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Nat2105 [25]
3 years ago
8

Help me solve this question please

Mathematics
1 answer:
viktelen [127]3 years ago
8 0

Answer:

10

Step-by-step explanation:

the answer is 10

bcoz there's a Pythagoras teiplate

6,8,10

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A solid right pyramid has a square base. The length of the base edge is4 cm and the height of the pyramid is 3 cm period what is
Illusion [34]

Answer:

The volume of this pyramid is 16 cm³.

Step-by-step explanation:

The volume V of a solid pyramid can be given as:

\displaystyle V = \frac{1}{3} \cdot b \cdot h,

where

  • b is the area of the base of the pyramid, and
  • h is the height of the pyramid.

Here's how to solve this problem with calculus without using the previous formula.

Imaging cutting the square-base pyramid in half, horizontally. Each horizontal cross-section will be a square. The lengths of these squares' sides range from 0 cm to 3 cm. This length will be also be proportional to the vertical distance from the vertice of the pyramid.

Refer to the sketch attached. Let the vertical distance from the vertice be x cm.

  • At the vertice of this pyramid, x = 0 and the length of a side of the square is also 0.
  • At the base of this pyramid, x = 3 and the length of a side of the square is 4 cm.

As a result, the length of a side of the square will be

\displaystyle \frac{x}{3}\times 4 = \frac{4}{3}x.

The area of the square will be

\displaystyle \left(\frac{4}{3}x\right)^{2} = \frac{16}{9}x^{2}.

Integrate the area of the horizontal cross-section with respect to x

  • from the top of the pyramid, where x = 0,
  • to the base, where x = 3.

\displaystyle \begin{aligned}\int_{0}^{3}{\frac{16}{9}x^{2}\cdot dx} &= \frac{16}{9}\int_{0}^{3}{x^{2}\cdot dx}\\ &= \frac{16}{9}\cdot \left(\frac{1}{3}\int_{0}^{3}{3x^{2}\cdot dx}\right) & \text{Set up the integrand for power rule}\\ &= \left.\frac{16}{9}\times \frac{1}{3}\cdot x^{3}\right|^{3}_{0}\\ &= \frac{16}{27}\times 3^{3} \\ &= 16\end{aligned}.

In other words, the volume of this pyramid is 16 cubic centimeters.

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