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neonofarm [45]
3 years ago
12

Find the total area for the regular pyramid. T.A. xxx(sqr) + xx (sqr)x

Mathematics
1 answer:
N76 [4]3 years ago
5 0
We are to solve the total area of the pyramid and this can be done through area addition. We first determine the area of the base using the Heron's formula.
                           A = √(s)(s - a)(s - b)(s - c)
where s is the semi-perimeter
                                 s = (a + b + c) / 2
 Substituting for the base,
                               s = (12 + 12 + 12)/ 2 = 18
                          A = (√(18)(18 - 12)(18 - 12)(18 - 12) = 62.35
Then, we note that the faces are just the same, so one of these will have an area of,
                                s = (10 + 10 + 12) / 2 = 16
                                  A = √(16)(16 - 12)(16 - 10)(16 - 10) = 48
Multiplying this by 3 (because there are 3 faces with these dimensions, we get 144. Finally, adding the area of the base,
                          total area = 144 + 62.35 = 206.35
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Given a=20 and b=a/4-3 what is the value of b
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Step-by-step explanation:

Subsitute a for 20.

b=20/4-3

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If the garden were in the shape of a square, all of its sides would be of equal length. so if you divide 240 by 4 sides, each one would be 60 feet. Then, to find the square garden's area, you multiply 60 x 60 = 360 feet².

If the garden was 50 by 70, you would multiply the two values together to find the area, which is 350 feet².

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3 years ago
Find the similarity ratio and the ratio of perimeters for two regular octagons with areas of 18in2 and 50in2
sweet-ann [11.9K]

Answer:

3. 3 : 5; 3: 5

Step-by-step explanation:

First, let's find the leght of the sides of each octagon.

A= 18in^{2}

The area of an octagon is defined by

A=2(1+\sqrt{2})l^{2}

Replacing the area

18=2(1+\sqrt{2})l^{2}\\\frac{18}{2(1+\sqrt{2})} =l^{2}\\l=\sqrt{\frac{18}{3.4} } \approx 2.3 \ in

Therefore, the side of the first octagon is 1.6 inches long.

Its perimeter is: P=8(2.3in)=18.4in

A=50 in^{2}

50=2(1+\sqrt{2})l^{2}\\\frac{50}{2(1+\sqrt{2})} =l^{2}\\l=\sqrt{\frac{50}{3.4} } \approx 3.8 \ in

Therefore, the side of the second octagon is 3.8 inches long.

Its perimeter is P=8(3.8in)=30.4in.

Now, let's divide to find each ratio:

\frac{3.8}{2.3} \approx 1.65 (the ratio between sides).

\frac{30.4}{18.4} \approx 1.65 (the ratio between perimeters).

Therefore, the closest ratio is 3. 3 : 5; 3: 5

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3 years ago
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