Composite Solid Volume Calculation
The problem asks for the total volume of a composite solid formed by placing a pyramid on top of a cube.
The total volume is the sum of the volume of the cube and the volume of the pyramid.
Vtotal = Vcube + Vpyramid
Volume of the Cube
The cube has a side length s = 10 cm.
The formula for the volume of a cube is:
Vcube = s3
Substituting the value:
Vcube = (10 \text{ cm})^3 = 1000 \text{ cm}^3
Volume of the Pyramid
The pyramid has an equilateral triangular base with side length a = 10 cm and a height hpyramid = 14 cm.
First, calculate the area of the equilateral triangular base (Base Area):
\text{Base Area} = \frac{\sqrt{3}}{4} a^2
\text{Base Area} = \frac{\sqrt{3}}{4} (10 \text{ cm})^2 = \frac{\sqrt{3}}{4} \times 100 \text{ cm}^2 = 25\sqrt{3} \text{ cm}^2
Using the approximation $\sqrt{3} \approx 1.73205$:
\text{Base Area} \approx 25 \times 1.73205 \approx 43.301 \text{ cm}^2
The formula for the volume of a pyramid is:
Vpyramid = \frac{1}{3} \times \text{Base Area} \times h_{pyramid}
Substituting the values:
Vpyramid} = \frac{1}{3} \times (25\sqrt{3} \text{ cm}^2) \times (14 \text{ cm})
Vpyramid} = \frac{350\sqrt{3}}{3} \text{ cm}^3
Calculating the approximate value:
Vpyramid} \approx \frac{350 \times 1.73205}{3} \approx \frac{606.2175}{3} \approx 202.07 \text{ cm}^3
Total Volume Calculation
Add the volume of the cube and the volume of the pyramid:
Vtotal = Vcube + Vpyramid
Vtotal} \approx 1000 \text{ cm}^3 + 202.07 \text{ cm}^3
Vtotal} \approx 1202.07 \text{ cm}^3
Similar Questions
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Important Questions from 3-D Mensuration
Find the total surface area of a closed cylinder having a base radius of 70 m and a height of 110 m. [Use π = \(22\over7\)]
- The curved surface area of a cylinder is half of its total surface area. If its height is 195 cm, then find its diameter (in cm).
The diameter of the base and slant height of a right circular cone are 30 cm and 113 cm, respectively. Find the volume (in cm³) of the given cone.
(Use $\pi = \frac{22}{7}$)- A cylindrical rod has an curved surface area of $4,900 \text{ cm}^2$. If the length of the rod is 97 cm, then the radius (in cm) of the rod, correct to two places of decimal, is:
Take $\pi = \frac{22}{7}$ - The volume (in $m^3$) of a cube, each of whose edges is 42 m, is:
