answer：Okay, so I have this problem where there's a sequence {a_n}, and the sum of the first n terms is given by S_n = n³. I need to find the value of a₅ + a₆. Hmm, let me think about how to approach this.First, I remember that the sum of the first n terms of a sequence is related to the nth term. Specifically, the nth term a_n can be found by subtracting the sum of the first (n-1) terms from the sum of the first n terms. So, in formula terms, that would be:a_n = S_n - S_{n-1}Yeah, that makes sense because S_n is the sum up to n, and S_{n-1} is the sum up to n-1, so their difference should give the nth term.So, given that S_n = n³, I can write:a_n = n³ - (n - 1)³Alright, let me compute that. Let's expand (n - 1)³ first.(n - 1)³ = n³ - 3n² + 3n - 1So, substituting back into the expression for a_n:a_n = n³ - (n³ - 3n² + 3n - 1) = n³ - n³ + 3n² - 3n + 1 = 3n² - 3n + 1Okay, so the general term a_n is 3n² - 3n + 1. That seems right. Let me check with n=1 to see if it makes sense.For n=1, a₁ should be S₁, which is 1³ = 1. Plugging into the formula:a₁ = 3(1)² - 3(1) + 1 = 3 - 3 + 1 = 1. Perfect, that works.Let me check n=2 as well. S₂ should be 8, so a₂ = S₂ - S₁ = 8 - 1 = 7. Using the formula:a₂ = 3(2)² - 3(2) + 1 = 12 - 6 + 1 = 7. That also checks out.Good, so the formula seems correct.Now, I need to find a₅ + a₆. So, let's compute a₅ and a₆ separately using the formula.First, a₅:a₅ = 3(5)² - 3(5) + 1 = 3(25) - 15 + 1 = 75 - 15 + 1 = 61Wait, hold on, 75 - 15 is 60, plus 1 is 61. Hmm, okay.Now, a₆:a₆ = 3(6)² - 3(6) + 1 = 3(36) - 18 + 1 = 108 - 18 + 1 = 91Wait, 108 - 18 is 90, plus 1 is 91. So, a₆ is 91.Therefore, a₅ + a₆ = 61 + 91 = 152.Wait, let me double-check my calculations because 61 + 91 is 152, but let me make sure I didn't make a mistake in computing a₅ and a₆.Starting with a₅:3*(5)^2 = 3*25 = 75 3*5 = 15 So, 75 - 15 + 1 = 61. That seems correct.a₆:3*(6)^2 = 3*36 = 108 3*6 = 18 So, 108 - 18 + 1 = 91. That also seems correct.Adding them together: 61 + 91. Let's compute that again. 60 + 90 is 150, and 1 + 1 is 2, so total is 152. Hmm, okay, that seems right.Alternatively, maybe I can compute S₆ and S₄, and then find a₅ + a₆ as S₆ - S₄. Let me try that method to cross-verify.Because S₆ is the sum up to 6 terms, and S₄ is the sum up to 4 terms. So, S₆ - S₄ would be a₅ + a₆.Given that S_n = n³, so S₆ = 6³ = 216, and S₄ = 4³ = 64.Therefore, S₆ - S₄ = 216 - 64 = 152. So, that's the same result as before.Great, so both methods give me 152. That gives me more confidence that the answer is correct.Just to recap:1. I found the general term a_n by subtracting S_{n-1} from S_n.2. Expanded and simplified to get a_n = 3n² - 3n + 1.3. Calculated a₅ and a₆ individually, got 61 and 91, added them to get 152.4. Cross-verified by computing S₆ - S₄, which is 216 - 64 = 152.So, both methods lead to the same answer, which is reassuring. I think 152 is the correct value for a₅ + a₆.**Final Answer**The value of (a_5 + a_6) is boxed{152}.

answer：To determine the height of the table, let's define the following variables:- Let ( T ) represent the height of the table.- Let ( C ) represent the height of the can.- Let ( B ) represent the height of the bottle.From the problem, we have two key pieces of information:1. The distance from the top of the can on the floor to the top of the bottle on the table is 150 cm. This can be expressed as: [ T + C = 150 ]2. The distance from the top of the bottle on the floor to the top of the can on the table is 110 cm. This can be expressed as: [ T + B = 110 ]To find the height of the table, we can add these two equations together:[(T + C) + (T + B) = 150 + 110][2T + (C + B) = 260]Since the sum of the heights of the can and the bottle (( C + B )) is not provided, we need to find another way to determine ( T ). Observing the two original equations:[T + C = 150 quad text{and} quad T + B = 110]We can subtract the second equation from the first to eliminate ( T ):[(T + C) - (T + B) = 150 - 110][C - B = 40]This tells us that the can is 40 cm taller than the bottle. However, without additional information about the individual heights of the can or the bottle, we cannot determine the exact value of ( T ) from the given equations alone.Given the answer choices, we can infer that the height of the table is likely one of the provided options, and based on the context, the most reasonable choice is 130 cm.

answer：Okay, so I have this geometry problem here, and I need to figure out if the sum of the areas of triangles NDL and AKL is half the area of triangle NBA. Hmm, let me try to visualize the problem first.We have triangle NBA. On side NA, points Q and F are marked such that NQ and FA are each equal to NA divided by 4. So, NA is divided into four equal parts. That means NQ = QF = FA = NA/4. Wait, no, actually, the problem says NQ = FA = NA/4. So, NQ is NA/4, FA is NA/4, which means QF must be NA - NQ - FA = NA - NA/4 - NA/4 = NA/2. So, QF is half of NA.So, NA is divided into three segments: NQ = NA/4, QF = NA/2, and FA = NA/4. Got it. So, Q is closer to N, and F is closer to A, each at a quarter distance from N and A respectively.Now, a point L is chosen on segment QF. So, L is somewhere between Q and F on that middle half of NA.Then, lines are drawn through points Q and F, parallel to BL. These lines intersect sides NB and AB at points D and K respectively. So, through Q, draw a line parallel to BL, which hits NB at D. Through F, draw a line parallel to BL, which hits AB at K.We need to determine if the sum of the areas of triangles NDL and AKL is half the area of triangle NBA.Alright, let me try to sketch this in my mind. Triangle NBA, with NA as the base. Points Q and F divide NA into four equal parts. Point L is somewhere on QF. Then, from Q and F, we draw lines parallel to BL, intersecting NB at D and AB at K.I think coordinate geometry might be a good approach here. Let me assign coordinates to the triangle to make calculations easier.Let’s place point N at (0, 0), point B at (0, b), and point A at (a, 0). So, triangle NBA has vertices at N(0,0), B(0,b), and A(a,0). Then, side NA is from (0,0) to (a,0). Points Q and F are on NA such that NQ = FA = NA/4.Since NA is from (0,0) to (a,0), its length is 'a'. So, NQ = a/4, which means Q is at (a/4, 0). Similarly, FA = a/4, so since A is at (a,0), F is at (a - a/4, 0) = (3a/4, 0). Therefore, Q is at (a/4, 0), F is at (3a/4, 0), and L is somewhere on QF, which is from (a/4, 0) to (3a/4, 0). Let's denote point L as (l, 0), where a/4 ≤ l ≤ 3a/4.Now, point B is at (0, b). So, BL is the line from B(0, b) to L(l, 0). Let me find the equation of BL. The slope of BL is (0 - b)/(l - 0) = -b/l. So, the equation is y = (-b/l)x + b.Now, we need to draw lines through Q and F parallel to BL. Since they are parallel, they will have the same slope, which is -b/l.First, let's find point D, which is the intersection of the line through Q parallel to BL with side NB.Side NB is from N(0,0) to B(0, b), which is the y-axis. So, the line through Q(a/4, 0) with slope -b/l will intersect NB at D. Let me find the equation of this line.The line through Q(a/4, 0) with slope -b/l is y - 0 = (-b/l)(x - a/4). So, y = (-b/l)x + (b a)/(4 l).This line intersects NB at x=0. So, substituting x=0, y = (b a)/(4 l). Therefore, point D is at (0, (b a)/(4 l)).Similarly, let's find point K, which is the intersection of the line through F parallel to BL with side AB.Side AB is from A(a, 0) to B(0, b). Let me find the equation of AB. The slope is (b - 0)/(0 - a) = -b/a. So, the equation is y = (-b/a)(x - a) = (-b/a)x + b.The line through F(3a/4, 0) with slope -b/l is y - 0 = (-b/l)(x - 3a/4). So, y = (-b/l)x + (3 a b)/(4 l).This line intersects AB at some point K. To find K, set the equations equal:(-b/l)x + (3 a b)/(4 l) = (-b/a)x + b.Let me solve for x.Multiply both sides by l a to eliminate denominators:(-b a)x + (3 a^2 b)/4 = (-b l)x + b l a.Bring all terms to one side:(-b a)x + (3 a^2 b)/4 + b l x - b l a = 0.Factor x:x(-b a + b l) + (3 a^2 b)/4 - b l a = 0.Factor b:b [x(-a + l) + (3 a^2)/4 - l a] = 0.Since b ≠ 0, we have:x(-a + l) + (3 a^2)/4 - l a = 0.Let me solve for x:x(-a + l) = l a - (3 a^2)/4.So,x = [l a - (3 a^2)/4] / (-a + l).Simplify numerator:= [a(l - 3a/4)] / (l - a).Factor a:= a(l - 3a/4) / (l - a).Hmm, let me write it as:x = [a(l - 3a/4)] / (l - a).Similarly, let's compute y-coordinate of K using the equation of AB: y = (-b/a)x + b.So, y = (-b/a) * [a(l - 3a/4)/(l - a)] + b.Simplify:= (-b) * (l - 3a/4)/(l - a) + b.= [ -b(l - 3a/4) + b(l - a) ] / (l - a).Wait, let me compute step by step:First term: (-b/a) * [a(l - 3a/4)/(l - a)] = (-b)(l - 3a/4)/(l - a).Second term: + b.So, y = (-b)(l - 3a/4)/(l - a) + b.Combine terms:= [ -b(l - 3a/4) + b(l - a) ] / (l - a).Wait, no, that's not quite right. Let me write it as:y = [ -b(l - 3a/4) + b(l - a) ] / (l - a).Wait, actually, no. Let me factor b:y = b [ - (l - 3a/4)/(l - a) + 1 ].= b [ (- (l - 3a/4) + (l - a)) / (l - a) ].Simplify numerator:- l + 3a/4 + l - a = ( - l + l ) + (3a/4 - a ) = 0 - a/4 = -a/4.So, y = b [ (-a/4) / (l - a) ] = b [ (-a/4) / (l - a) ].Simplify:= ( -a b / 4 ) / (l - a ) = (a b / 4 ) / (a - l ).So, point K is at ( x, y ) where x = [a(l - 3a/4)] / (l - a ) and y = (a b / 4 ) / (a - l ).Wait, let me check the x-coordinate again.x = [a(l - 3a/4)] / (l - a ) = [a(l - 3a/4)] / (l - a ) = [a(l - 3a/4)] / (l - a ).Alternatively, factor out a negative:= [a(l - 3a/4)] / -(a - l ) = - [a(l - 3a/4)] / (a - l ).Hmm, maybe it's better to leave it as is for now.So, point K is at ( [a(l - 3a/4)] / (l - a ), (a b / 4 ) / (a - l ) ).Hmm, seems a bit messy, but perhaps manageable.Now, we need to find the areas of triangles NDL and AKL.First, triangle NDL: points N(0,0), D(0, (b a)/(4 l)), and L(l, 0).Wait, hold on, point L is on NA, which is the base. So, L is at (l, 0). So, triangle NDL has vertices at N(0,0), D(0, (b a)/(4 l)), and L(l, 0).Wait, actually, no. Point D is on NB, which is the y-axis, so D is at (0, (b a)/(4 l)). Point L is on NA, which is the x-axis, at (l, 0). So, triangle NDL is formed by points N(0,0), D(0, (b a)/(4 l)), and L(l, 0).Similarly, triangle AKL has points A(a,0), K, and L(l, 0). So, point K is somewhere on AB, and L is on NA.Let me compute the area of triangle NDL first.Triangle NDL has vertices at (0,0), (0, (b a)/(4 l)), and (l, 0). The area can be calculated using the formula for the area of a triangle with coordinates.But since two points are on the y-axis and one on the x-axis, it's a right triangle with base l and height (b a)/(4 l). So, area is (1/2)*base*height = (1/2)*l*(b a)/(4 l) = (1/2)*(b a)/4 = (b a)/8.Wait, that's interesting. The area of triangle NDL is (b a)/8, regardless of l? Hmm, let me verify.Yes, because the base is l and the height is (b a)/(4 l). Multiplying them gives (b a)/4, and half of that is (b a)/8. So, area of NDL is (b a)/8.Now, let's compute the area of triangle AKL.Points A(a,0), K, and L(l, 0). So, point K is on AB, and L is on NA.We need coordinates of K and L to compute the area.We have point K at ( [a(l - 3a/4)] / (l - a ), (a b / 4 ) / (a - l ) ).Wait, that seems complicated. Maybe there's a better way.Alternatively, perhaps using vectors or ratios.Wait, another approach: since the lines through Q and F are parallel to BL, which is from B(0,b) to L(l,0). So, the direction vector of BL is (l, -b). So, the lines through Q and F have the same direction vector.Therefore, the line through Q(a/4, 0) in direction (l, -b) intersects NB at D. Similarly, the line through F(3a/4, 0) in direction (l, -b) intersects AB at K.Wait, maybe using parametric equations.Let me parametrize the line through Q(a/4, 0) with direction (l, -b). So, parametric equations:x = a/4 + t*ly = 0 - t*bThis intersects NB, which is the y-axis x=0. So, set x=0:0 = a/4 + t*l => t = -a/(4 l)Then, y = 0 - (-a/(4 l))*b = (a b)/(4 l). So, point D is at (0, (a b)/(4 l)), which matches what I found earlier.Similarly, parametrize the line through F(3a/4, 0) with direction (l, -b):x = 3a/4 + t*ly = 0 - t*bThis intersects AB. Let me find where this intersects AB.Equation of AB is y = (-b/a)x + b.So, set y = (-b/a)x + b equal to y = -t b.So, -t b = (-b/a)(3a/4 + t l) + b.Simplify:- t b = (-b/a)(3a/4) - (b/a)(t l) + b= (-3b/4) - (b l /a) t + b= ( -3b/4 + b ) - (b l /a ) t= (b/4) - (b l /a ) tSo, -t b = b/4 - (b l /a ) tMultiply both sides by a to eliminate denominators:- t b a = (b a)/4 - b l tBring all terms to left:- t b a - (b a)/4 + b l t = 0Factor t:t(-b a + b l ) - (b a)/4 = 0Factor b:b [ t(-a + l ) - a /4 ] = 0Since b ≠ 0,t(-a + l ) - a /4 = 0So,t = (a /4 ) / ( -a + l ) = (a /4 ) / ( l - a )So, t = a / [4(l - a ) ]Therefore, coordinates of K:x = 3a/4 + t l = 3a/4 + [ a / (4(l - a )) ] * l = 3a/4 + (a l ) / [4(l - a ) ]Similarly, y = - t b = - [ a / (4(l - a )) ] * b = - (a b ) / [4(l - a ) ]So, point K is at:x = 3a/4 + (a l ) / [4(l - a ) ] = [ 3a(l - a ) + a l ] / [4(l - a ) ] = [ 3a l - 3a² + a l ] / [4(l - a ) ] = (4a l - 3a² ) / [4(l - a ) ] = a(4 l - 3a ) / [4(l - a ) ]Similarly, y = - (a b ) / [4(l - a ) ] = (a b ) / [4(a - l ) ]So, point K is at ( a(4 l - 3a ) / [4(l - a ) ], (a b ) / [4(a - l ) ] )Hmm, that seems consistent with earlier calculations.Now, to find the area of triangle AKL.Points A(a,0), K( a(4 l - 3a ) / [4(l - a ) ], (a b ) / [4(a - l ) ] ), and L(l, 0).We can use the shoelace formula for the area.Let me denote:Point A: (a, 0)Point K: (x, y ) = ( a(4 l - 3a ) / [4(l - a ) ], (a b ) / [4(a - l ) ] )Point L: (l, 0 )So, the coordinates are:A(a, 0), K(x, y), L(l, 0 )Using shoelace formula:Area = (1/2)| (a*(y - 0) + x*(0 - 0) + l*(0 - y) ) |= (1/2)| a y - l y | = (1/2)| y (a - l ) |.Since y = (a b ) / [4(a - l ) ], plug in:Area = (1/2)| (a b ) / [4(a - l ) ] * (a - l ) | = (1/2)*(a b ) /4 = (a b ) /8.So, area of triangle AKL is (a b ) /8.Wait, that's the same as the area of triangle NDL.So, the area of NDL is (a b ) /8, and the area of AKL is also (a b ) /8.Therefore, the sum of their areas is (a b ) /8 + (a b ) /8 = (a b ) /4.Now, what is the area of triangle NBA?Triangle NBA has base NA of length a and height b (from point B to base NA). So, area is (1/2)*a*b.Therefore, half the area of triangle NBA is (1/2)*(1/2)*a*b = (a b ) /4.Wait, no. Wait, half of the area of NBA is (1/2)*( (1/2)*a*b )? No, wait. The area of NBA is (1/2)*a*b. So, half of that is (1/4)*a*b.But the sum of the areas of NDL and AKL is (a b ) /4, which is exactly half the area of NBA.Wait, hold on, no. Wait, the area of NBA is (1/2)*a*b. So, half of that is (1/4)*a*b. But the sum of NDL and AKL is (a b ) /4, which is equal to half the area of NBA.Wait, but hold on, actually, no. Wait, the area of NBA is (1/2)*a*b, so half of that is (1/4)*a*b. The sum of NDL and AKL is (a b ) /4, which is equal to half of NBA's area.Wait, but hold on, in my calculation, both NDL and AKL have areas of (a b ) /8 each, so their sum is (a b ) /4, which is half of NBA's area, since NBA is (1/2)*a*b. So, yes, the sum is half of NBA's area.Wait, but hold on, is that correct? Because in my coordinate system, the area of NBA is (1/2)*a*b, which is correct. Then, the sum of NDL and AKL is (a b ) /4, which is half of NBA's area.But wait, in my coordinate system, the area of NDL was (a b ) /8 and AKL was (a b ) /8, so together (a b ) /4. So, yes, that is half of NBA's area.Wait, but hold on, in my coordinate system, I set N at (0,0), B at (0,b), and A at (a,0). So, the area of NBA is indeed (1/2)*a*b. So, half of that is (1/4)*a*b.But in my calculation, the sum of the areas is (a b ) /4, which is exactly half of NBA's area.Wait, but hold on, in my coordinate system, the area of NDL was (a b ) /8 and AKL was (a b ) /8, which adds up to (a b ) /4, which is half of NBA's area.So, yes, the sum is indeed half of the area of NBA.But wait, let me think again. Is this result dependent on the position of L on QF? Because in my calculation, the area of NDL and AKL each turned out to be (a b ) /8 regardless of l. So, regardless of where L is on QF, the areas of NDL and AKL remain constant?Wait, that seems counterintuitive. If L moves closer to Q or closer to F, wouldn't the areas of NDL and AKL change?Wait, in my calculation, the area of NDL was (a b ) /8, which was independent of l. Similarly, the area of AKL was (a b ) /8, independent of l. So, regardless of where L is on QF, the sum is always (a b ) /4, which is half of NBA's area.But that seems surprising. Let me check my calculations again.Starting with triangle NDL: points N(0,0), D(0, (b a)/(4 l)), and L(l,0). The area is (1/2)*base*height. The base is l, the height is (b a)/(4 l). So, area is (1/2)*l*(b a)/(4 l) = (1/2)*(b a)/4 = (b a)/8. So, that's correct.For triangle AKL: points A(a,0), K(x,y), L(l,0). Using shoelace formula, I found the area to be (a b ) /8. So, that's correct as well.So, regardless of where L is on QF, the areas of NDL and AKL remain constant? That seems to be the case here.But let me think about it geometrically. Since the lines through Q and F are parallel to BL, which is a line from B to some point L on NA. So, as L moves, BL changes, but the lines through Q and F remain parallel to BL, so the intersections D and K move accordingly.But in this specific setup, with Q and F dividing NA into four equal parts, the areas of NDL and AKL remain constant. So, their sum is always half of NBA's area.Therefore, the answer is yes, the sum of the areas of triangles NDL and AKL is half the area of triangle NBA.**Final Answer**boxed{dfrac{1}{2}}

answer：To express 37/80 as a terminating decimal, I first need to determine if the denominator has only the prime factors 2 and 5. The denominator is 80, which factors into 2^4 multiplied by 5^1. Since it only contains the prime factors 2 and 5, the fraction will indeed be a terminating decimal.Next, I'll convert the fraction by adjusting the denominator to a power of 10. To do this, I'll multiply both the numerator and the denominator by 5 to make the denominator 100.After multiplying, the fraction becomes 185/100. Dividing this out gives me 1.85.Therefore, 37/80 as a terminating decimal is 0.4625.

answer：To solve the first expression, I will handle the addition and subtraction of the integers step by step. Starting with -3 and -9, adding these gives -12. Then, subtracting +10 results in -22. Finally, subtracting -18 is equivalent to adding 18, which brings the total to -4.For the second expression, I will perform the division and multiplication operations in sequence. Dividing -81 by 9/4 is the same as multiplying by 4/9, which gives -36. Next, multiplying by -4/9 results in 16. Finally, dividing by -16 yields -1.

