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Goryan [66]
4 years ago
15

Please help immediately

Mathematics
1 answer:
STALIN [3.7K]4 years ago
8 0

So basically all you have to do is find the area of one of the smaller semi circles by using the formula for the area of a circle (A=πr^2). You know that the length of the larger semi circle's radius is equivalent to 6 cm because the radius of the smaller ones are 3 cm, meaning the diameter would have to be 6 cm and in this case, the length of the smaller semi circles' diameters is equal to the radius of the big semi circle. Then you would find the area of the big semi circle again by using the area of a circle formula, but after getting the answer you would half it, obviously because it's a semi circle. Subtract the are of the smaller semi circle you found earlier from the answer you just got and that's it ;) (you wouldn't have to half the area since there are two smaller semi circles and 1/2 + 1/2 = 1 but u knew that)

Put simply, the answer would be about 88.2644 cm because circles.

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1. S(–4, –4), P(4, –2), A(6, 6) and Z(–2, 4) a) Apply the distance formula for each side to determine whether SPAZ is equilatera
Aleksandr [31]

Answer:

a) SPAZ is equilateral.

b) Diagonals SA and PZ are perpendicular to each other.

c) Diagonals SA and PZ bisect each other.

Step-by-step explanation:

At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.

a) If figure is equilateral, then SP = PA = AZ = ZS:

SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}

SP \approx 8.246

PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}

PA \approx  8.246

AZ =\sqrt{(-2-6)^{2}+(4-6)^{2}}

AZ \approx 8.246

ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}

ZS \approx 8.246

Therefore, SPAZ is equilateral.

b) We use the slope formula to determine the inclination of diagonals SA and PZ:

m_{SA} = \frac{6-(-4)}{6-(-4)}

m_{SA} = 1

m_{PZ} = \frac{4-(-2)}{-2-4}

m_{PZ} = -1

Since m_{SA}\cdot m_{PZ} = -1, diagonals SA and PZ are perpendicular to each other.

c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:

M_{SA} = \frac{1}{2}\cdot S(x,y) + \frac{1}{2}\cdot A(x,y)

M_{SA} = \frac{1}{2}\cdot (-4,-4)+\frac{1}{2}\cdot (6,6)

M_{SA} = (-2,-2)+(3,3)

M_{SA} = (1,1)

M_{PZ} = \frac{1}{2}\cdot P(x,y) + \frac{1}{2}\cdot Z(x,y)

M_{PZ} = \frac{1}{2}\cdot (4,-2)+\frac{1}{2}\cdot (-2,4)

M_{PZ} = (2,-1)+(-1,2)

M_{PZ} = (1,1)

Then, the diagonals SA and PZ bisect each other.

8 0
3 years ago
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