answer:1. Consider the pyramid (SABC) which is a regular triangular pyramid with - (SA = l), - (SO perp (ABC)), - and the angle between the side edge (SA) and the base plane (ABC) is (alpha). 2. In the right triangle (triangle SAO), we identify the following relationships: - The height from the apex (S) to the base plane (ABC) is (SO). - We can use trigonometric functions to express (SO) and (AO): [ SO = l sin alpha, ] [ AO = l cos alpha. ] 3. Since (A) and (B) are adjacent vertices of the equilateral triangle base (ABC), the length (AB) can be found using the relationship in an equilateral triangle: [ AB = AO sqrt{3} = l cos alpha cdot sqrt{3} = l sqrt{3} cos alpha. ] 4. Calculate the area of the equilateral triangle base (ABC). Using the formula for the area of an equilateral triangle ( frac{sqrt{3}}{4} s^2 ) where ( s ) is the side length: [ S_{triangle ABC} = frac{(AB)^2 sqrt{3}}{4} = frac{(l sqrt{3} cos alpha)^2 sqrt{3}}{4} = frac{3 l^2 cos^2 alpha cdot sqrt{3}}{4} = frac{3 sqrt{3} l^2 cos^2 alpha}{4}. ] 5. The volume (V) of the pyramid (SABC) is given by: [ V = frac{1}{3} cdot text{Base Area} cdot text{Height}, ] where the base area is (S_{triangle ABC}) and the height is (SO): [ V = frac{1}{3} cdot S_{triangle ABC} cdot SO = frac{1}{3} cdot frac{3 sqrt{3} l^2 cos^2 alpha}{4} cdot l sin alpha. ] Simplifying this expression, [ V = frac{1}{3} cdot frac{3 sqrt{3} l^2 cos^2 alpha}{4} cdot l sin alpha = frac{l^3 sqrt{3} cos^2 alpha sin alpha}{4}. ] 6. Recognize that ( sin 2alpha = 2 sin alpha cos alpha ), thus our expression for volume can be rewritten as: [ V = frac{l^3 sqrt{3} cos^2 alpha sin alpha}{4} = frac{l^3 sqrt{3} cos alpha}{4} cdot cos alpha sin alpha = frac{l^3 sqrt{3} cos alpha cdot sin 2 alpha}{8}. ] # Conclusion: The volume of the pyramid is: [ boxed{frac{l^3 sqrt{3} sin 2 alpha cos alpha}{8}} ]

answer:To solve this, we notice that the configuration of the smaller circles changes slightly from the original setup. With three smaller circles aligned in both the east-to-west and north-to-south directions, the diameter of the larger circle needs to accommodate the diameters of three smaller circles. 1. Since three smaller circles fit exactly along the diameter of the larger circle, their combined diameter (which is three times the diameter of one smaller circle) equals the diameter of the larger circle. 2. Therefore, the diameter of the larger circle is 2 times 10 meters = 20 meters (since the diameter is twice the radius). 3. Thus, the diameter of one smaller circle is frac{20}{3} meters. 4. Hence, the radius of one smaller circle is half of its diameter, which equals frac{20}{3 times 2} = frac{10}{3} meters. Therefore, the radius of each of the six smaller circles is boxed{frac{10}{3}} meters.

answer:1. Given: Point (C) divides the diameter (AB) in the ratio (AC:BC = 2:1). Point (P) is selected on the circle. 2. Observe that (angle APB = 90^circ) because it subtends the diameter (AB) (angle subtended by a diameter in a semicircle is a right angle, known as the Thales' theorem). 3. Drop a perpendicular from (C) to (AP) and let (H) be the foot of this perpendicular. Notice that (CH) is parallel to (PB) since both are perpendicular to (AP). - Because (CH parallel PB), corresponding angles (angle ACH) and (angle APB) are equal. 4. We now focus on the similar triangles (triangle APB) and (triangle AHC): [ triangle APB sim triangle AHC ] by the AA (angle-angle) similarity criterion since (angle APB = angle AHC = 90^circ) and (angle PBA = angle CHA) (corresponding angles). 5. Determine the similarity ratio of these triangles. Since (AC:BC = 2:1), we can analyze the proportionality in the triangles: [ frac{AC}{AB} = frac{2}{3} ] Using this ratio, the side lengths of (triangle AHC) and (triangle APB) are proportional by (frac{2}{3}). 6. Calculate the tangent values of the angles: [ text{Let } tan angle PAC = k implies left(text{Let } text{height per length of } PA text{ be } kright)text{ (since the tangent of an angle is the opposite side over the adjacent side)} ] Since [ triangle APB propto triangle AHC text{ with ratio } frac{2}{3} ] [ Rightarrow tan angle APC = frac{HC}{PC} ] Given [ PC = 2PB ;;text{(since } AB = AC + CB text{ and ratio}, 2:1 implies BC text{ is half of } AB) ] [ Rightarrow tan angle APC = frac{2PC}{AP} ] Therefore, [ tan angle APC =2 cdot tan angle PAC ] 7. We need to express the ratio (frac{tan angle PAC}{tan angle APC}): [ frac{tan angle PAC}{tan angle APC} = frac{k}{2k} = frac{1}{2} ] # Conclusion: The minimum value for the ratio (frac{tan angle PAC}{tan angle APC}) is (boxed{frac{1}{2}}).

answer:Let's break down the given problem into its subparts and provide a detailed solution for each. Part (a) 1. **Calculate the area of the square:** The area of a square with side length 10 , text{cm} is given by [ text{Area of square} = 10^2 = 100 , text{cm}^2. ] 2. **Calculate the area of one isosceles triangle:** Since it's given that the triangle is isosceles with two sides equal to the length of the square's side, both sides of the triangle are also 10 , text{cm}. The base of the triangle is another 10 , text{cm}. Therefore, we can use the formula: [ text{Area of triangle} = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 10 , text{cm} times 10 , text{cm} = 50 , text{cm}^2. ] 3. **Calculate the visible area when one triangle is folded over the square:** When one triangle is folded over the square, the visible area of the square becomes: [ text{Visible area} = text{Area of square} - text{Area of one triangle} = 100 , text{cm}^2 - 50 , text{cm}^2 = 50 , text{cm}^2. ] Thus, the visible area when one triangle is folded over is: [ boxed{50 , text{cm}^2} ] Part (b) 1. **Dividing the figure into 8 triangles of the same area:** To simplify the problem, we can draw a segment connecting the midpoints of the vertical sides of the square. This divides the square into 8 smaller triangles of the same area. Each of these triangles is one-eighth of the total area of the square. The area of each smaller triangle is: [ frac{text{Total area of square}}{text{Number of triangles}} = frac{100 , text{cm}^2}{8} = 12.5 , text{cm}^2. ] 2. **Identify the triangles that remain visible:** When both isosceles triangles are folded over the square, they cover 6 out of the 8 smaller triangles (3 from each side), leaving 2 out of the 8 smaller triangles uncovered. The area of the two visible triangles is: [ text{Visible area} = 2 times 12.5 , text{cm}^2 = 25 , text{cm}^2. ] Thus, the visible area when both triangles are folded over is: [ boxed{25 , text{cm}^2} ]