find the center of the circle that can be circumscribed about triangle EFG, E(6,4) F(6,2) and G(10,2)

Mathematics

1 answer:

Sindrei [870]3 years ago

6 0

Letter A) should do, i believe

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Semicircles and quarter circles are types of arc lengths. Recall that an arc is simply part of a circle. We learned about the de

attashe74 [19]

The question is incomplete. Here is the complete question.

Semicircles and quarter circles are types of arc lengths. Recall that an arc is simply part of a circle. we learned about the degree measure of an ac, but they also have physical lengths.

a) Determine the arc length to the nearest tenth of an inch.

b) Explain why the following proportion would solve for the length of AC below: $\frac{x}{12\pi } = \frac{130}{360}$

c) Solve the proportion in (b) to find the length of AC to the nearest tenth of an inch.

Note: The image in the attachment shows the arc to solve this question.

Answer: a) 9.4 in

c) x = 13.6 in

Step-by-step explanation:

a) $\frac{arclength}{2\pi.r } = \frac{mAB}{360}$ , where:

r is the radius of the circumference

mAB is the angle of the arc

arc length = $\frac{mAB.2.\pi.r }{360}$

arc length = $\frac{90.2.3.14.6}{360}$

arc length = 9.4

The arc lenght for the image is 9.4 inches.

b) An <u>arc</u> <u>length</u> is a fraction of the circumference of a circle. To determine the arc length, the ratio of the length of an arc to the circumference is equal to the ratio of the measure of the arc to 360°. So, suppose the arc length is x, for the arc in (b):

$\frac{x}{2.6.\pi } = \frac{130}{360}$