of the altitude drawn to the base?\nHint: The rectangle and the triangle both have equal areas.
213732
The area of the rectangle is:\n A of rectangle = L X W = 12 x 8 = 96 sq in\n\nThe area of the triangle is:\nA of triangle = 1/2 (A)(B) = 1/2 (A)(32 in) where A is the altitude to the base B of the triangle.\n\nsince both areas are equal:\n96 sq. in = 1/2(A)(32 in)\n96 sq. in = A (16 in)\nA = (96 sq. in)/16 in\nA = 6 inches
117140
Let's break the area into many vertical rectangular strips between x=1 and x=2, each strip so narrow that we can assume that the top of the strip is horizontal. Thus each strip will be a very narrow and tall rectangle.\n\nLet's say the number of vertical strips we use is n. The first strip will have its left-hand side at x = 1, the next will have its left-hand side at x = 1 + 1/n, the third will have its left-hand side at x = 1 + 2/n, and in general the kth strip will have its left-hand side at x = 1 + (k-1)/n.\n\nNow, what is the y value at each of these x points? Using f(x)=x² we get, for the kth point,\n\ny = (1 + (k-1)/n)² = 1 + 2(k-1)/n + ((k-1)/n)² \n\nNow the area of each strip is equal to its base times its height. The base of each strip is 1/n. So the area of the kth strip is\n\nA = 1/n * (1 + 2(k-1)/n + ((k-1)/n)²)\n\nNow we have to add the areas of all these strips from k = 1 to k = n and rearrange the terms.\n\n.....k=n\nA = Σ (1/n + 2(k-1)/n² + (k-1)²/n^3)\n.....k=1   \n\n.....k=n\nA = Σ (1/n + (2k - 2)/n² + (k² - 2k + 1)/n^3)\n.....k=1\n\nGiving all fractions the common denominator of n^3,\n\n.....k=n\nA = Σ ((n² + 2kn - 2n + k² - 2k + 1)/n^3)\n.....k=1\n\n.....k=n\nA = Σ ((n² - 2n + 1)/n^3  + (2kn + k² - 2k)/n^3)\n.....k=1\n\nThe first term in the summation sign appears n times in the sum, so we can take it out by multiplying it by n. \n\nSum going from k=1 to k=n:\nA =  (n² - 2n + 1)/n² + Σ (2kn + k² - 2k)/n^3\n\nNow let's clear the parentheses in the summation:\n\nSum going from k=1 to k=n:\nA =  (n² - 2n + 1)/n² + Σ 2k/n² + Σ k²/n^3 - Σ 2k/n^3)\n\nThe first summation contains k=1, 2, 3, ... n, so we use the formula for the sum of the arithmetic series:\n\nS = n(n+1)/2\n\nso the first summation becomes\n\nn(n+1)/n² = (n² + n)/n² = 1 + 1/n\n\nThe last summation also contains k=1, 2, 3, ... n, so we use the same formula for the sum of a series, and we get\n\nn(n+1)/n^3 = (n² + n)/n^3 = 1/n + 1/n²\n\nFor the middle summation, we have to compute the sum of k² for k=1, 2, 3, ... n, or that is the sum of 1, 4, 9, 16, ... n².\n\nThe site\n\nhttp://mathworld.wolfram.com/PowerSum.html\n\ngives the formula for the sum of the squares of a series as\n\nS = (2n^3 + 3n² + n)/6\n\nso the middle summation becomes\n\n1/3 + 1/2n + 1/6n²\n\nNow, we are almost done.\n\nLet's let the number of strips get very large, go to infinity as a limit. Let's see what each term becomes:\n\n(n² - 2n + 1)/n² becomes 1 as n gets very large,\n1 + 1/n becomes 1 as n gets very large,\n1/n + 1/n² becomes 0 as n gets very large, and\n1/3 + 1/2n + 1/6n² becomes 1/3 as n gets very large.\n\nThe area under the curve is 1 + 1 + 1/3, or 7/3.\n\nYou can see why integral calculus was invented. Using integral calculus, this problem can be solved in two steps. The integral of f(x)=x² is x^3/3. Inserting the limits of x=1 and x=2 gives\n\nA = 2^3/3 - 1^3/3 = 7/3
508171
Angle of elevation should be that raised from the ground until it is vertical (90 degrees), correct?  The wall which the ladder is placed against will have a vertical face (90 degrees).  The ladder, therefore, represents the hypotenuse of the triangle so formed.\nCosine is the length of the adjacent sided divided by the hypotenuse.  So cos 70 = X/40 or 40*(cos 70) = X.\nWith me so far?\nOnce you have X, you have the hypotenuse and 1 side of a right angle triangle and finally a use for Pythagoras's theorem the square on the hypotenuse equals the sum of the squares on the other 2 sides.\nCan you do the rest of the arithmetic?  I hope this problem was in Trigonometry, otherwise I have given you the wrong way to solve it. But it is the easier way.
448503
From the Question, it should be a triangle....\nso take take 28 X 25 / 2 should be the answer\n350 ft!
155438
Square and Rectangle= lenght*height\ntriangle=base*height divided by 2
355681
Since the triangle is Isoceles, if you split the triangle down the middle you can turn in into a right triangle. You can use the Pythagorean theorem. (a^2 + b^2 = c^2.)\n\n13 will be the hypotenuse and 5 will be the base. To find the altitude you just fill everything in. \n\n5^2 + b^2= 13^2\n25 + b^2 = 169\nb^2 = 144\nb = 12
561616
Draw a picture of the two congruent sections, which are connected at B. The exact angle between them doesn't matter so long as it's not greater than 90 degrees. You will see that if AD=CD, then angles A and C are equal. So what happens as angle A gets larger and angle C gets smaller? You should be able to see the answer for yourself.
