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aleksandrvk [35]
2 years ago
13

Given a circle of radius = 10 m, find the area of a 60 degree sector. Please round to 1 d.p.

Mathematics
1 answer:
LUCKY_DIMON [66]2 years ago
7 0

Answer:

Area ~= 52.4 m^2

OR (see image why)

Area ~= 52.3 m^2

Step-by-step explanation:

Use A = pi•r^2 to find the area of the whole circle. For a sector, times that by degrees/360.

Your problem has a 60° sector. 60/360 is 1/6.

Use either, 60/360 or 1/6 times 100pi.

If you use the pi button on the calculator the answer is .1 different than if you use 3.14 for pi in your calculation. The pi button answer is more accurate. See image.

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Answer:

Second statement is true.

The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

Step-by-step explanation:

for first part of statement

The lengths 7, 40 and 41 can not be sides of a right triangle.

If the square of long side is equal to the sum of square of other two sides

then the given length can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

c^{2} =a^{2} +b^{2}----------(1)

Let c=41 and a = 7 and b=40

Put all the value in equation 1.

41^{2} =7^{2} +40^{2}

1681=49+1600

1681=1649

Therefore, the square of long side is not equal to the sum of square of other two sides, So given lengths 7, 40 and 41 can not be sides of a right triangle.

for second part of statement.

The lengths 12, 16, and 20 can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

Let c=20 and a = 12 and b=16

20^{2} =12^{2} +16^{2}

400=144+256

400=400

Therefore, the square of long side is equal to the sum of square of other two sides, So given the lengths 12, 16, and 20 can be sides of a right triangle.

Therefore, The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

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