answer:First note that the area of the region determined by the triangle topped by the semicircle of diameter 1 is frac{1}{2}cdotfrac{sqrt{3}}{2} + frac{1}{2}pidisplaystyleleft(frac{1}{2}displaystyleright)^2 = frac{sqrt{3}}{4} + frac{1}{8}pi. The area of the lune results from subtracting from this the area of the sector of the larger semicircle, frac{1}{6}pi(1)^2 = frac{1}{6}pi. So the area of the lune is frac{sqrt{3}}{4} + frac{1}{8}pi -frac{1}{6}pi=frac{sqrt{3}}{4} - frac{1}{24}pi. asy fill((0,2.73)..(1,1.73)--(-1,1.73)..cycle,gray(0.5)); draw((0,2.73)..(1,1.73)--(-1,1.73)..cycle,linewidth(0.7)); fill((0,2)..(2,0)--(-2,0)..cycle,white); fill((0,2)..(1,1.73)--(-1,1.73)..cycle,gray(0.7)); fill((0,0)--(1,1.73)--(-1,1.73)--cycle,gray(0.9)); draw((0,2)..(2,0)--(-2,0)..cycle,linewidth(0.7)); draw((-1,1.73)--(1,1.73),dashed); label("2",(0,0),S); label("1",(0,1.73),SW); draw((0,0)--(0,1.73),dashed); label("1",(-0.5,0.87),SW); label("1",(0.5,0.87),SE); label("frac{sqrt{3}}{2}",(0,0.87),E); /asy Note that the answer does not depend on the position of the lune on the semicircle.,The area of a circle is given by the formula pi r^2, and since both semicircles have diameter 1 and 2, we can calculate their radii as 1/2 and 2/2=1 respectively. The area of the smaller semicircle (s) is pi cdot (1/2)^2=1/4pi and the area of the larger semicircle (l) is pi cdot 1^2 = pi. Therefore, the lune has area l - s = pi - 1/4pi = pi * (4/4 - 1/4), reducing to pi/2. , The area of the lune is boxed{frac{3-sqrt{3}}{2}pi}. Here's the solution: Consider the larger semicircle with diameter 2. Its radius is 1, so its area is picdot 1^2=pi. Now consider the smaller semicircle with diameter 1. Its radius is frac12, so its area is picdot left(frac12right)^2=frac{pi}{4}. Now let's find the combined area of the two triangles that make up the rest of the lune. Note that the height of each triangle is frac{sqrt{3}}{2} and the base is 1. Therefore, the area of each triangle is frac12cdot 1cdot frac{sqrt{3}}{2}=frac{sqrt{3}}{4}. Since there are two such triangles, their combined area is 2cdot frac{sqrt{3}}{4}=frac{sqrt{3}}{2}. Subtracting the areas of the two triangles from the area of the smaller semicircle, we get area of lune = text{area of smaller semicircle} - text{area of two triangles} = frac{pi}{4} - frac{sqrt{3}}{2} = boxed{frac{3-sqrt{3}}{2}pi}. Note: The exact numerical value of the area of the lune is approximately 0.62349 pi approx 1.9635 square units.

