT Class X Chapter 12: Surface Areas And Volumes Example 1 Question: Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere. The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Use \( \pi = \frac{22}{7} \)) Given: Total height of top, \( H = 5 \) cm Diameter of top \( = 3.5 \) cm Radius of top, \( r = \frac{3.5}{2} = 1.75 \) cm \( \pi = \frac{22}{7} \) To Find: The total surface area of the top to be coloured. Formula: Curved surface area of a cone: \( \pi r l \) Curved surface area of a hemisphere: \( 2 \pi r^2 \) Slant height of cone: \( l = \sqrt{r^2 + h^2} \) Solution: Step 1: Find the height of the cone part. The hemisphere sits on top of the cone, so the height of the cone is: $$ h = H - r = 5 - 1.75 = 3.25 \text{ cm} $$ Step 2: ...

NCERT Class X Chapter 12: Surface Areas and Volumes Question: The decorative block is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. (Use \( \pi = \frac{22}{7} \)) Given: Edge of cube, \( a = 5 \) cm Diameter of hemisphere, \( d = 4.2 \) cm Radius of hemisphere, \( r = \frac{4.2}{2} = 2.1 \) cm To Find: Total surface area of the block Formula: Surface area of cube: \( 6a^2 \) Curved surface area of hemisphere: \( 2\pi r^2 \) Area of base of hemisphere: \( \pi r^2 \) Total surface area of block: \( \text{Surface area of cube} + \text{Curved surface area of hemisphere} - \text{Area of base of hemisphere} \) Solution: Step 1: Find the surface area of the cube. $$ \text{Surface area of cube} = 6a^2 = 6 \times (5)^2 = 6 \times 25 = 150 \ \text{cm}^2 $$ Step 2: Find the curved surface area ...

NCERT Class X Chapter 12: Surface Areas And Volumes Example 3 Question: A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in Fig. 12.8. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Use \( \pi = 3.14 \)) Given: Height of rocket, \( H = 26 \) cm Height of conical part, \( h_c = 6 \) cm Diameter of conical base, \( d_c = 5 \) cm \( \Rightarrow r_c = 2.5 \) cm Diameter of cylindrical base, \( d_{cy} = 3 \) cm \( \Rightarrow r_{cy} = 1.5 \) cm \( \pi = 3.14 \) To Find: Area to be painted orange (curved surface area of conical part) Area to be painted yellow (curved surface area of cylindrical part) Formula: Curved surface area of con...