Acircle has radius of 18 cm. find the area of the smaller of the two regions determined by a chord with length of 18√2 cm . hint: (look for right triangles using pythagorean theorem converse)

The area is 81*pi-162. Look at the triangle formed by the chord and the two radii connected to the endpoints of the chord. Since 18^2+18^2=(18*\sqrt2)^2, this is an isosceles right triangle by Pythagorean theorem. The area of the quarter circle containing the region we look for is pi*18^2/4=81*pi. The area of the right triangle is 18*18/2=162. The difference between these two areas, or 81*pi-162, is the area of the smaller region.

Abit confused about what you mean, but i think you might be referring to slope intercept form: y = mx + bin that case b, or the y intercept is already given.(0,-1)find the slope by using the formulay2 - y1/x2 - x1the final slope will be 4/2, equivalent to 2.now your equation should look like thisy = 4/2x -1sorry if this didn’t you out!

Answer:

13.46

step-by-step explanation:

16.80 x .20 = 13.46