$1413\pi$
The toy consists of two parts: a solid cone and a frustum. The total surface area of the toy is the sum of the curved surface area of the cone, the curved surface area of the frustum, and the area of the larger base of the frustum.
1. Calculate Cone Details
Given:
- Cone radius (\(r_1\)) = 5 cm
- Cone height (\(h_1\)) = 12 cm
Calculate the slant height (\(l_1\)) of the cone using the Pythagorean theorem:
\(l_1 = \sqrt{r_1^2 + h_1^2}\)
\(l_1 = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm}\)
Calculate the Curved Surface Area (CSA) of the cone:
\(CSA_{\text{cone}} = \pi r_1 l_1\)
\(CSA_{\text{cone}} = \pi \times 5 \times 13 = 65\pi \text{ cm}^2\)
2. Calculate Frustum Details
Given:
- Frustum smaller base radius (\(r_1\)) = 5 cm
- Frustum larger base radius (\(r_2\)) = 22 cm
- Frustum slant height (\(l_2\)) = 32 cm
Calculate the Curved Surface Area (CSA) of the frustum:
\(CSA_{\text{frustum}} = \pi (r_1 + r_2) l_2\)
\(CSA_{\text{frustum}} = \pi (5 + 22) \times 32 = \pi \times 27 \times 32 = 864\pi \text{ cm}^2\)
Calculate the Area of the larger base of the frustum:
\(Area_{\text{base}} = \pi r_2^2\)
\(Area_{\text{base}} = \pi \times 22^2 = 484\pi \text{ cm}^2\)
3. Calculate Total Surface Area
The total surface area (TSA) of the toy is:
\(TSA = CSA_{\text{cone}} + CSA_{\text{frustum}} + Area_{\text{base}}\)
\(TSA = 65\pi + 864\pi + 484\pi\)
\(TSA = (65 + 864 + 484)\pi\)
\(TSA = 1413\pi \text{ cm}^2\)
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Important Questions from Mensuration
In a circular garden of radius 15 m, a path of 2 m wide has to be made inside the garden at the rate of ₹ 24 per sq. m. The cost of making the path is: (Take π = \(\frac{22}{7}\) )
A hollow cylinder with outer radius 4 cm and height 2 cm is made up of 1 cm thick metal sheet. What is the volume of metal used? (Take π = \(\frac{22}{7}\))
The sides of a triangular park are 60 m, 297 m and 303 m. Its area is equal to the area of a square-shaped garden. What is the perimeter (in m) of the garden?
The length and breadth of a rectangular field are in the ratio 4 : 3. If the cost of cultivating the field at ₹2 per m 2is ₹600, then the length of the field is:
A cylindrical tank has a capacity of 5632 m3. If the diameter of its base is 8 m, what is the depth of the cylindrical tank? (Use π = \(\frac{22}{7}\))
