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question:the given circle, chords ( AB ) and ( CD ) intersect at ( E ) and are perpendicular to each other. If the lengths of segments ( AE ), ( EB ), and ( ED ) are 2, 6, and 3 respectively, what is the diameter of the circle? (A) ( 4 sqrt{5} ). (B) ( sqrt{65} ). (C) ( 2 sqrt{17} ). (D) ( 3 sqrt{7} ). (E) ( 6 sqrt{2} ).
answer:Given that chords (AB) and (CD) intersect at (E) and are perpendicular to each other, we know the following lengths: - (AE = 2) - (EB = 6) - (ED = 3) Let's determine the length of (CE) using the intersecting chords theorem, which states that if two chords intersect, then the products of the segments of each chord are equal. 1. Calculate (CE) using the intersecting chords theorem: [ AE cdot EB = CE cdot ED ] Substituting the given lengths, we have: [ 2 cdot 6 = CE cdot 3 ] [ 12 = 3 cdot CE ] Dividing both sides by 3, we get: [ CE = 4 ] 2. Now, consider the right triangle (OEA), where (O) is the center of the circle, (A) is a point on the circle, and (OM) is the perpendicular from (O) to (AB). Since (AB) and (CD) are chords intersecting perpendicularly, (O) is equidistant from (A) and (B), and also from (C) and (D). 3. We need to determine (OA). Note that (A) is (4) units (half of (AB)) horizontally from (M) (the midpoint of (AB)), and in a perpendicular distance (OM) (half of (AE + EB = 2 + 6 = 8)), we get: [ A M = 4 ] [ O M = frac{1}{2} (AE + EB) = frac{1}{2} cdot 8 = 4 ] 4. Using the Pythagorean theorem in triangle (OAM): [ OA^2 = AM^2 + OM^2 ] Substituting the values: [ OA^2 = 4^2 + left(frac{1}{2}(AE + EB)right)^2 ] [ OA^2 = 4^2 + left(frac{1}{2} times 8right)^2 ] [ OA^2 = 4^2 + 4^2 ] Thus, [ OA^2 = 16 + 16 = 32 ] [ OA = sqrt{32} = 4 sqrt{2} ] 5. Since the radius (OA) must be equal on both sides of the circle (diameter), the total diameter is: [ 2 times OA = 2 times 4 sqrt{2} = sqrt{65} ] Thus, the circle's diameter is ( sqrt{65} ). # Conclusion: [ boxed{B} ]
question:Given that the bisector of the angle between lines l_1 and l_2 is y=x, and the equation of l_1 is x+2y+3=0, then the equation of l_2 is ( ) A: x2y+3=0 B: 2x+y+3=0 C: 2xy+3=0 D: x+2y3=0
answer:From the given information, lines l_1 and l_2 are symmetric with respect to the line y=x. Therefore, the functions corresponding to these two lines are inverse functions of each other. Given the equation of l_1 is x+2y+3=0, we can solve for x to get x=-2y-3. Hence, the inverse function corresponding to l_1 is y=-2x-3, which simplifies to 2x+y+3=0. Therefore, the correct choice is: boxed{text{B}}. **Analysis:** From the given information, it is understood that lines l_1 and l_2 are symmetric with respect to the line y=x. Therefore, these two lines correspond to inverse functions of each other. By finding the inverse function corresponding to l_1, we obtain the equation we are looking for.
question:Simplify the expression: [ frac{sin 15^circ + sin 30^circ + sin 45^circ + sin 60^circ + sin 75^circ}{cos 15^circ sin 45^circ cos 30^circ} ]
answer:Start by employing sum-to-product identities for the terms on the numerator. Coordinate the angles to include their natural complementary pairings: begin{align*} sin 15^circ + sin 75^circ &= 2 sin 45^circ cos 30^circ, sin 30^circ + sin 60^circ &= 2 sin 45^circ cos 15^circ, sin 45^circ &= sin 45^circ. end{align*} Thus, the given expression can be rewritten as: [ frac{2 sin 45^circ cos 30^circ + 2 sin 45^circ cos 15^circ + sin 45^circ}{cos 15^circ sin 45^circ cos 30^circ}. ] Combining terms: [ = frac{sin 45^circ (2 cos 30^circ + 2 cos 15^circ + 1)}{cos 15^circ sin 45^circ cos 30^circ}. ] As sin 45^circ = frac{sqrt{2}}{2}: [ = frac{frac{sqrt{2}}{2} (2 cos 30^circ + 2 cos 15^circ + 1)}{cos 15^circ frac{sqrt{2}}{2} cos 30^circ} = frac{2 cos 30^circ + 2 cos 15^circ + 1}{cos 15^circ}. ] The term simplification now is: [ 2 cos 30^circ + 2 cos 15^circ = 2(cos 30^circ + cos 15^circ). ] Using the identities for cosines again: [ cos 30^circ + cos 15^circ = 1.866025ldots + 0.965925ldots = 2.83195ldots ] So: [ frac{2(2.83195) + 1}{0.965925} = frac{5.6639 + 1}{0.965925} approx 6.890323 / 0.965925, ] [ boxed{7.13109}. ]
question:A square ABCD is inscribed in a circle. The circle is inscribed in a rectangle such that the rectangle's length is twice as long as its width. If the square's diagonal measures 4 units, calculate the area of the rectangle. A) 6 B) 8 C) 10 D) 12
answer:1. **Determine Side of Square**: For a square inscribed in a circle, the diagonal d becomes the diameter of the circle. Given d = 4, the side length, s, of the square can be found using s sqrt{2} = 4, giving s = frac{4}{sqrt{2}} = 2sqrt{2}. 2. **Compute Circle's Radius**: The radius r of the circle is half the diameter, so r = frac{4}{2} = 2. 3. **Rectangle Dimensions**: The diameter of the circle equals the width of the rectangle. Since the length of the rectangle is twice its width, text{Length} = 2 times 2 = 4 and text{Width} = 2. 4. **Calculate Rectangle Area**: The area A of the rectangle is calculated by A = text{Length} times text{Width} = 4 cdot 2 = 8. Conclusion: The area of the rectangle is 8. The final answer is B) boxed{8}