answer:1. **Verify the Existence of an Incircle:** By Pitot's Theorem, a quadrilateral has an incircle if and only if the sum of the lengths of its opposite sides are equal. We need to check if (AB + CD = AD + BC). [ AB + CD = 14 + 7 = 21 ] [ AD + BC = 12 + 9 = 21 ] Since (AB + CD = AD + BC), the quadrilateral (ABCD) has an incircle. 2. **Set Up the Problem:** Let the circle be tangent to (AB), (BC), (CD), and (DA) at points (P), (Q), (R), and (S), respectively. Let the radius of the circle be (r) and the distances from the vertices to the points of tangency be (x), (y), (z), and (w). 3. **Express the Side Lengths in Terms of Tangent Segments:** [ AP = AS = x, quad BP = BQ = y, quad CQ = CR = z, quad DR = DS = w ] [ AB = x + y = 14 ] [ BC = y + z = 9 ] [ CD = z + w = 7 ] [ DA = w + x = 12 ] 4. **Solve for the Tangent Segments:** Adding all the equations: [ (x + y) + (y + z) + (z + w) + (w + x) = 14 + 9 + 7 + 12 ] [ 2(x + y + z + w) = 42 ] [ x + y + z + w = 21 ] Using the individual equations: [ x + y = 14 ] [ y + z = 9 ] [ z + w = 7 ] [ w + x = 12 ] Subtracting the first equation from the sum: [ (x + y + z + w) - (x + y) = 21 - 14 ] [ z + w = 7 ] Subtracting the second equation from the sum: [ (x + y + z + w) - (y + z) = 21 - 9 ] [ x + w = 12 ] Subtracting the third equation from the sum: [ (x + y + z + w) - (z + w) = 21 - 7 ] [ x + y = 14 ] Subtracting the fourth equation from the sum: [ (x + y + z + w) - (w + x) = 21 - 12 ] [ y + z = 9 ] 5. **Calculate the Radius:** The area (A) of the quadrilateral can be found using Brahmagupta's formula for a cyclic quadrilateral: [ A = sqrt{(s-a)(s-b)(s-c)(s-d)} ] where (s) is the semiperimeter: [ s = frac{AB + BC + CD + DA}{2} = frac{14 + 9 + 7 + 12}{2} = 21 ] Substituting the side lengths: [ A = sqrt{(21-14)(21-9)(21-7)(21-12)} = sqrt{7 cdot 12 cdot 14 cdot 9} ] [ A = sqrt{7 cdot 12 cdot 14 cdot 9} = sqrt{10584} ] The radius (r) of the incircle is given by: [ r = frac{A}{s} = frac{sqrt{10584}}{21} ] Simplifying: [ r = frac{sqrt{24 cdot 441}}{21} = frac{21 sqrt{24}}{21} = sqrt{24} = 2sqrt{6} ] Therefore, the radius of the largest possible circle that fits inside or on the boundary of such a quadrilateral is (2sqrt{6}). The final answer is (boxed{2sqrt{6}})

answer:Since the set M={0, 2}, the number of proper subsets of M is: 2^2 - 1 = 3. Therefore, the correct answer is boxed{C}. If a set A has n elements, then set A has 2^n - 1 proper subsets. This question tests the method of calculating the number of proper subsets of a set, which is a basic question. When solving the problem, it is important to carefully read the question and properly apply the properties of sets.

answer:The formula for the area of an equilateral triangle is ( frac{s^2 sqrt{3}}{4} ). According to the problem, this must be twice the length of a side, which gives us: [ frac{s^2 sqrt{3}}{4} = 2s ] Solving for ( s ) involves the following steps: [ s^2 sqrt{3} = 8s ] [ s^2 = frac{8s}{sqrt{3}} ] [ s = frac{8}{sqrt{3}} ] [ s = frac{8sqrt{3}}{3} ] The perimeter of the triangle is: [ 3s = 3 left( frac{8sqrt{3}}{3} right) = 8sqrt{3} ] Thus, the perimeter of the triangle is ( boxed{8sqrt{3}} ) units.

answer:1. Calculate the radii of the circular bases: - Lower base: ( pi r^2 = 144pi ) implies ( r = 12 ) cm. - Upper base: ( pi r^2 = 36pi ) implies ( r = 6 ) cm. 2. Using the properties of similar triangles (or cones in this case), the ratio of the radii ( frac{6}{12} = frac{1}{2} ) suggests that the height of the smaller cone is ( frac{1}{2} ) of the total height of the original cone. 3. Let ( H ) be the total height of the original cone. The height of the frustum is 18 cm, which is ( frac{1}{2} ) of ( H ), so ( H = 36 ) cm. The height of the small cone is ( H - 18 ) cm: [ H - 18 = 36 - 18 = 18 text{ cm} ] Therefore, the altitude of the small cone is ( boxed{18} ) centimeters.