answer：Okay, so I've got this geometry problem here about an isosceles triangle (ABC) with base (AC). Let me try to visualize it first. Since it's isosceles with base (AC), that means sides (AB) and (BC) are equal? Wait, no, actually, in an isosceles triangle, the two equal sides are called the legs, and the unequal side is the base. So if the base is (AC), then sides (AB) and (BC) are equal? Hmm, no, wait, that can't be because if (AC) is the base, then the two equal sides must be (AB) and (CB). Wait, no, that would make (B) the apex. So, actually, (AB = BC), making (AC) the base. Hmm, maybe I should draw a rough sketch.So, triangle (ABC), with (AC) as the base. Let me label the vertices: (A) and (C) at the base, and (B) at the top. So, (AB = BC), making it isosceles. Now, point (D) divides side (BC) in the ratio (3:1) starting from vertex (B). So, (BD:DC = 3:1). That means (D) is closer to (B) than to (C). Then, point (E) is the midpoint of segment (AD). So, (E) is halfway between (A) and (D).Given that (BE = sqrt{7}) and (CE = 3). We need to find the radius of the circumcircle of triangle (ABC).Alright, so let me think about how to approach this. Since it's an isosceles triangle, maybe coordinate geometry could help here. Assign coordinates to the points and use distance formulas. Alternatively, maybe using vectors or coordinate geometry. Let me try coordinate geometry.Let me place the triangle in a coordinate system. Since (AC) is the base, let me place points (A) and (C) on the x-axis for simplicity. Let me denote point (A) as ((-a, 0)) and point (C) as ((a, 0)), so that (AC) is centered at the origin. Then, since the triangle is isosceles with base (AC), point (B) must lie somewhere on the y-axis. Let's denote (B) as ((0, b)), where (b > 0).So, coordinates:- (A = (-a, 0))- (C = (a, 0))- (B = (0, b))Now, point (D) divides (BC) in the ratio (3:1) from (B). So, using the section formula, the coordinates of (D) can be found.The section formula says that if a point divides a line segment joining ((x_1, y_1)) and ((x_2, y_2)) in the ratio (m:n), then the coordinates are (left(frac{m x_2 + n x_1}{m + n}, frac{m y_2 + n y_1}{m + n}right)).Here, (D) divides (BC) in the ratio (3:1), so (m = 3), (n = 1). Coordinates of (B = (0, b)), coordinates of (C = (a, 0)).So, coordinates of (D) would be:(x = frac{3 cdot a + 1 cdot 0}{3 + 1} = frac{3a}{4})(y = frac{3 cdot 0 + 1 cdot b}{3 + 1} = frac{b}{4})So, (D = left(frac{3a}{4}, frac{b}{4}right)).Now, point (E) is the midpoint of (AD). So, coordinates of (A = (-a, 0)), coordinates of (D = left(frac{3a}{4}, frac{b}{4}right)).Midpoint formula: average of x-coordinates and average of y-coordinates.So, coordinates of (E):(x = frac{-a + frac{3a}{4}}{2} = frac{-frac{a}{4}}{2} = -frac{a}{8})(y = frac{0 + frac{b}{4}}{2} = frac{b}{8})So, (E = left(-frac{a}{8}, frac{b}{8}right)).Now, we are given that (BE = sqrt{7}) and (CE = 3).Let me compute the coordinates of (E) and then find the distances (BE) and (CE).First, coordinates of (B = (0, b)), coordinates of (E = left(-frac{a}{8}, frac{b}{8}right)).Distance (BE) is:[BE = sqrt{left(0 - left(-frac{a}{8}right)right)^2 + left(b - frac{b}{8}right)^2} = sqrt{left(frac{a}{8}right)^2 + left(frac{7b}{8}right)^2}]Simplify:[BE = sqrt{frac{a^2}{64} + frac{49b^2}{64}} = sqrt{frac{a^2 + 49b^2}{64}} = frac{sqrt{a^2 + 49b^2}}{8}]Given that (BE = sqrt{7}), so:[frac{sqrt{a^2 + 49b^2}}{8} = sqrt{7}]Multiply both sides by 8:[sqrt{a^2 + 49b^2} = 8sqrt{7}]Square both sides:[a^2 + 49b^2 = 64 times 7 = 448]So, equation (1): (a^2 + 49b^2 = 448)Now, let's compute distance (CE).Coordinates of (C = (a, 0)), coordinates of (E = left(-frac{a}{8}, frac{b}{8}right)).Distance (CE):[CE = sqrt{left(a - left(-frac{a}{8}right)right)^2 + left(0 - frac{b}{8}right)^2} = sqrt{left(frac{9a}{8}right)^2 + left(-frac{b}{8}right)^2}]Simplify:[CE = sqrt{frac{81a^2}{64} + frac{b^2}{64}} = sqrt{frac{81a^2 + b^2}{64}} = frac{sqrt{81a^2 + b^2}}{8}]Given that (CE = 3), so:[frac{sqrt{81a^2 + b^2}}{8} = 3]Multiply both sides by 8:[sqrt{81a^2 + b^2} = 24]Square both sides:[81a^2 + b^2 = 576]So, equation (2): (81a^2 + b^2 = 576)Now, we have two equations:1. (a^2 + 49b^2 = 448)2. (81a^2 + b^2 = 576)We can solve this system of equations for (a^2) and (b^2).Let me denote (x = a^2) and (y = b^2). Then, the equations become:1. (x + 49y = 448)2. (81x + y = 576)Now, let's solve for (x) and (y).From equation (2): (81x + y = 576), we can express (y = 576 - 81x).Substitute this into equation (1):(x + 49(576 - 81x) = 448)Compute:(x + 49 times 576 - 49 times 81x = 448)Calculate constants:First, (49 times 576). Let's compute that:(49 times 576 = 49 times (500 + 76) = 49 times 500 + 49 times 76 = 24500 + 3724 = 28224)Next, (49 times 81 = 3969)So, equation becomes:(x + 28224 - 3969x = 448)Combine like terms:(x - 3969x + 28224 = 448)(-3968x + 28224 = 448)Subtract 28224 from both sides:(-3968x = 448 - 28224)Compute (448 - 28224 = -27776)So, (-3968x = -27776)Divide both sides by -3968:(x = frac{-27776}{-3968} = frac{27776}{3968})Simplify the fraction:Divide numerator and denominator by 16:(27776 ÷ 16 = 1736)(3968 ÷ 16 = 248)So, (x = frac{1736}{248})Simplify further:Divide numerator and denominator by 8:(1736 ÷ 8 = 217)(248 ÷ 8 = 31)So, (x = frac{217}{31})Wait, 217 divided by 31 is 7, because 31*7=217.So, (x = 7)So, (x = a^2 = 7), so (a = sqrt{7}) (since (a > 0))Now, substitute (x = 7) into equation (2): (81x + y = 576)So, (81*7 + y = 576)Compute 81*7: 567So, 567 + y = 576Thus, y = 576 - 567 = 9So, (y = b^2 = 9), so (b = 3) (since (b > 0))So, now we have (a = sqrt{7}) and (b = 3).So, the coordinates of the triangle are:- (A = (-sqrt{7}, 0))- (C = (sqrt{7}, 0))- (B = (0, 3))Now, we need to find the radius of the circumcircle of triangle (ABC).For a triangle with coordinates known, the circumradius can be found using the formula:[R = frac{abc}{4Delta}]Where (a), (b), (c) are the lengths of the sides, and (Delta) is the area of the triangle.Alternatively, since we have coordinates, we can use the circumradius formula for a triangle with vertices at ((x_1, y_1)), ((x_2, y_2)), ((x_3, y_3)):[R = frac{sqrt{(|AB|^2 |BC|^2 |CA|^2)}}{4 Delta}]But maybe it's easier to compute the lengths of the sides and then use the formula.First, let's compute the lengths of sides (AB), (BC), and (AC).Compute (AB):Points (A = (-sqrt{7}, 0)), (B = (0, 3)).Distance (AB):[AB = sqrt{(0 - (-sqrt{7}))^2 + (3 - 0)^2} = sqrt{(sqrt{7})^2 + 3^2} = sqrt{7 + 9} = sqrt{16} = 4]Compute (BC):Points (B = (0, 3)), (C = (sqrt{7}, 0)).Distance (BC):[BC = sqrt{(sqrt{7} - 0)^2 + (0 - 3)^2} = sqrt{7 + 9} = sqrt{16} = 4]Compute (AC):Points (A = (-sqrt{7}, 0)), (C = (sqrt{7}, 0)).Distance (AC):[AC = sqrt{(sqrt{7} - (-sqrt{7}))^2 + (0 - 0)^2} = sqrt{(2sqrt{7})^2} = sqrt{4 times 7} = sqrt{28} = 2sqrt{7}]So, sides:- (AB = 4)- (BC = 4)- (AC = 2sqrt{7})So, triangle (ABC) is isosceles with two sides of length 4 and base (2sqrt{7}).Now, let's compute the area (Delta) of triangle (ABC).Since it's an isosceles triangle, we can compute the area as:[Delta = frac{1}{2} times text{base} times text{height}]We know the base (AC = 2sqrt{7}). The height can be found from point (B) perpendicular to base (AC). Since in our coordinate system, point (B) is at ((0, 3)), and the base (AC) is on the x-axis, the height is simply the y-coordinate of (B), which is 3.So, area:[Delta = frac{1}{2} times 2sqrt{7} times 3 = frac{1}{2} times 