Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.
Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.
And so the area of the inscribed triangle is 12 cm2.
Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.
Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.
And so the area of the inscribed triangle is 12 cm2.
Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.
Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.
And so the area of the inscribed triangle is 12 cm2.
Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.
Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.
And so the area of the inscribed triangle is 12 cm2.
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
radius
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
This is true. The answer is obvious if you think about it the following way: an equilateral triangle has three equal sides, and every point on the circumference of a circle is the same distance from the center of the circle. Therefore, it is safe to assume that the circle will touch the midpoint of each side of the triangle. It also means that the center of the circle will be in the center of the triangle. Therefore, the radius of the circle will travel from the center of the triangle to the midpoint of one of the sides. This will cover the distance of half the triangle's median.
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
radius
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
This is true. The answer is obvious if you think about it the following way: an equilateral triangle has three equal sides, and every point on the circumference of a circle is the same distance from the center of the circle. Therefore, it is safe to assume that the circle will touch the midpoint of each side of the triangle. It also means that the center of the circle will be in the center of the triangle. Therefore, the radius of the circle will travel from the center of the triangle to the midpoint of one of the sides. This will cover the distance of half the triangle's median.
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.
FALSE
True
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.
false
The answer in 6.... draw an angular bisector from one of the angles to the centre of circle then draw a perpendicular from the centre of circle. Those to lines will form a triangle... use trigonometry and find the length of the perpendicular, which is also a radius... double the radius and u will get the diagonal for the square... using formula :- (Side)^2 + (Side)^2 = (Diagonal)^2, find the side of square and square the answer, which will give you your final answer
You cannot show it in general since it need not be true!