r = ( BC2 - BC2/AB )½ / (1+√2)
Proof">Proof">ProofIt is easy to see that the radius of the inscribed circle must touch the edges of the triangle at the hypotenuse, height, and base. Furthermore, radius lines that extend to the base (AC) and to the height (BC) of the triangle must be at right angles to one another. Finally, the radius that extends to the hypotenuse (AC) if continued outward will reach point C of the triangle, dividing the angle in half.Let the point at which the radius of the inscribed circle touches the hypotenuse (AC) of the triangle be called Z.
This divides the original triangle into two similar triangles, CBZ and ACZ.
Applying the principle of similar triangles, we know that:
(1) BC / AB = ZB / BC solving for ZB,
(2) ZB = BC2 / AB
The length of the line ZC can be found using the Pythagorean theorem:
(3) ZC = √( BC2 - ZB2 )
However, since the radius of the inscribed circle passes through ZC we can write ZC in terms of that radius. The formula below becomes more evident when you draw a picture to illustrate the problem.
(4) ZC = r ( 1 + √2 )
We can now combine (4) and (3) to solve for r.
(5) √( BC2 - ZB2 ) = r ( 1 + √2 ) solving for r yields:
(6) r = √( BC2 - ZB2 ) / (1+√2) substituting (2) for ZB yields the final formula
(7) r = ( BC2 - BC2/AB )½ / (1+√2)
The answer makes the following statement:
Finally, the radius that extends to the hypotenuse (AC) if continued outward will reach point C of the triangle, dividing the angle in half.
This is not necessarily a true statement. It is only true for a 45 degree, 45 degree, 90 degree triangle. The formula given does not give the correct answer of 2 for a 6, 8, 10 triangle.
For a formula that works on all rt. triangles:
if c is the hypotenuse and a and b are the other 2 sides
ab/a+b+c or
a+b-c/2
you should get the same answer
The radius of a circle inscribed in a regular hexagon equals the length of one side of the hexagon.
The answer is 72, i think!
Make a sketch of the situation. From a corner of the equilateral triangle draw a radius of the large circle, and from an adjacent side draw a radius of the smaller circle. You should have formed a small right-angled triangle with a known side of 10cm. and known angles of 30o, 60o and 90o. (The interior angles of an equilateral triangle are each 60o.) The hypotenuse is the unknown radius of the larger circle. But since cos 60 = 0.5, it is evident that the hypotenuse is 20cm. long.
Half of the chord, the distance of the midpoint from the center, and a radius, form a right triangle, with the radius as its hypotenuse. (4.5)2 + (6)2 = (radius)2 (20.25) + (36) = 56.25 = R2 Radius = 7.5 inches
Use the standard formula to calculate the area of the circle, then you can compare with the area given for the triangle. Note 1: The formula for a circle is: area = pi x radius squared Note 2: The radius is half the diameter
radius
Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius a in C programming
It has at least two radii, the radius of the circle going through the vertices and the radius of the inscribed circle touching all the sides.
True
There are different formula for: Height, Area, Perimeter, Angle, Length of Median Radius of inscribed circle Perimeter of inscribed circle Area of inscribed circle etc.
True
FALSE
false
yes
It is 2*r^2.
Where the side of the equilateral triangle is s and the radius of the inscribed circle is r:s = 2r * tan 30° = 48.50 cm
First you half all the sides, so 4cm, them you multiply by pi, giving the radius as 12pi, or 12.56637061