answer:The area of rectangle ABCD is 25/4 square inches. Here's the reasoning: 1. Let AD=10x inches, then AB=5x inches. 2. Since triangle EFG is a right triangle with overline{EF} perpendicular to overline{EG}, we can find the height of triangle EFD using the altitude from F to side EG: EF=7 inches. 3. Using the similar triangles triangle EFD and triangle EAB, we can write the proportion: frac{EF}{AD}=frac{AB}{AE}. 4. Substitute the known values: frac{7}{10x}=frac{5x}{10}. 5. Cross-multiply: 7(10)=5x^2Rightarrow x^2=frac{70}{5}=14. 6. Since the side lengths of the rectangle must be positive, take the positive square root of 14: x=sqrt{14}approx 3.74. 7. Calculate the area of the rectangle ABCD=ABtimes AD=5xtimes 10x=50x^2approx 50(14)=700. 8. However, since x=sqrt{14} is an approximation, let's find the exact value for the area using rational numbers. - We know that x^2=14Rightarrow x=sqrt{14}=frac{10}{sqrt{2}}. - Multiply both numerator and denominator by sqrt{2} to get rid of the square root in the denominator: x=frac{10sqrt{2}}{2}=frac{5sqrt{2}}{1}=5sqrt{2}. 9. Now, calculate the area exactly using rational numbers: A=50x^2=frac{50(5sqrt{2})^2}{1^2}=frac{50(50times 2)}{1}=frac{50(100)}{1}=boxed{frac{25}{4}}. Therefore, the area of rectangle ABCD is 25/4 square inches.,I understand that we are given a rectangle ABCD inscribed in a triangle with the corresponding dimensions. To find the area of rectangle ABCD, we need to first find the length of each side of the rectangle. We can start by finding the length of AB, which is given as half the length of AD. From the diagram, we have: AB = (1/2) * AD To find the length of AD, we need to use the Pythagorean theorem on the right triangle ADF. Since, FD = (AB^2 + AD^2) = 7, we have: AD^2 = AD^2 + (1/2)AD^2 2AD^2 = 7AD^2 1/2 = 7/4 8/4 = 7/4, which simplifies to 4 = 7, so we can see that things are not correctly set up in the statement of the problem. So, unfortunately, the problem doesn't make sense as it is written, since its solutions lead to contradictions.,Suppose that the altitude from F to EG intersects EG at point H. Then triangle EAB sim triangle EHF, and we have that frac{HE}{HF} = frac{AE}{AB}. Also, triangle GDC sim GHF, and frac{HG}{HF} = frac{DG}{DC}. Adding these equalities, we find that frac{HE + HG}{HF} = frac{AE + DG}{AB}, since AB = DC. But HE + HG = EG = 10, HF = 7, and finally AE + DG = EG - AD = 10 - 2AB. Plugging in, we find that frac{10}{7} = frac{10-2AB}{AB}, or AB = frac{35}{12}. Thus the area of ABCD is frac{35}{12}cdotfrac{35}{6} = frac{1225}{72}.

answer:To compute cos 135^circ, we can first recall that a 360^circ rotation in standard position sends the unit circle to itself, and the coordinates of a point on the unit circle coincide with its x-coordinate and cos theta and its y-coordinate and sin theta. Thus, cos 135^circ = xtext{ coordinate of point at }135^circ Now, let's find this point. We begin by dividing the circle into four quadrants. Since boxed{360^circ div 4 = 90^circ}, cos 135^circ corresponds to the x-coordinate of the point at boxed{135^circ = 90^circ + 45^circ}. Therefore, it falls in the second quadrant, meaning it has a negative x-coordinate and a positive y-coordinate. Now, recall the boxed{45-45-90} right triangle, where the legs are equal: boxed{a = a = sqrt{2}/2}. Here, the hypotenuse has length boxed{c = 1}, since we are on the unit circle. Since cos theta corresponds to the adjacent side over the hypotenuse, we have cos 135^circ = -frac{a}{c} As mentioned, a = sqrt{2}/2, and c = 1. Thus, we have cos 135^circ = -frac{sqrt{2}/2}{1} = -frac{sqrt{2}}{2} That is, boxed{cos 135^circ = -frac{sqrt{2}}{2}},To compute cos 135^circ, we know that the cosine function is periodic with a period of 360^circ. Therefore, we can express the cosine of 135 degrees in terms of cosine of 0 degrees or 90 degrees, as follows: cos 135^circ = cos(90^circ + 45^circ) Using the cosine sum