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solong [7]
2 years ago
8

A figure is broken into a rectangle and a triangle. The triangle has a base of 2 and two-thirds feet and height of 3 feet. The r

ectangle has a base of 5 feet and height of 1 and one-third feet.
The irregular figure can be broken into a triangle and a rectangle as shown with the dashed line.

The length of b, the base of the triangle, is
ft.
The area of the triangle is

The area of the rectangle is

The area of the irregular figure is

Mathematics
2 answers:
Ksenya-84 [330]2 years ago
8 0

Answer:

  • See below

Step-by-step explanation:

The length of b, <u>the base</u> of the triangle, is

  • 5 - 2 1/3 = 3 - 1/3 = 2 2/3 ft

The <u>area of the triangle</u> is

  • A = 1/2*3*(2 2/3) = (3/2)(8/3) = 8/2 = 4 ft²

The <u>area of the rectangle</u> is

  • 1 1/3 * 5 = 4/3 * 5 = 20/3 = 6 2/3 ft²

The <u>area of the irregular figure</u> is

  • 4 + 6 2/3 = 10 2/3 ft²
ohaa [14]2 years ago
5 0

Answer:

b=2\frac23\:\sf ft

Area of triangle = 4 ft²

Area of rectangle = 6\frac23\: \sf ft^2

Area of irregular figure = 10\frac23\: \sf ft^2

Step-by-step explanation:

\begin{aligned}\sf Base\:of\:triangle\:(b) & = 5-2\frac13\\\\& = \dfrac{15}{3}-\dfrac73\\\\ & = \dfrac83\\\\ & = 2\frac23\:\sf ft \end{aligned}

\begin{aligned}\sf Area\:of\:a\:triangle & =\dfrac12 \sf \times base \times height\\\\& = \dfrac12 \times b \times 3\\\\ & = \dfrac12 \times \dfrac83 \times \dfrac31\\\\ & = \dfrac{24}{6}\\\\ & = 4\: \sf ft^2\end{aligned}

\begin{aligned}\sf Area\:of\:rectangle& =\sf length \times width\\\\& = 5 \times 1\frac13\\\\ & = \dfrac51\times \dfrac43\\\\& = \dfrac{20}{3}\\\\ & = 6\frac23\: \sf ft^2\end{aligned}

\begin{aligned} \sf Area\:of\:irregular\:figure & = \sf area\:of\:triangle+area\:of\:rectangle\\\\ & = 4 + 6\frac23\\\\ & = 10\frac23\: \sf ft^2\end{aligned}

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The base of a prism has n sides. Find the numbers of faces, edges, and
Harman [31]

Answers:

  • faces = n+2
  • edges = 3n
  • vertices = 2n

===========================================================

Explanation:

Think of a hexagonal room with n = 6 walls, i.e. the floor is a hexagon with n = 6 sides. The floor and ceiling are parallel to each other, and congruent hexagons. That's 2 faces so far. Then we have another 6 faces to account for the walls. This gives 2+6 = 8 faces of a hexagonal prism.

In more general terms, a prism with a base of n sides will have 2 parallel and congruent base faces, and n walls or lateral faces. This gives n+2 total faces.

------------------------

Let's go back to the hexagonal prism. The floor has 6 sides to it, and so does the ceiling. We have 6+6 = 12 edges so far. Then we have another 6 edges where each of the rectangular walls meet up. That gives 12+6 = 18 edges total of this hexagonal prism room.

For any more general case, each base has n sides. That gives 2n sides so far for just the bases. Then add on another n for the lateral edges and we get 3n total edges.

--------------------------

Once again lets revisit the room with the hexagonal floor and ceiling. The floor has 6 vertices and the ceiling has the same vertex count. Therefore, this prism has 6+6 = 12 vertices.

For the general case, each base has n vertices. There are 2 such identical bases giving 2n vertices total.

---------------------------

One way to check the answer:

We could use Euler's Polyhedron Formula which is

F+V-E = 2

where,

  • F = number of faces
  • V = number of vertices
  • E = number of edges

For the hexagonal prism we found

  • F = 8
  • V = 12
  • E = 18

Then notice how

F+V-E = 2

8+12-18 = 2

20-18 = 2

2 = 2

This confirms the formula works for a hexagonal prism.

Now let's check it for the more general case

We found earlier that,

  • F = n+2
  • V = 2n
  • E = 3n

So,

F+V-E = 2

n+2+2n-3n = 2

3n-3n+2 = 2

0n+2 = 2

0+2 = 2

2 = 2

This helps confirm the answer for any prism with the base of n sides.

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Answer:

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Step-by-step explanation:

A subset simply means a set of which all of its elements are contained in a larger set :

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