💡First, I recognize that the area of a square is calculated by squaring its side length. If the side length is increased by 30%, the new side length becomes 1.3 times the original side length.Next, I calculate the new area by squaring the new side length. This gives me (1.3)^2, which equals 1.69. This means the new area is 1.69 times the original area.To find the percentage increase in the area, I subtract 1 from 1.69, resulting in 0.69. Multiplying this by 100% gives a 69% increase in the area.Therefore, increasing the side of a square by 30% leads to a 69% increase in its area.

💡First, I will simplify each fraction individually. For the fraction (frac{2}{12}), I can divide both the numerator and the denominator by 2 to get (frac{1}{6}).Next, for the fraction (frac{4}{24}), I can divide both the numerator and the denominator by 4 to simplify it to (frac{1}{6}).Similarly, the fraction (frac{6}{36}) can be simplified by dividing both the numerator and the denominator by 6, resulting in (frac{1}{6}).Now, I have three simplified fractions: (frac{1}{6}), (frac{1}{6}), and (frac{1}{6}).I will add these fractions together. Since they all have the same denominator, I can simply add the numerators:[frac{1}{6} + frac{1}{6} + frac{1}{6} = frac{1 + 1 + 1}{6} = frac{3}{6}]The fraction (frac{3}{6}) can be simplified by dividing both the numerator and the denominator by 3, resulting in (frac{1}{2}).Finally, I will convert (frac{1}{2}) to a decimal by dividing 1 by 2, which equals 0.5.

💡Let ( E ) be the points scored by the Eagles and ( H ) be the points scored by the Hawks.1. **Given Information:** - The total score of the game is 82 points: ( E + H = 82 ). - The Eagles won by a margin of 18 points: ( E = H + 18 ). - The Eagles scored 6 three-pointers, which is ( 6 times 3 = 18 ) points.2. **Setting Up the Equations:** - From the margin of victory: ( E = H + 18 ). - Total score: ( E + H = 82 ).3. **Substituting ( E ) in the Total Score Equation:** [ (H + 18) + H = 82 ] [ 2H + 18 = 82 ] [ 2H = 82 - 18 ] [ 2H = 64 ] [ H = frac{64}{2} = 32 ]4. **Conclusion:** - The Hawks scored 32 points.The final answer is (boxed{text{(B) } 32}).

💡Let ( x ) be the total number of pages in the novel.1. **First Day:** - Nina reads ( frac{1}{6}x + 10 ) pages. - Remaining pages: [ x - left(frac{1}{6}x + 10right) = frac{5}{6}x - 10 ]2. **Second Day:** - Nina reads ( frac{1}{5} ) of the remaining pages plus 20 pages. - Pages read: [ frac{1}{5}left(frac{5}{6}x - 10right) + 20 ] - Remaining pages after the second day: [ left(frac{5}{6}x - 10right) - left(frac{1}{5}left(frac{5}{6}x - 10right) + 20right) = frac{4}{5}left(frac{5}{6}x - 10right) - 20 = frac{2x}{3} - 28 ]3. **Third Day:** - Nina reads ( frac{1}{4} ) of the remaining pages plus 25 pages. - Pages read: [ frac{1}{4}left(frac{2x}{3} - 28right) + 25 ] - Remaining pages after the third day: [ left(frac{2x}{3} - 28right) - left(frac{1}{4}left(frac{2x}{3} - 28right) + 25right) = frac{3}{4}left(frac{2x}{3} - 28right) - 25 = frac{1}{2}x - 46 ]4. **Final Remaining Pages:** - It is given that after the third day, there are 80 pages left: [ frac{1}{2}x - 46 = 80 implies frac{1}{2}x = 126 implies x = 252 ]Conclusion: The total number of pages in the novel is ( 252 ).

