How is the radius formula for a triangle inscribed circle derived?

Let rt abc, c 90 degrees, bc a, ac b, ab c conclude that the radius of the inscribed circle r (a b c) 2

There are generally two ways to prove this:

So the quadrilateral cdoe is a square.

So cd cer

So ad b r, be a r, because ad af, ce cf, so af b r, cf a r

Because af cf ab r

So b r a r r

The radius of the inscribed circle r (a b c) 2

That is, the diameter of the inscribed circle l a b c

Apparently there are od ac, oe bc, of ab

So s abc s oac s obc s oab so ab 2 br 2 ar 2 cr 2 so r ab (a b c).

ab(a b c) (a b c)(a b c) ab(a b c) [(a b) 2 c 2] because a 2 b 2 c 2

So the radius of the inscribed circle r (a b c) 2

That is, the diameter of the inscribed circle l a b c

The formula for the radius of the inscribed circle of the triangle: r=2s (a+b+c)Derivation: Let the radius of the inscribed circle be r, the center of the circle O, and connect OA, ob and OC to obtain three triangles OAB, OBC and OAC

Then, the heights on the sides of the three triangles ab, bc, and ac are inscribed circle radius r, so: s=s abc=s oab+s obc+s oac(1 2)ab*r+(1 2)bc*r+(1 2)*ac*r(1 2)(ab+bc+ac)*r

1/2)(a+b+c)*r

So, r=2s (a+b+c)

The formula for the radius of the inscribed circle of the triangle is: r=2s (a+b+c).

Derivation: Let the radius of the inscribed circle be r, the center of the circle O, and connect OA, ob, and OC to obtain three triangles oab, obc, and oac.

Then, the heights on the sides ab, bc, and ac of these three triangles are the radius r of the inscribed circle.

So: s=s abc=s oab+s obc+s oac

1/2)ab*r+(1/2)bc*r+(1/2)*ac*r

1/2)(ab+bc+ac)*r

1/2)(a+b+c)*r

So, r=2s (a+b+c)

Scalloped inscribed circles.

The circle tangent to the arc AB and the two radii OA and OB of the sector AOB is called the inscribed circle of the fan.

The center of the inscribed circle o is on the bisector of the angle of the central angle AOB of the sector oo = r-r (r is the radius of the sector and r is the radius of the inscribed circle).

Over o as o a oa, perpendicular foot a, right triangle oao, o oa=30°, o a=r, oo =r-r.

r=(r-r)*sin30°，r=1/2(r-r)，r=3r。

Inscribed circle area = r 2.

Formula derivation. First draw a triangle and the inner circle of the triangle, connect the center of the circle and the three vertices of the triangle respectively (at this time the visible triangle is divided into three triangles), bury the banquet and then connect the center of the circle and the three tangent points respectively (at this time the visible triangle is divided into six small triangles), these three line segments can be obtained and the three sides of the triangle are perpendicular to a, b and c respectively, then the triangle area can be found by three small triangles, that is, a*r 2+b*r 2+c*r 2=(a+b+c)*r 2=s

So, r=2s (a+b+c)

Right triangle: The radius of the inscribed circle is r=(a+b-c) 2 (a, b is the right-angled side, c is the hypotenuse).

General triangle: The radius of the inscribed circle is r=2s (a+b+c), and s is the area formula of the triangle.

The circle tangent to all three sides of the triangle is called the inscribed circle of the triangle, the center of the circle is called the inner part of the triangle, and the resistant triangle is called the inscribed triangle of the circle. The heart of the triangle is the intersection of the bisector of the three corners of the triangle.

The formula for the radius of the inscribed circle of a right triangle: r=(a+b-c) 2.

The circle tangent to all three sides of the triangle is called the inscribed circle of the triangle, the center of the circle is called the inner part of the triangle, the triangle is called the inscribed triangle of the circle, and the inner part of the triangle is the intersection of the bisector of the three corners of the triangle.

The inscribed circle of ABC is A'b'c'circumscribed circle. And a'a、b'b and c'C The intersection of the three lines at one point is the Lemoine Point (or Gergonne Point), or a similar center of gravity, i.e., the intersection of three similar middle lines. The inscribed circle is tangent to the nine-point circle, and the tangent point is called the Feuerbach point (see Nine-point circle).

If the inscribed circle of the triangle is used as the inversion circle, the three sides of the triangle and the circumscribed circle become four circles of equal radius (the radius is equal to half of the radius of the inscribed circle).

The formula for the radius of the inscribed circle of a triangle is: r=(a+b-c) 2. A circle tangent to all sides of a polygon is called an inscribed circle of a polygon.

In particular, a circle tangent to all three sides of a triangle is called the inscribed circle of the triangle, the center of the circle is called the inner part of the triangle, and the triangle is called the inscribed triangle of the circle. The heart of the triangle is the intersection of the bisector of the three corners of the triangle.

A triangle is a closed figure composed of three line segments in the same plane that are not on the same straight line, which are connected sequentially and have applications in mathematics and architecture. Common triangles are divided into ordinary triangles (the three sides are not equal) and isosceles triangles (isosceles triangles with unequal waists and bases, and isosceles triangles with equal waists and bottoms, that is, equilateral triangles); According to the angle, there are right triangles, acute triangles, obtuse triangles, etc., of which acute triangles and obtuse triangles are collectively referred to as oblique triangles.

Let the three sides of ABC be A, B, and C, the area is S, and the radius of the inscribed circle is R, then:

1/2ar＋1/2br＋1/2cr＝s

r＝2s/(a＋b＋c)

This is the formula for calculating the radius of the inscribed circle in a triangle, i.e. the radius of the inscribed circle in the triangle is equal to 2 times the area divided by the perimeter.

In a triangle, the intersection of the angular bisector of the three corners is the center of the inscribed circle, and the perpendicular segments from the center of the circle to each side of the triangle are equal.

Right triangle: The radius of the inscribed circle is r=(a+b-c) 2 (a, b is the right-angled side, c is the hypotenuse).

or r=ab (a+b+c) (a, b is the right-angled edge, c is the hypotenuse) <>

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The radius of the inscribed circle of any triangle: s=1 2lr (s represents the area of the triangle, l represents the perimeter of the triangle).

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The formula for the radius of the inscribed circle of a right triangle: r=(a+b-c) 2.

Let RT ABC, C 90 degrees, BC A, AC B, AB C

The conclusion is that the radius of the cutting circle of the inner mill is r (a b c) 2

There are generally two ways to prove this:

Let the center of the inscribed circle be O, and the three tangent points are D, E, and F, and connect OD and OE

Apparently there is od ac, oe bc, od oe so the quadrilateral cdoe is a square.

So cd ce r so ad b r, be a r, celery.

Because ad af, ce cf so af b r, cf a r

Because af cf ab r so b r a r r inscribed circle radius r (a b c) 2

That is, the diameter of the inscribed circle l a b c

Meaning. Right triangle: divided into two cases, there are ordinary right triangles, and isosceles right triangles (special cases) In the right angle triangle shape, the two sides adjacent to the right angle are called right angles, and the sides opposite the right angles are called hypotenuses.

The sides of a right triangle are also called "chords". If the two right-angled sides are not the same length, the short side is called the "hook" and the long side is called the "strand".