formula, we have: cos(90^circ + 45^circ) = cos 90^circ cos 45^circ - sin 90^circ sin 45^circ Recall that: cos 90^circ = 0, sin 90^circ = 1, cos 45^circ = sqrt{2}/2, and sin 45^circ = sqrt{2}/2 Substituting these values, we get: cos 135^circ = (0 + sin 90^circ)(cos 45^circ + sin 45^circ) - sin 90^circ cos 45^circ cos 135^circ = 1(sqrt{2}/2) - 1(sqrt{2}/2) cos 135^circ = -sqrt{2}/2 So, cos 135^circ = -sqrt{2}/2.,Let P be the point on the unit circle that is 135^circ counterclockwise from (1,0), and let D be the foot of the altitude from P to the x-axis, as shown below. asy pair A,C,P,O,D; draw((0,-1.2)--(0,1.2),p=black+1.2bp,Arrows(0.15cm)); draw((-1.2,0)--(1.2,0),p=black+1.2bp,Arrows(0.15cm)); A = (1,0); O= (0,0); label("x",(1.2,0),SE); label("y",(0,1.2),NE); P = rotate(135)*A; D = foot(P,A,-A); draw(O--P--D); draw(rightanglemark(O,D,P,2)); draw(Circle(O,1)); label("O",O,SE); label("P",P,NW); //label("A",A,SE); label("D",D,S); /asy Triangle POD is a 45-45-90 triangle, so DO = DP = frac{sqrt{2}}{2}. Therefore, the coordinates of P are left(-frac{sqrt{2}}{2}, frac{sqrt{2}}{2}right), so cos 135^circ = -frac{sqrt{2}}{2}.

answer:The perimeter of the rectangle is 160 units. Here's the reasoning: 1. Since the diagram shows a dissection of the rectangle into nine non-overlapping squares, we can determine the dimensions of the rectangle by looking at the side lengths of these squares. 2. There are two squares with side length 36 units, which form the largest 2x2 square within the rectangle. 3. To the right of this 2x2 square, there is a square with side length 3 units, and above it, there are two squares with side lengths 8 units and 3 units, respectively. 4. Now, let's calculate the dimensions of the rectangle: width = 36 + 3 = 39 units, height = 36 + 8 + 3 = 47 units. 5. The given information states that the width and height are relatively prime positive integers, which they are (gcd(39,47) = 1). 6. To find the perimeter of the rectangle, we use the formula P = 2(width + height) = 2(39 + 47) = 160 units. Therefore, the perimeter of the rectangle is 160 units.,Call the squares' side lengths from smallest to largest a_1,ldots,a_9, and let l,w represent the dimensions of the rectangle. The picture shows thatbegin{align*} a_1+a_2 &= a_3 a_1 + a_3 &= a_4 a_3 + a_4 &= a_5 a_4 + a_5 &= a_6 a_2 + a_3 + a_5 &= a_7 a_2 + a_7 &= a_8 a_1 + a_4 + a_6 &= a_9 a_6 + a_9 &= a_7 + a_8.end{align*} Expressing all terms 3 to 9 in terms of a_1 and a_2 and substituting their expanded forms into the previous equation will give the expression 5a_1 = 2a_2. We can guess that a_1 = 2. (If we started with a_1 odd, the resulting sides would not be integers and we would need to scale up by a factor of 2 to make them integers; if we started with a_1 > 2 even, the resulting dimensions would not be relatively prime and we would need to scale down.) Then solving gives a_9 = 36, a_6=25, a_8 = 33, which gives us l=61,w=69. These numbers are relatively prime, as desired. The perimeter is 2(61)+2(69)=260.,Let the width of the rectangle be a and the height be b, where a and b are relatively prime positive integers. From the diagram, we can see that the length of each horizontal square is a and the length of each vertical square is b. Since the rectangle is dissected into nine non-overlapping squares, the sum of the length of the horizontal squares should equal the width of the rectangle (6 a) and the sum of the length of the vertical squares should equal the height of the rectangle (6 b). Additionally, the four corner squares have a width of (a/2) and a height of (b/2). Since these are relatively prime positive integers, we can say that: a/2 = 1 and b/2 = 1 Simplifying the equations, we have: a = 2 and b = 2 Therefore, the width and height of the rectangle are both 2. The perimeter of the rectangle is the sum of all four sides, which is twice the sum of the width and height, or 2 * (2 + 2) = 8 units.