answer:- **Midpoint Calculation**: D is the midpoint of BC. Hence, CD = DB = BC/2 = 40/2 = 20. - **Triangle Properties**: In this scenario, triangles formed by the perpendicular bisector and the sides of ABC will involve a 45^circ-45^circ-90^circ triangle. Since CD is a leg of this right triangle, and DE is the other leg, both are equal. - **Calculate DE**: Since angle CDE = 45^circ and angle DCE = 90^circ, DE = CD = 20. Thus, the length of DE is boxed{20}.

answer:Let s = x + y and p = xy. The first equation can be rewritten as s + p = 119 and the second equation can be written as [ x^2y + xy^2 = (x+y)xy = sp = 1870. ] Thus, s and p are the roots of the quadratic equation [ t^2 - 119t + 1870 = 0. ] Factoring the quadratic, we have: [ t^2 - 119t + 1870 = (t-34)(t-85) = 0. ] Thus, s and p are 34 and 85, respectively, in some order. Given x < y, let's evaluate both cases: 1. s = 34 and p = 85, then [ x^2+y^2 = (x+y)^2 - 2xy = s^2 - 2p = 34^2 - 2 cdot 85 = 1156 - 170 = 986. ] 2. s = 85 and p = 34 (this set does not satisfy x < y since x+y would be greater and not result in a valid pair given xy = 34). Hence, the valid scenario is when s = 34 and p = 85. The solution is: [ boxed{986} ]

answer:1. **Introduction**: Let us consider a convex polyhedron and a line segment inside it. We aim to prove that the length of any segment inside the polyhedron does not exceed the length of the greatest segment between two vertices of the polyhedron. 2. **Useful Planimetric Statement**: We start by using the following geometric assertion: Given a triangle ( triangle ABC ), if point ( X ) lies on the side ( BC ), then either ( AB geq AX ) or ( AC geq AX ). 3. **Fundamental Concept**: For point ( X ) lying on ( BC ): - Either ( angle BXA geq 90^circ ) or ( angle CAX geq 90^circ ). - If ( angle BXA geq 90^circ ), then ( AB geq AX ). - If ( angle CAX geq 90^circ ), then ( AC geq AX ). 4. **Extension of the Segment**: Let’s extend the given internal segment to intersect the faces of the polyhedron at points ( P ) and ( Q ). By extending, the length of the segment can only increase. 5. **Intermediate Analysis**: Let ( MN ) be any segment with endpoints on the edges of the polyhedron intersecting at ( P ): - We know that either ( MQ geq PQ ) or ( NQ geq PQ ). 6. **Vertex Segment Replacement**: For the given condition, assume without loss of generality that ( MQ geq PQ ). Points ( M ) and ( Q ) lie on different edges, say ( AB ): - Either ( AQ geq MQ ) or ( BQ geq MQ ). 7. **Finding a Greater Segment**: By iterative substitution: - Replace ( PQ ) with a greater segment, one of whose endpoints lies on a vertex of the polyhedron. - Apply similar reasoning for subsequent segments until segment endpoints are vertices of the polyhedron. 8. **Conclusion**: Through the iterative replacement and comparison steps, we eventually replace the initial segment with a segment whose endpoints are vertices of the polyhedron. This maximal segment is necessarily between two vertices of the polyhedron, proving the statement. [ boxed{text{The length of the segment inside the polyhedron cannot exceed the length of the greatest segment between two vertices of the polyhedron.}} ]

answer:To find the total combined weight of the three batches of dried grapes after the drying process, we need to calculate the weight of the non-water content (solid content) in each batch of fresh grapes and then determine what that weight would be when the water content is reduced to the specified percentages in the dried grapes. Let's start with Batch A: Batch A has 70% water, so it has 30% solid content. The weight of the solid content in Batch A is 30% of 100 kg = 0.30 * 100 kg = 30 kg. When dried, Batch A will have 20% water content, which means it will have 80% solid content. Since the solid content doesn't change, the dried weight of Batch A will be: Let the dried weight of Batch A be ( x ) kg, then ( 0.80 * x = 30 kg ) ( x = 30 kg / 0.80 ) ( x = 37.5 kg ) Now, let's calculate for Batch B: Batch B has 75% water, so it has 25% solid content. The weight of the solid content in Batch B is 25% of 150 kg = 0.25 * 150 kg = 37.5 kg. When dried, Batch B will have 25% water content, which means it will have 75% solid content. The dried weight of Batch B will be: Let the dried weight of Batch B be ( y ) kg, then ( 0.75 * y = 37.5 kg ) ( y = 37.5 kg / 0.75 ) ( y = 50 kg ) Finally, let's calculate for Batch C: Batch C has 80% water, so it has 20% solid content. The weight of the solid content in Batch C is 20% of 200 kg = 0.20 * 200 kg = 40 kg. When dried, Batch C will have 30% water content, which means it will have 70% solid content. The dried weight of Batch C will be: Let the dried weight of Batch C be ( z ) kg, then ( 0.70 * z = 40 kg ) ( z = 40 kg / 0.70 ) ( z = 57.14 kg ) Now, we add the weights of the dried grapes from all three batches to get the total combined weight: Total dried weight = ( 37.5 kg + 50 kg + 57.14 kg ) Total dried weight = ( 144.64 kg ) Therefore, the total combined weight of the three batches of dried grapes after the drying process is ( boxed{144.64} kg ).