💡Okay, so I have this problem: In triangle ABC, I need to prove that if p + q = 1, then p sin²A + q sin²B > pq sin²C. Hmm, okay, let me try to figure this out step by step.First, I know that in any triangle, the sum of the angles is 180 degrees, so A + B + C = 180°. Also, I remember the Law of Sines, which says that a/sin A = b/sin B = c/sin C = 2R, where R is the circumradius of the triangle. Maybe that could be useful here.Since p + q = 1, I can express one variable in terms of the other. Let's say q = 1 - p. That might simplify things a bit. So the inequality becomes p sin²A + (1 - p) sin²B > p(1 - p) sin²C.Hmm, maybe I can use the Law of Sines to express sin A, sin B, and sin C in terms of the sides of the triangle. If I let a, b, and c be the lengths opposite angles A, B, and C respectively, then sin A = a/(2R), sin B = b/(2R), and sin C = c/(2R). Substituting these into the inequality, I get:p (a²/(4R²)) + (1 - p) (b²/(4R²)) > p(1 - p) (c²/(4R²))Since 1/(4R²) is a common factor on all terms, I can multiply both sides by 4R² to simplify:p a² + (1 - p) b² > p(1 - p) c²Okay, so now the inequality is in terms of the sides of the triangle. Maybe I can use the Law of Cosines here. The Law of Cosines states that c² = a² + b² - 2ab cos C. Let me substitute that into the inequality:p a² + (1 - p) b² > p(1 - p) (a² + b² - 2ab cos C)Expanding the right-hand side:p a² + (1 - p) b² > p(1 - p) a² + p(1 - p) b² - 2p(1 - p) ab cos CNow, let's bring all terms to the left-hand side:p a² + (1 - p) b² - p(1 - p) a² - p(1 - p) b² + 2p(1 - p) ab cos C > 0Factor out common terms:p(1 - (1 - p)) a² + (1 - p)(1 - p) b² + 2p(1 - p) ab cos C > 0Wait, that doesn't seem right. Let me check the factoring again.Looking back, I have:p a² - p(1 - p) a² + (1 - p) b² - p(1 - p) b² + 2p(1 - p) ab cos C > 0Factor out p a² and (1 - p) b²:p a² [1 - (1 - p)] + (1 - p) b² [1 - p] + 2p(1 - p) ab cos C > 0Simplify the coefficients:p a² [p] + (1 - p) b² [1 - p] + 2p(1 - p) ab cos C > 0So that becomes:p² a² + (1 - p)² b² + 2p(1 - p) ab cos C > 0Hmm, this looks familiar. It resembles the expansion of (pa + (1 - p)b)², which is p² a² + 2p(1 - p)ab + (1 - p)² b². But in our case, we have an extra cos C term. So, actually, it's:p² a² + (1 - p)² b² + 2p(1 - p) ab cos C > 0Which is equivalent to:(pa + (1 - p)b)² + 2p(1 - p) ab (cos C - 1) > 0Wait, maybe that's not helpful. Let me think differently.I know that cos C is less than or equal to 1, so 2p(1 - p) ab cos C is less than or equal to 2p(1 - p) ab. But I'm not sure if that helps directly.Alternatively, maybe I can consider the expression p² a² + (1 - p)² b² + 2p(1 - p) ab cos C and see if it's always positive.Since p and (1 - p) are positive (assuming p is between 0 and 1), and a, b are positive lengths, the entire expression is a sum of positive terms plus a term involving cos C.But cos C can be negative if angle C is obtuse. Hmm, so that term could be negative. However, the other terms are always positive. So, is the entire expression still positive?Let me see. The first two terms, p² a² and (1 - p)² b², are definitely positive. The last term, 2p(1 - p) ab cos C, could be positive or negative depending on angle C. But even if it's negative, the sum of the first two terms might still make the entire expression positive.Wait, but how can I be sure? Maybe I can use the fact that in a triangle, the sides satisfy certain inequalities. For example, by the triangle inequality, a + b > c, but I don't know if that's directly applicable here.Alternatively, maybe I can consider the case when angle C is 90 degrees, making cos C = 0. Then the inequality becomes p² a² + (1 - p)² b² > 0, which is obviously true.If angle C is acute, cos C is positive, so the last term is positive, making the entire expression even larger.If angle C is obtuse, cos C is negative, so the last term is negative. But then, is the sum still positive?Let me test with specific values. Suppose p = 0.5, so q = 0.5. Let’s take a simple triangle, say a = b = 1, and angle C = 120 degrees, so cos C = -0.5.Then, the left-hand side of the inequality becomes:0.25 * 1 + 0.25 * 1 + 2 * 0.5 * 0.5 * 1 * 1 * (-0.5) = 