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The outer center of an acute triangle is______ The outer center of an obtuse triangle is______ The outer center of a right triangle is______ .
The outer center of the acute triangle is inside the triangle;
The outer center of the obtuse triangle is outside the triangle;
The outer center of a right triangle is at the midpoint of the hypotenuse
So the answer is: inside the triangle, outside the triangle, the midpoint of the hypotenuse
The triangles in the figure are taken from left to right as acute angle triangles, right angle triangles and obtuse angle triangles. The circumscribed circles of these triangles are drawn respectively, and the characteristics of the new positions of these triangles are described
The acute angle is inside the triangle, the right triangle is at the midpoint of the hypotenuse, and the obtuse angle is outside the triangle,
As shown in the figure, there are______ There are triangles______ There are sharp triangles______ An obtuse triangle with______ A right triangle
According to the graph, there are 4 single triangles, 3 of them are acute angle triangles, and the remaining one is right triangle; there are 3 triangles composed of 2 triangles, 2 of which are acute angle triangles, and the other is right triangle; there are 2 triangles composed of 3 triangles, 1 of which is a right triangle and 1 is a sharp triangle
Radius formula of triangle inscribed circle
R = (a + B-C) / 2 and R = AB / (a + B + C)
Area formula of inscribed circle of triangle
According to Helen's formula: (P = (a + B + C) / 2) s triangle = under the root sign [p * (P-A) * (P-B) * (P-C)] connecting the inner part to the three vertices, the s triangle = a * r / 2 + b * r / 2 + C * r / 2 = R (a + B + C) / 2 = P * r, so: r = under the radical sign [p * (P-A) * (P-B) * (P-C)] / p the area of the circle = pi * (P-A) * (P-B) * (P-C) / P
Area formula of right triangle of inscribed circle The edge length of a right triangle is ab BC AC (hypotenuse). E is the tangent point of the inscribed circle on the AC side. It is proved that AE * EC is equal to the area of the triangle
Radius of inscribed circle r = (a + C-B) / 2
s=ac/2
=2ac/4
=(b^2-a^2-c^2+2ac)/4
=[b^2-(a-c)^2]/4
=[b-(a-c)][b+(a-c)]/4
=[(c+b-a)/2][(a+b-c)/2]
=AE*EC
The length of the three sides of a triangle is given, and the area of the inscribed circle of the triangle is calculated Three times were 3,4,5
Radius of inscribed circle: r = (a + B-C) △ 2, only for right triangle, C is hypotenuse;
For any triangle, the formula is as follows:
Triangle three sides a, B, C, half perimeter P (P = (a + B + C) / 2)
Area: S = √ [P (P - a) (P - b) (P - C)] (Helen's formula)
The radius h of the inscribed circle can be obtained by 2S = (a + B + C) * H
Why is the radius formula of inscribed circle of right triangle r = (a + B-C) / 2 and how to deduce it
AB = (a + B + C) r can be obtained from equal area
That is (a + b) ^ 2-A ^ 2-B ^ 2 = 2 (a + B + C) r
(a+b)^2-c^2=2(a+b+c)r
(a+b+c)(a+b-c)=2(a+b+c)r
r=(a+b-c)/2
In a circle, what is the derivation process of radius formula of inscribed circle of right triangle: r = (a + B-C) △ 2? What is the derivation of R = (a + B-C) △ 2 What is the "^" in "2Ab = (a + b) ^ 2-C ^ 2"? What's the point? Do not understand,
Firstly, a formula is proposed
Area s = 0.5 * (a + B + C) * r, R is the radius of inscribed circle
The proof can be obtained by connecting each vertex with the center of the inscribed circle
Let C be the hypotenuse
∵S=0.5*（a+b+c)*r=0.5ab
∴r=ab/（a+b+c)
Therefore, we only need to prove AB / (a + B + C) = (a + B-C) / 2
That is, 2Ab = (a + B + C) * (a + B-C)
That is, 2Ab = (a + b) ^ 2-C ^ 2
That is, C ^ 2 = a ^ 2 + B ^ 2
Because C is an oblique edge, the above formula holds
So r = (a + B-C) △ 2
That symbol represents the number of times, C ^ 2 = C * C
It is known that the circumference of an obtuse triangle is 48 cm, and the sum of the two sides is twice the length of the third side
48/(2+1)=16
The length of the third side is 16 cm
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