In △ ABC, a, B and C are sides, satisfying a & # 178; - AB BC AC = 0

In △ ABC, a, B and C are sides, satisfying a & # 178; - AB BC AC = 0

(a²－ab)－(bc－ac)=0
a(a－b)－c(b－a)=0
a(a－b)＋c(a－b)=0
(a－b)(a＋c)=0
Then:
A-B = 0 or a + C = 0
A = b
This triangle is isosceles triangle

Given that there are points a (1,1), B (2,0) C (0, - 1) in the coordinate plane, the area of triangle ABC is?
Given that there are points a (1,1), B (2,0) C (0, - 1) in the coordinate plane, the area of triangle ABC is?
Did not learn trigonometric function, just learned plane rectangular coordinates

The area of ABC with X-axis as the base is the sum of two triangles, and the analytic formula of line AC is y = 2x-1
So the point of intersection with X axis (1 / 2,0) is s = 1 / 2 * (2-1 / 2) * 1 = 3 / 4
S=1/2*（2-1/2）*1=3/4
So s △ ABC = 3 / 4 + 3 / 4 = 3 / 2

If the lengths of the two sides of an isosceles triangle are 4 and 6, what is the height of the bottom?

The two sides of an isosceles triangle are 4 and 6, and the third side is 4 or 6
If the third edge is 4, then 6 is the bottom edge. Using Pythagorean theorem, the height of the bottom edge equals the root sign 16-9 equals the root sign 7
If the third edge is 6, then 4 is the bottom edge. Using Pythagorean theorem, the height of the bottom edge is equal to 4 times the root sign 2