0.25 + 0.25 - 0.25 = 0.25Which is still positive. So even when cos C is negative, the expression remains positive.Another test case: p = 0.25, q = 0.75, a = 3, b = 4, angle C = 150 degrees, cos C = -√3/2 ≈ -0.866.Compute p² a² + (1 - p)² b² + 2p(1 - p) ab cos C:(0.0625)(9) + (0.5625)(16) + 2*(0.25)*(0.75)*3*4*(-0.866)= 0.5625 + 9 + 2*(0.1875)*12*(-0.866)= 0.5625 + 9 + (0.225)*(-10.392)= 9.5625 - 2.343≈ 7.2195, which is still positive.So it seems that even when cos C is negative, the expression remains positive. Therefore, the inequality holds.But I need a more general proof, not just specific examples. Maybe I can consider the expression p² a² + (1 - p)² b² + 2p(1 - p) ab cos C and show that it's always positive.Let me denote x = pa and y = (1 - p)b. Then the expression becomes x² + y² + 2xy cos C.Wait, that looks like the Law of Cosines for a triangle with sides x and y and angle C between them. So, x² + y² + 2xy cos C = (x + y cos C)² + (y sin C)², which is always positive because it's a sum of squares.Therefore, the expression is always positive, which means our original inequality holds.So, putting it all together:Starting from p sin²A + q sin²B > pq sin²C, with p + q = 1, we transformed it using the Law of Sines and Law of Cosines into an expression involving the sides of the triangle. By recognizing the structure of the expression, we saw that it's equivalent to the Law of Cosines for a certain triangle, ensuring it's positive. Therefore, the original inequality is proven.**Final Answer**boxed{p sin^2 A + q sin^2 B > pq sin^2 C}

💡Alright, so I have this geometry problem here, and I need to figure out if the area of quadrilateral ABCD can be an integer given certain side lengths. Let me try to visualize the figure first based on the description.We have quadrilateral ABCD with AB perpendicular to BC, and BC perpendicular to CD. So, AB is perpendicular to BC, and BC is perpendicular to CD. That means AB is parallel to CD because they're both perpendicular to BC. So, ABCD is a trapezoid with AB and CD as the two parallel sides, and BC as one of the legs.Additionally, BC is tangent to a circle with center O and diameter AD. Hmm, okay. So, the circle has diameter AD, which means the center O is the midpoint of AD. And BC is tangent to this circle. Tangent means that BC touches the circle at exactly one point.I think I need to draw this out to get a better sense. Let me sketch it mentally: points A, B, C, D. AB is vertical, BC is horizontal, CD is vertical again, and DA is the other horizontal side. The circle is inside the trapezoid with diameter AD, so it's sitting on the base DA.Since BC is tangent to the circle, the distance from the center O to BC must be equal to the radius of the circle. The radius is half of AD, so if I can find the length of AD, I can find the radius. But I don't know AD yet.Wait, maybe I can relate the lengths AB and CD to AD somehow. Since AB and CD are the two parallel sides, and BC is the leg, maybe there's a relationship involving the tangent condition.I remember that if a line is tangent to a circle, then the distance from the center of the circle to the tangent line is equal to the radius. So, in this case, the distance from O to BC is equal to the radius, which is AD/2.Let me denote the coordinates to make it clearer. Let's place point A at (0, 0). Since AB is perpendicular to BC, let's assume AB is vertical. So, point B would be at (0, AB). Then, BC is horizontal, so point C would be at (BC, AB). CD is vertical again, so point D would be at (BC, AB - CD). Wait, is that right?Hold on, if AB is vertical from A(0,0) to B(0, AB), then BC is horizontal to the right to point C(BC, AB). Then CD is vertical downward to point D(BC, AB - CD). Then DA connects back to A(0,0). So, DA is the line from D(BC, AB - CD) to A(0,0). The diameter AD is the line segment from A(0,0) to D(BC, AB - CD). So, the center O is the midpoint of AD, which would be at ((BC)/2, (AB - CD)/2).Now, the circle has center O at ((BC)/2, (AB - CD)/2) and radius equal to half of AD. Let me compute AD's length. Using the distance formula, AD is sqrt[(BC)^2 + (AB - CD)^2]. So, the