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Kitty [74]
1 year ago
9

A special shape is 2.02 m long and 1.22 m wide. A double row of tiles makes a border . Each tile is 2cm by 2cm. How many tiles a

re needed to make the border.?
Mathematics
1 answer:
Alex787 [66]1 year ago
3 0

Answer:

6.48

Step-by-step explanation:

The computation of tiles are needed to make the border is shown below:-

= 2 \times (length + breadth)

where,

Length is 2.02 m

And, the breadth is 1.22

Now placing these values to the above formula

So, the number of tiles needed to make the border is

= 2 \times (2.02 + 1.22)

= 2 \times 3.24

= 6.48

Therefore for computing the tiles are needed to make the border we simply applied the above formula.

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The perimeter of the base of a regular quadrilateral prism is 60 cm, the area of one of the lateral faces is 105 cm2.
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For a better understanding of the solution, please follow the diagram in the attached file.

A regular quadrilateral is basically a square.

So, if the base of the prism has a perimeter of 60 cm, then the length of the side of the square will be \frac{60}{4}=15 cm. It is shown of the diagram.

Now, from the diagram, it is clear that the lateral face area, which is given as 105 cm^2, is the product of the side of the square, which is known, and the unknown height, let us call it h. Thus, we will get the following equation:

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This is depicted on the diagram.

Now, all our required parameters are in place. Thus, let us find what has been asked.

<u>SURFACE AREA</u>

Surface Area (SA) will be the sum of the areas of the two bases (squares) and the areas of the four lateral faces.

Since the side of one square base is 15 cm, therefore, the area of one square base will be 15^2.

Likewise, the area of one lateral surface is actually the area of a rectangle with length 15 cm and height 7 cm. Thus, it's area will be given as: 15\times 7.

Thus, our equation will be:

SA=2\times 15^2+4\times 15\times 7=870 cm^2

Therefore, Surface Area=870 cm^2

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The volume of the prism will simply be the area of the base times the height of the prism.

Thus, the volume is:

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