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Two identical acute angle triangles, right angle triangles, obtuse angle triangles, spell one, there are several cases, what kind of map
Any quadrilateral, if it's a right angle, a triangle is also possible
The relationship among acute angle triangle, right angle triangle and obtuse angle triangle is shown by graph (Fig.)
Acute angle right angle obtuse angle
Area formula of right triangle If the three sides of a right triangle are 10 cm 8 cm 6 cm, the area of this right triangle is () cm 2
“liu630799935”：
The area of a right triangle is the product of two right angles divided by two
8 cm × 6 cm △ 2 = 24 square cm
Are you right? Good bye
How to calculate the area of a right triangle and which formula is it?
Base times height divided by 2
Area formula of right triangle
S=ab/2
Where AB is the length of two right angles respectively!
Radius formula of inscribed circle of right triangle
Radius of inscribed circle of right triangle r = 1 / 2 (AB + ac-bc) (formula 1) r = AB * AC / (AB + triangle with BC as oblique side 1. R = 1 / 2 (AB + ac-bc) (formula 1) uses the property of tangent line
What is the area formula of right triangle
Multiply two right angles and divide by two
There are two radius formulas for inscribed circle of right triangle. How to deduce each other? (1)r=a+b-c/2 (2)r=ab/a+b+c
(1) (a+b-c)/2=｛〔(a+b-c)/2〕*(a+b+c)｝/(a+b+c)
=｛〔(a+b)^2-c^2〕/2｝/(a+b+c)
Because (a + b) ^ 2 = a ^ 2 + B ^ 2 + 2Ab = C ^ 2 + 2Ab,
So (a + B-C) / 2 = {[C ^ 2 + 2ab-c ^ 2] / 2} / (a + B + C)
=ab/(a+b+c )
(2)r=ab/(a+b+c) =ab(a+b-c)/〔(a+b+c)(a+b-c)〕
=ab(a+b-c)/〔(a+b)^2-c^2〕
=ab(a+b-c)/2ab
=(a+b-c)/2
Why is the radius of inscribed circle of right triangle (a + B-C) / 2? 2S / (a + B + C) I understand Isn't it (AB) / (a + B + C) instead of a right triangle? How does (AB) / (a + B + C) equal to (a + B-C) / 2? Trouble to teach
(ab)/(a+b+c)
=[(a+b)^2-a^2-b^2]/2(a+b+c)
=[(a+b)^2-c^2]/2(a+b+c)
=(a+b+c)(a+b-c)/2(a+b+c)
=(a+b-c)/2
Formula of radius of inscribed circle and circumscribed circle of right triangle
1. The radius of inscribed circle is r = (a + B-C) / 2. The radius of circumscribed circle is r = C / 2Ab is right angle side and C is oblique edge. Firstly, we put forward a formula: Area s = 0.5 * (a + B + C) * r, R is the radius of inscribed circle. It is proved that it can be obtained by connecting the vertex and the center of the inscribed circle. Let C be the hypotenuse ∵ s = 0.5 * (a + B + C) * r = 0.5ab ᙨ r = AB / (...)
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