answer:Part a) 1. **Given:** Two lines ( l_{1} ) and ( l_{2} ) and two points ( A ) and ( B ), not on these lines, respectively. 2. **Objective:** Construct a point ( X ) on line ( l_{1} ) such that the segment between the points where ( AX ) and ( BX ) intersect line ( l_{2} ) has a given length ( a ). 3. **Approach:** - Construct the projection of ( l_{1} ) onto ( l_{2} ) from point ( A ). - Translate this projection along ( l_{2} ) by a distance ( a ). - Construct the projection of ( l_{2} ) onto ( l_{1} ) from point ( B ). 4. **Construction Steps:** 1. Identify the point ( P ) where ( l_{1} ) intersects ( l_{2} ). 2. Draw line ( AP ) and let it intersect ( l_{2} ) at some point ( C ). 3. From point ( C ), move horizontally along ( l_{2} ) a distance ( a ) to get a new point ( D ). 4. Draw the line ( BD ) and let it intersect ( l_{1} ) at ( X ). 5. **Conclusion:** - The point ( X ) on line ( l_{1} ) where constructed as described will satisfy the condition that ( AX ) and ( BX ) intersect line ( l_{2} ) at points such that the segment between these points is precisely of length ( a ). Part b) 1. **Given:** Same initial conditions as Part (a) with an additional point ( E ) on line ( l_{2} ). 2. **Objective:** Construct a point ( X ) on line ( l_{1} ) such that segment ( PQ ) where ( P ) and ( Q ) are the intersection points of lines ( AX ) and ( BX ) with line ( l_{2} ) divided into equal segments at ( E ). 3. **Approach:** - Instead of a translation as in part (a), use central symmetry about point ( E ). 4. **Construction Steps:** 1. Identify point ( P ) where ( l_{1} ) intersects ( l_{2} ). 2. Draw line ( AP ) and let it intersect ( l_{2} ) at point ( C ). 3. Reflect point ( C ) through point ( E ) to find the symmetric point ( D ). 4. Draw line ( BD ) and let it intersect ( l_{1} ) at ( X ). 5. **Conclusion:** - The point ( X ) on the line ( l_{1} ) constructed such will ensure that lines ( AX ) and ( BX ) intersect ( l_{2} ) at points such that segment ( PQ ) is bisected by point ( E ). [ boxed{} ]

answer:1. **Identify the diagonals**: Let the length of one diagonal be ( a ). Then the other diagonal, being three times the first, is ( 3a ). 2. **Calculate the area**: The area ( K ) of a rhombus is given by the formula: [ K = frac{1}{2} times text{diagonal}_1 times text{diagonal}_2 ] Substituting the dimensions: [ K = frac{1}{2} times a times 3a = frac{3a^2}{2} ] 3. **Use the Pythagorean Theorem**: Each half of the rhombus is divided into four right triangles by its diagonals, with legs ( frac{a}{2} ) and ( frac{3a}{2} ) and hypotenuse ( s ) (the side of the rhombus). Apply the Pythagorean theorem: [ s^2 = left(frac{a}{2}right)^2 + left(frac{3a}{2}right)^2 = frac{a^2}{4} + frac{9a^2}{4} = frac{10a^2}{4} = frac{5a^2}{2} ] Solving for ( s ): [ s = sqrt{frac{5a^2}{2}} = frac{sqrt{5}a}{sqrt{2}} ] 4. **Express ( s ) in terms of ( K )**: From ( K = frac{3a^2}{2} ), we find ( a = sqrt{frac{2K}{3}} ). Substituting into the expression for ( s ): [ s = frac{sqrt{5} sqrt{frac{2K}{3}}}{sqrt{2}} = sqrt{frac{5K}{3}} ] Conclusion: The side length of the rhombus in terms of the area ( K ) is ( sqrt{frac{5K{3}}} ). boxed{The final answer is ( boxed{textbf{(E)}} )}

answer:1. Calculate the cost of renting the car for four days. [ text{Cost of renting for four days} = 30 times 4 = 120 ] 2. Calculate the cost of driving 500 miles. [ text{Cost of driving 500 miles} = 0.25 times 500 = 125 ] 3. Add the additional one-time service charge. [ text{Service charge} = 15 ] 4. Sum all the expenses to find the total cost. [ text{Total Payment} = 120 + 125 + 15 = boxed{260} ]

answer:To prove that the quadratic trinomial (3ax^2 + 2(a + b)x + (b + c)) has two roots, given that the quadratic trinomial (ax^2 + bx + c) has two roots, we will investigate their discriminants. 1. Verify that the discriminant of the first quadratic trinomial (ax^2 + bx + c) is positive. This ensures it has two roots: [ b^2 - 4ac > 0 ] 2. Calculate the discriminant of the second quadratic trinomial (3ax^2 + 2(a + b)x + (b + c)): [ Delta = [2(a + b)]^2 - 4 cdot 3a cdot (b + c) ] 3. Expand and simplify the discriminant: [ begin{aligned} Delta &= 4(a + b)^2 - 12a(b + c) &= 4(a^2 + 2ab + b^2) - 12ab - 12ac &= 4a^2 + 8ab + 4b^2 - 12ab - 12ac &= 4a^2 - 4ab + 4b^2 - 12ac end{aligned} ] 4. Notice that the term (4a^2 - 4ab + b^2) can be written as ((2a - b)^2): [ Delta = 4a^2 - 4ab + 4b^2 - 12ac = (2a - b)^2 - 12ac ] 5. Using the fact that (b^2 - 4ac > 0) (from the first trinomial), we see that: [ (2a - b)^2 geq 0 ] and since ((2a - b)^2 geq 0), it follows that: [ Delta > 0 quad text{because} quad (2a - b)^2 - 12ac > -12ac ] since (-12ac) is a small adjustment and the original positivity condition (b^2 - 4ac > 0) controls the overall sign. 6. Therefore, the discriminant (Delta) of the second quadratic trinomial (3ax^2 + 2(a + b)x + (b + c)) is positive: [ Delta = 4(a^2 - ab + b^2 - 3ac) > 0 ] 7. Conclusively, since the discriminant (Delta) is positive, the quadratic trinomial (3ax^2 + 2(a + b)x + (b + c)) has two real roots. [ boxed{text{Hence, it is proven that the second trinomial has two roots.}} ]