6sqrt{7} = 3sqrt{7}]Alternatively, using coordinates, we can compute the area via the shoelace formula.Coordinates:- (A = (-sqrt{7}, 0))- (B = (0, 3))- (C = (sqrt{7}, 0))Shoelace formula:[Delta = frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|]Plugging in:[Delta = frac{1}{2} | (-sqrt{7})(3 - 0) + 0(0 - 0) + sqrt{7}(0 - 3) |][= frac{1}{2} | (-sqrt{7} times 3) + 0 + (sqrt{7} times (-3)) |][= frac{1}{2} | -3sqrt{7} - 3sqrt{7} | = frac{1}{2} | -6sqrt{7} | = frac{1}{2} times 6sqrt{7} = 3sqrt{7}]Same result, so that's good.Now, going back to the formula for circumradius:[R = frac{abc}{4Delta}]Where (a), (b), (c) are the lengths of the sides.In our case, sides are 4, 4, and (2sqrt{7}). So:[R = frac{4 times 4 times 2sqrt{7}}{4 times 3sqrt{7}} = frac{32sqrt{7}}{12sqrt{7}} = frac{32}{12} = frac{8}{3}]Simplify:[R = frac{8}{3}]So, the radius of the circumcircle is (frac{8}{3}).Wait, let me double-check my calculations because sometimes I might have messed up.Compute numerator: (4 times 4 times 2sqrt{7} = 32sqrt{7})Denominator: (4 times 3sqrt{7} = 12sqrt{7})So, (32sqrt{7} / 12sqrt{7} = 32/12 = 8/3). Yes, that's correct.Alternatively, since the triangle is isosceles, maybe we can use another formula for the circumradius.In an isosceles triangle with equal sides (AB = BC = 4), base (AC = 2sqrt{7}), the circumradius can also be calculated using the formula:[R = frac{a}{2 sin A}]Where (a) is the length of a side, and (A) is the angle opposite that side.But in this case, since two sides are equal, maybe it's better to use the formula for the circumradius in terms of sides and area, which we already did.Alternatively, since we have coordinates, we can find the circumcircle equation and compute the radius.Let me try that approach as a verification.The general equation of a circle is:[x^2 + y^2 + Dx + Ey + F = 0]We can plug in the coordinates of points (A), (B), and (C) into this equation to find (D), (E), and (F), then compute the radius.So, plugging in point (A = (-sqrt{7}, 0)):[(-sqrt{7})^2 + 0^2 + D(-sqrt{7}) + E(0) + F = 0][7 - Dsqrt{7} + F = 0 quad text{(Equation 3)}]Point (B = (0, 3)):[0^2 + 3^2 + D(0) + E(3) + F = 0][9 + 3E + F = 0 quad text{(Equation 4)}]Point (C = (sqrt{7}, 0)):[(sqrt{7})^2 + 0^2 + D(sqrt{7}) + E(0) + F = 0][7 + Dsqrt{7} + F = 0 quad text{(Equation 5)}]Now, we have three equations:3. (7 - Dsqrt{7} + F = 0)4. (9 + 3E + F = 0)5. (7 + Dsqrt{7} + F = 0)Let me subtract Equation 3 from Equation 5:Equation 5 - Equation 3:[(7 + Dsqrt{7} + F) - (7 - Dsqrt{7} + F) = 0 - 0][7 + Dsqrt{7} + F - 7 + Dsqrt{7} - F = 0][2Dsqrt{7} = 0][D = 0]So, (D = 0). Now, plug (D = 0) into Equation 3:Equation 3: (7 - 0 + F = 0) => (7 + F = 0) => (F = -7)Now, plug (F = -7) into Equation 4:Equation 4: (9 + 3E - 7 = 0) => (2 + 3E = 0) => (3E = -2) => (E = -frac{2}{3})So, the equation of the circle is:[x^2 + y^2 + 0x - frac{2}{3}y - 7 = 0][x^2 + y^2 - frac{2}{3}y - 7 = 0]To find the radius, we can rewrite this equation in standard form.Complete the square for y:[x^2 + left(y^2 - frac{2}{3}yright) = 7]Take the y terms:[y^2 - frac{2}{3}y = left(y - frac{1}{3}right)^2 - left(frac{1}{3}right)^2 = left(y - frac{1}{3}right)^2 - frac{1}{9}]So, substitute back:[x^2 + left(y - frac{1}{3}right)^2 - frac{1}{9} = 7][x^2 + left(y - frac{1}{3}right)^2 = 7 + frac{1}{9} = frac{63}{9} + frac{1}{9} = frac{64}{9}]So, the standard form is:[x^2 + left(y - frac{1}{3}right)^2 = left(frac{8}{3}right)^2]Thus, the center is at ((0, frac{1}{3})) and the radius is (frac{8}{3}).So, that confirms our earlier result.Therefore, the radius of the circumcircle of triangle (ABC) is (frac{8}{3}).**Final Answer**The radius of the circumcircle of triangle (ABC) is boxed{dfrac{8}{3}}.