If the images of y = - X-B * 2 and y = 2x + 4 intersect at a point of x-axis, a and y-axis intersect at B and C respectively, how to find the angle ABC area

If the images of y = - X-B * 2 and y = 2x + 4 intersect at a point of x-axis, a and y-axis intersect at B and C respectively, how to find the angle ABC area

S triangle ABC = 6
For example, the key to this problem is the combination of text and graphics. If the images of y = - X-B * 2 and y = 2x + 4 intersect at a point on the x-axis, then a (- 2,0). Then b * 2 = 2 obtained by y = - X-B * 2 and y = 2X + 4 intersect at B and C respectively, which is the value of B in y = KX + B. B (0, - 2) C (0,4)

The surface area of the geometric figure obtained by rotating the rectangle with side lengths of 2cm and 4cm around one side is

2π×2×4+2π×2²=24π㎝²
Or 2 π × 2 × 4 + 2 π × 4 & # 178; = 48 π CM & # 178;

A rectangular paperboard with two sides of 2cm and 4cm in length is used to calculate the surface area of the geometry formed by one revolution around the straight line on one side

The surface area of the lock is 2 * 4 * 2 π + 2 π * 4 ^ 2 = 48 π
The surface area of the lock is 2 * 4 * 2 π + 2 π * 2 ^ 2 = 24 π