559013
Let the breadth of each red rectangle be x\n\nSo, area of red portion = 4x + x(3-x)\n\nSo, area of white portion is (4-x)(3-x)\n\nSo, 4x+x(3-x) = (4-x)(3-x)\nor 4x + 3x -3x^2 = x^2 -7x+12\nor 4x^2-14x+12 = 0\nor 2x^2-7x+6 = 0\nor 2x^2-6x-x+6=0\nor(2x-1)(x-3)=0\n\nx=1/2 or 0.5 foot is the reasonable answer.
463762
1. So, you have two sides and included angle.  Find the 3rd side and either use Pythagorean theorem to calculate the altitude or use the tangent of 1/2 the angle and 1/2 the base.\n\n2a.  You have two sides and the included angle\n2b.  You have two angles and the included side
551725
area of the base times the height.
431092
If the bases are in the ratio of k:1, then we can say they have lengths of kx and x (kx and x have ratio k:1). Let's say the trapezoid has height h.\n\nThe mid-segment of the trapezoid cuts the trapezoid into two similar trapezoids, and has length (1/2)(kx + x)\n\nThe area of a trapezoid is (1/2)(height)(b1 + b2).\n\nThe top trapezoid has height h/2, and bases kx and (1/2)(kx + x).  It has an area \nA1 = (1/2)(h/2)(kx + (1/2)(kx + x))\n     = (h/4)(kx + kx/2 + x/2)\n     = (h/4)(3kx/2 + x/2)\n     = (hx/8)(3k + 1)\n\nThe bottom trapezoid has height h/2 and bases (1/2)(kx + x) and x.  It has an area\nA2 = (1/2)(h/2)((1/2)(kx + x) + x)\n     = (h/4)(kx/2 + x/2 + x)\n     = (h/4)(kx/2 + 3x/2)\n     = (hx/8)(k + 3)\n\nIf A1:A2 = 5:2, then \n[(hx/8)(3k + 1)]/[(hx/8)(k + 3)] = 5/2\n(3k+1)/(k+3) = 5/2\n2(3k+1) = 5(k+3)\n6k + 2 = 5k + 15\nk = 13\n(C)
451814
Several ways. The first that comes to mind is to figure the area with Heron's formula (area = sqrt(s(s - a)(s - b)(s - c)), where s = (a + b + c)/2). The area can also be figured as 1/2 base times altitude, so you know the altitudes of the triangle. The altitudes give you right triangles where you know the hypotenuse (a side of the triangle), and a leg, so you can compute the trig functions of all the angles. Then the angles can be computed as arcsin or arccos of the angle (arcsin(altitude/side))
258085
you can cross a line from one of them (A) equal (just like the line in the down ) and have the triangle just equal (B) and also for C . then yu have them in a line and we know that a line is 180 .
400396
.............6\n........______\n....../.|...........|&#92;\n...../..|...........|..&#92; 2\n..../_|_____|__&#92;\n\n\nThe triangle on the right is a 30-60-90, so the shorter side must be 2/1 = 1.  So the length of the bottom part is 6+1+1 or 8.  The height must be 1√3, so the area is 1/2(6+8)√3 = 7√3.\n\nGopal screwed up when he did 2*2cos60.  That's 2*2*1/2 = 2, not 4.
348886
If you drop a vertical line h representing the balloon's altitude, you have two right triangles.  Let the base of the 83 degree triangle be x.  Now here's where you run into a question and if the book shows a diagram, it might answer that question.  Are the stations looking toward a balloon in between them, or are they both looking in the same direction at a balloon very much farther away, and very much higher?  If it's between them, the base of the 57 degree triangle will be 15 km-x, but if it's on the same side of them both, that base will be 15 km+x.  Do you know the trick to remembering how the sides relate to the trig functions?  Ooh--aah--oh ah--ah oh--ha--ho.  You have to sound this out and it makes a distinctive but stupid sounding little chant.  For the sides opposite an angle, adjacent to the angle and the hypotenuse in a right triangle, sin=o/h, cos=a/h, tan=o/a, cot=a/o, sec=h/a, csc=h/o.  In this case we want the ratio of the height to the base, the opposite side being the height and the adjacent side being the base, so we want the tangents of the angles.  The altitude, opposite side to the angles, will be equal for both triangles.  Call it y.  x will be the base for the 83 degree triangle in both cases, but for the 57 degree triangle the base will be (15 km-x) if the balloon is between the stations or (15 km+x) if it's to the same side of both stations.  Tan 83=h/x and tan 57=h/(15-x) (between) or tan 57=h/(15+x) (same side).  Setting the heights of the triangles equal since the balloon can only be at one height: x tan 83=(15-x) tan 57 if between, x tan 83=(15+x) tan 57 if to the same side.  Solve for x.  Then plug x back into the earlier equations to get y, the altitude.  Work out the answer on your own, but I'll give you this: if the balloon is between the stations, x is between 2 and 3 and y is between 15 and 20.  If the balloon is to one side of both the stations, x is between 3 and 4 and y is between 25 and 30.
464377
The area of a trapezoid equals one half the height times the sum of the bases.
282115
Correct! It's a right triangle! Use the pythagorean theorem!\nc^2 = a^2 + b^2\nWhere c is the length of the hypotenuse (opposite the right angle), a and b are the lengths of the legs of the triangle.