answer:(1) Given the eccentricity of the ellipse is frac{sqrt{2}}{2}, we have: frac{c}{a} = frac{sqrt{2}}{2}, which implies a=sqrt{2}c. Also, 2a = |AF_{1}|+|AF_{2}|=4sqrt{2}, which gives us a=2sqrt{2} and c=2. Then, b^{2}=a^{2}-c^{2}=4, so the equation of the ellipse is: boxed{frac{x^{2}}{8} + frac{y^{2}}{4} = 1}; (2) Let the equation of line AB be y=kx+m. Let A be (x_{1}, y_{1}), and B be (x_{2}, y_{2}). By substituting y=kx+m into the ellipse equation, we obtain: (1+2k^{2})x^{2}+4kmx+2m^{2}-8=0 The discriminant Delta must be positive since there are two intersection points: Delta = (4km)^{2} - 4(1+2k^{2})(2m^{2}-8) = 8(8k^{2}-m^{2}+4) > 0 Now, find x_{1}+x_{2} and x_{1}x_{2} using the quadratic equation: x_{1}+x_{2} = frac{-4km}{1+2k^{2}}, quad x_{1}x_{2} = frac{2m^{2}-8}{1+2k^{2}} Given that k_{OA} cdot k_{OB} = -frac{b^{2}}{a^{2}} = -frac{1}{2}, it follows that: frac{y_{1}y_{2}}{x_{1}x_{2}} = -frac{1}{2} Now, we have: y_{1}y_{2} = -frac{1}{2}x_{1}x_{2} = -frac{1}{2}frac{2m^{2}-8}{1+2k^{2}} = -frac{m^{2}-4}{1+2k^{2}} Also, y_{1}y_{2} = (kx_{1}+m)(kx_{2}+m) = k^{2}x_{1}x_{2} + km(x_{1}+x_{2}) + m^{2} Substituting the expressions for x_{1}+x_{2} and x_{1}x_{2}: -frac{m^{2}-4}{1+2k^{2}} = k^{2}frac{2m^{2}-8}{1+2k^{2}} + kmfrac{-4km}{1+2k^{2}} + m^{2} Simplifying, we obtain: 4k^{2} + 2 = m^{2} Let d be the distance between the origin and the line AB. The area of triangle OAB can be calculated as follows: S_{triangle OAB} = frac{1}{2}|AB|cdot d = frac{1}{2} sqrt{1+k^{2}}cdot |x_{2}-x_{1}|cdot frac{|m|}{sqrt{1+k^{2}}} Substituting the expressions for x_{1}+x_{2} and x_{1}x_{2}: S_{triangle OAB} = frac{|m|}{2} sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}} = frac{|m|}{2} sqrt{left(frac{-4km}{1+2k^{2}}right)^{2}-4frac{2m^{2}-8}{1+2k^{2}}} Simplifying, we get: S_{triangle OAB} = 2sqrt{4k^{2}-m^{2}+4} = 2sqrt{2} When the slope of the line does not exist, we have A(2,sqrt{2}), B(2,-sqrt{2}), and d=2. Therefore, the area of triangle OAB is: S_{triangle OAB} = frac{1}{2} cdot 2 cdot 2sqrt{2} = 2sqrt{2} Thus, the area of triangle OAB is constant and equal to boxed{2sqrt{2}}.

answer:- **No 1times1, 2times2, or 3times3 squares contain seven black squares**. - **Consider 4times4 squares**: These can have a maximum of 8 black squares (every alternate square in a 4x4 grid being black). However, not every position of the 4x4 square will guarantee 7 black squares. The upper left corner of such a square must be black, so we can choose the upper left corner of a 4times4 square in 7cdot7=49 ways (since the black squares are in a checker pattern, half of these will be valid). This gives 49/2 = 24.5, but since we can't have half a square, check the layout manually to verify 25 valid positions. - **For 5times5 and larger squares**: All will contain at least 7 black squares due to the increasing area. - 5times5 squares: Positioned in 6cdot6=36 ways. - 6times6 squares: Positioned in 5cdot5=25 ways. - 7times7 squares: Positioned in 4cdot4=16 ways. - 8times8 squares: Positioned in 3cdot3=9 ways. - 9times9 squares: Positioned in 2cdot2=4 ways. - 10times10 square: Positioned in 1cdot1=1 way. - **Summing up**: 25+36+25+16+9+4+1=116. Conclusion: There are a total of boxed{116} squares containing at least 7 black squares on the 10x10 checkerboard.

answer:To find the average age of the school, we need to find the total age of all the students and then divide it by the total number of students. First, let's find the number of boys in the school. We know the total number of students and the number of girls, so we can subtract the number of girls from the total number of students to find the number of boys: Number of boys = Total number of students - Number of girls Number of boys = 632 - 158 Number of boys = 474 Now, let's find the total age of the boys and the total age of the girls: Total age of boys = Average age of boys * Number of boys Total age of boys = 12 years * 474 Total age of boys = 5688 years Total age of girls = Average age of girls * Number of girls Total age of girls = 11 years * 158 Total age of girls = 1738 years Now, let's find the total age of all the students: Total age of all students = Total age of boys + Total age of girls Total age of all students = 5688 years + 1738 years Total age of all students = 7426 years Finally, let's find the average age of the school: Average age of the school = Total age of all students / Total number of students Average age of the school = 7426 years / 632 students Average age of the school ≈ 11.75 years Therefore, the average age of the school is approximately boxed{11.75} years.

answer:1. Assume that triangle (ABC) is constructed where (AM) is the given median, (AH) and (BB_1) are the given heights. 2. We start by dropping a perpendicular (MD) from the midpoint (M) of side (BC) to side (AC). This step ensures that (MD) is perpendicular to (AC). 3. Notice that since (M) is the midpoint of (BC), (M D = frac{1}{2} B B_1), because (BB_1) is the height from vertex (B) to side (AC). 4. Next, consider the triangles (AMD) and (AMH). Both triangles can be constructed independently as follows: - Construct triangle (AMD) using points (A), (M), and the calculated point (D). - Construct triangle (AMH) using points (A), (M), and the given height (H) from (A) to (BC). Here, (D) and (H) are locations such that (MD perp AC) and (AH perp BC), respectively. 5. With (D) and (H) identified: - The vertex (C) is determined as the intersection of the perpendiculars from (A) and (M): (AD) (where (D) is the foot of the perpendicular from (M) on (AC)) and (MH). 6. To locate point (B): - Extend line (CM) and construct cut-off segment (CB = 2 cdot CM). The location of (B) can either be such that it forms triangle (AMD) on one side of line (AM) or on the opposite side. 7. The task provides two solutions: - One where triangle (AMD) and (AMH) are on one side of the median (AM). - The other where triangles (AMD) and (AMH) are on opposite sides of the line (AM). Thus, we have symmetry on either side of the median. In both configurations, the final structure of the triangle depends on the accurate placement of the given heights and the lengths deduced through perpendicularity conditions. [ boxed{text{Successful construction of the triangle}} ]