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coldgirl [10]
3 years ago
13

AREA ADDITION AND SUBTRACTION URGENT?

Mathematics
2 answers:
qaws [65]3 years ago
6 0

Okay, so we know that the radius of the circle is 7.53

The area of the circle is 178.13, so we'll remember that.

Because the diameter is 15.06, we actually realize that the square is now two triangles, with the hypotenuse being 15.06.

a^2 + b^2 = 15.06^2

Now, because a^2 and b^2 are of the same value, we can basically turn it into this:

2a^2 = 226.8036

a^2 = 113.4018

Doing this actually solves for the square's area:

178.13 - 113.4018 = 64.7282

The yellow region is 64.7282 cm^2

Aleksandr [31]3 years ago
4 0

The area of a circle:

A_C=\pi r^2\\\\r=7.53\ cm\\\\A_C=\pi\cdot(7.53)^2=56.7009\pi\ cm^2\approx56.7009\cdot3.14\approx178.04\ cm^2

The area of a square (like an area of a rhombus):

A_S=\dfrac{d\cdot d}{2}=\dfrac{d^2}{2}\\\\d=2\cdot7.53=15.06\ cm\\\\A_S=\dfrac{15.06)^2}{2}=\dfrac{226.8036}{2}=113.4018\ cm^2\approx113.40\ cm^2

The area of the yellow region:

A=A_O-A_S\\\\A=178.04-113.40=64.64\ cm^2\approx64.6\ cm^2

Answer: 64.4 cm²

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URGENT!!!!!<br> Work out the image attached<br> Include Workings
Ann [662]

9514 1404 393

Answer:

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Step-by-step explanation:

The first attachment shows the sort of geometry involved. The line of sight from the top of the lighthouse makes a tangent with the earth's surface at the horizon (neglecting optical effects due to the atmosphere).

The hypotenuse of the relevant triangle is (r+h), and the angle is adjacent to the radius, r. Then the cosine relation applies:

  Cos = Adjacent/Hypotenuse

  cos(α) = r/(r+h)

We note that the radius is in km, and the length h is given in meters, so we need to do a conversion to put the numbers in this formula. We don't care so much about the angle itself, but we need its value to find the length of the arc along the earth's surface. That is ...

  s = r·α

so, the distance of interest is found from ...

  cos(α) = 6367/(6367 +0.025)

  r·α = 6367·arccos(6367/6367.025) ≈ 17.84 . . . km

The horizon is about 17.8 km from the observer.

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The second problem requires we find the height value that makes the distance 32 km.

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Dividing by the earth's radius and taking the cosine, we have ...

  cos(32/6367) = 6367/(6367 +h)

  h = 6367 -6367/cos(32/6367) ≈ 6367.0804 -6367 = 0.0804 . . . km

The cliffs must be about 80.4 meters high for it to be possible to see that distance.

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