The shape of a workpiece is shown in the figure, the degree of BC is 60 ° AB = 6cm, the distance from point B to point C is equal to AB, ∠ BAC = 30 °, then the area of the workpiece is equal to () A. 4πB. 6πC. 8πD. 10π

The shape of a workpiece is shown in the figure, the degree of BC is 60 ° AB = 6cm, the distance from point B to point C is equal to AB, ∠ BAC = 30 °, then the area of the workpiece is equal to () A. 4πB. 6πC. 8πD. 10π

As shown in the figure, in the circle O, ∵ BAC = 30 °, the degree of BC is 60 °, the ∵ BOC = 60 °, the △ BOC and △ ABO are equilateral triangles, and the area of the workpiece is equal to = 16 π × 62 = 6 π

Given a (9,1), B (3,4), I (4,1), whether there is a point C, so that the distance from I to AC and BC is equal to the distance from I to ab

First, the slope of AB is obtained: k = (4-1) / (3-9) = - 1 / 2
So AB equation: y-4 = (- 1 / 2) (x-3)
That is: x + 2y-11 = 0
Then the radius of the inscribed circle r = 5 / √ 5 = √ 5 is obtained
Let BC equation be y-4 = K (x-3)
That is: kx-y-3k + 4 = 0
The distance from the heart to each side is the radius of the inscribed circle
So | 4k-1-3k + 4 | / √ (1 + K & sup2;) = √ 5
Solution: k = 2 or - 1 / 2
Obviously k = - 1 / 2 is the slope of ab
So BC: y-4 = 2 (x-3)
That is, 2x-y-2 = 0
Similarly, AC: x-2y-7 = 0
The simultaneous AC and BC equations give C (1, - 2)