radius is (1/2)*sqrt[(BC)^2 + (AB - CD)^2].But BC is tangent to the circle, so the distance from O to BC must equal the radius. Let's find the distance from O to BC. Since BC is the horizontal line y = AB, and O is at ((BC)/2, (AB - CD)/2). The distance from O to BC is the vertical distance from O to the line y = AB. That distance is AB - (AB - CD)/2 = (AB + CD)/2.So, the distance from O to BC is (AB + CD)/2, and this must equal the radius, which is (1/2)*sqrt[(BC)^2 + (AB - CD)^2]. Therefore, we have:(AB + CD)/2 = (1/2)*sqrt[(BC)^2 + (AB - CD)^2]Multiply both sides by 2:AB + CD = sqrt[(BC)^2 + (AB - CD)^2]Now, square both sides to eliminate the square root:(AB + CD)^2 = (BC)^2 + (AB - CD)^2Let me expand both sides:Left side: (AB)^2 + 2*AB*CD + (CD)^2Right side: (BC)^2 + (AB)^2 - 2*AB*CD + (CD)^2Subtract right side from left side:(AB)^2 + 2*AB*CD + (CD)^2 - [(BC)^2 + (AB)^2 - 2*AB*CD + (CD)^2] = 0Simplify:(AB)^2 + 2*AB*CD + (CD)^2 - (BC)^2 - (AB)^2 + 2*AB*CD - (CD)^2 = 0Combine like terms:(2*AB*CD + 2*AB*CD) - (BC)^2 = 0So, 4*AB*CD - (BC)^2 = 0Therefore, (BC)^2 = 4*AB*CDTaking square roots:BC = 2*sqrt(AB*CD)Okay, so BC is twice the square root of the product of AB and CD.Now, the area of trapezoid ABCD is given by the formula:Area = (1/2)*(AB + CD)*BCSubstitute BC from above:Area = (1/2)*(AB + CD)*(2*sqrt(AB*CD)) = (AB + CD)*sqrt(AB*CD)So, the area is (AB + CD)*sqrt(AB*CD)We need this area to be an integer. So, (AB + CD)*sqrt(AB*CD) must be an integer.Let me denote AB = a and CD = b for simplicity. Then, the area becomes (a + b)*sqrt(a*b). We need this to be an integer.So, (a + b)*sqrt(a*b) must be integer.Let me see. For this to be integer, sqrt(a*b) must be rational because (a + b) is integer (given a and b are integers). So, sqrt(a*b) must be rational, which implies that a*b must be a perfect square.Therefore, a*b must be a perfect square.So, AB*CD must be a perfect square.Looking back at the options:A) AB=4, CD=2: 4*2=8, which is not a perfect square.B) AB=8, CD=2: 8*2=16, which is a perfect square.C) AB=12, CD=3: 12*3=36, which is a perfect square.D) AB=16, CD=4: 16*4=64, which is a perfect square.E) AB=20, CD=5: 20*5=100, which is a perfect square.So, for options B, C, D, E, AB*CD is a perfect square, so sqrt(AB*CD) is integer, and thus the area (AB + CD)*sqrt(AB*CD) is integer.But let's check option A just to be thorough. AB=4, CD=2, so sqrt(8) is irrational, so the area would be (4 + 2)*sqrt(8) = 6*2.828..., which is approximately 16.97, not an integer.So, only options B, C, D, E result in integer areas.Wait, but the question says "Determine whether the area of quadrilateral ABCD can be an integer, given the following choices." So, it's asking which of the options can result in an integer area. So, the answer should be all options except A.But let me double-check my reasoning because sometimes there might be a miscalculation.We found that BC = 2*sqrt(AB*CD). Then, area = (AB + CD)*sqrt(AB*CD). So, if AB*CD is a perfect square, sqrt(AB*CD) is integer, so the area is integer. If AB*CD is not a perfect square, then sqrt(AB*CD) is irrational, and since (AB + CD) is integer, the area would be irrational, hence not integer.Therefore, only when AB*CD is a perfect square can the area be integer. So, the correct options are B, C, D, E.But let me check if there's another way this could be approached, maybe using similar triangles or something else, just to confirm.Since BC is tangent to the circle with diameter AD, by the property of tangents, the tangent at a point is perpendicular to the radius at that point. So, if we let the point of tangency be T, then OT is perpendicular to BC.But OT is the radius, which is AD/2. So, OT is perpendicular to BC, which is consistent with our earlier reasoning that the distance from O to BC is equal to the radius.Alternatively, maybe using coordinate geometry as I did earlier is another way, but I think the key insight is recognizing that AB*CD must be a perfect square for the area to be integer.So, yeah, I think my conclusion is correct. The area can be integer for options B, C, D, E.**Final Answer**boxed{B}, boxed{C}, boxed{D}, boxed{E}