I need help figuring out the area of the triangle with the following vertices.?
921650
The area of a triangle is half base times height.\n\nCould you properly define the length of the base, height and hypotenuse?
904847
The total area or the can is give by:\nArea of curved surface = 2π r h\nArea of top of can = π r²\nArea of bottom of can = π r²\n-----------------------------------------\nTotal area (A) of can = 2π r h + 2 π r²\n A = 2π r h + 2 π r²\n A - 2 π r² = 2π r h\n (A - 2 π r²) / 2π r = h\n h = (A / 2π r) - r\n\nIt's the best I can do for you based on the information given.
716347
Of three points, A,B,C\n Check the slope of A-B,  B-C, A-C.  If any of the slopes are the same, you have parallel lines and thus no triangle.  So, to do (2), do (4) first.\n  If you know the distance between A-B, B-C, A-C, you can calculate the area with the formula, so (5) also depends on (4).\n\n You write your own code.  yA isn't a site to do your homework, since if you don't do it yourself, you don't pass the class!
38263
5.196 (3 times root 3)\nDive the triangle into 3 equal triangles.  Then divide each one of those triangles into 2 triangles.  The hypotnuse of each triangle has a length of 2, and by trig. (it is a 30-60-90 triangle) the base has a length of the square root of 3 and the height is 1. Area of a triangle =(base times height)/2=(radical 3)/2.  There are 6 smaller triangles total, so multiply the area by 6=total area=3 times radical 3 (5.196)\nHope this helps, it would be a lot easier if I had a picture that I could post and it would make more sense.
917824
The triangle area formula is\n\nA = 1/2 b h\n\nA = 1/2(8)(12)\n\nA = 8/2(12)\n\nA = 96/2\n\nA = 48\n\nThe area of the triangle is 48 inches\n\n- - - - - - -s-
926515
Let x be the length for the square and 14-x for the triangle.\n\nThe area of the square is (x/4)^2 = (x^2)/16.\n\nThe height of the triangle is sqrt(3)(14-x)/6 and the base of the triangle is (14-x)/3, so the area of the triangle is\n\n1/2 bh = [(14-x)/6] [sqrt(3)(14-x)/6] = (14-x)^2 sqrt(3)/36\n\nTotal area A = (x^2)/16 + (14-x)^2 sqrt(3)/36\n\nA = [9x^2 + 4(x^2 - 28x + 196) sqrt(3)] / 144\n\nA = [(9 + 4 sqrt 3) x^2 - (112 sqrt 3) x + (784 sqrt 3)] / 144\n\ndA/dx = [2(9 + 4 sqrt 3) x - (112 sqrt 3)] / 144 = 0\n\n(9 + 4 sqrt 3) x = 56 sqrt 3\n\nx = (56 sqrt 3) / (9 + 4 sqrt 3) = 6.0895 meters  (Answer)\n\nThe area of the square is 2.32 m^2.  The area of the triangle is 3.01 m^2, and the total area is 5.33 m^2.
522898
Imagine you unraveled the cylinder, the middle would turn into a rectangle, right? so to find the surface area, take the height of the cylinder and multiply it by the circumference of the circles at the bases, then add in the area of the circles. Circumference is 2(pi)r and area is (pi)r^2.
552905
The AREA of an isosceles triangle is found by taking 1/2 the sum of the long and short parallel sides, times the height. One way to compare this with the parallelogram is to take two copies, turn one 180 degrees and glue them along the equal (nonparallel) edges. \nYou now have a larger parallelogram - with one side composed of the sum of the original short and long parallel edges. Use the usual formula for the area of this parallelogram, then divide by 2 (since it is composed of two equal copies).
156289
Yes, follow the suggestion of x60157. In addition, if you can determine the:\n\n1. height ,h, of each triangle you can use the formula for area: A = 0.5bh where b = side prependicular to the height\n\n2. sides of the triangle, then this formula can be used:\nA = Sq. root of (s(s-a)(s-b)(s-c)) where a,b,c are the sides of each triangle; s = semi-perimeter = 0.5(a+b+c)
669681
1.  Area of rt triangle whose leg is 6ft and whose\n    hypotenuse is 10ft?\n\n    Area = 24 sq. ft.  \n\n  Explanation:\n\n    First we need to find the other leg's length...\n    Using Pythagorean Theorem,\n    \n    c**2=a**2+b**2 \n       \n    where c is the hypotenuse, a is one leg, and \n    b is the other leg.\n\n    So, 100 = 36 + b**2;  b**2 = 100-36 = 64; therefore b=8.\n    Now, Area = 1/2(base*height) since the base is one leg\n    (we'll say it is a) and the height is the other leg \n    (therefore we'll name it b) \n    the Area = 1/2(8*6) = 1/2(48)= 24 sq. ft.\n\n2.   The hypotenuse is approx. 4.9 meters.\n    \n  Explanation:\n     \n     Using Pythagorean Theorem: c**2=a**2+b**2 we find that\n     c**2 = (3.5)(3.5) + (3.5)(3.5)\n     c**2 = 12.25 + 12.25\n     c**2 = 24.25\n     c = sq. rt. (24.25)\n     c = 4.9244 meters\n     hypotenuse is approx. 4.9 meters.\n\n3. hypotenuse is approx. 20 inches long.\n\n  Explanation: Area = 1/2(x)**2 \n               where x is length of one leg.\n\n     x**2 = 2*Area;\n     x**2 = 2*98\n     x**2 = 196\n     x = sq. rt. (196)\n     x = 14 inches\n\n     Now, to find length of hypotenuse, we use the\n     Pythagorean Theorem:\n\n     c**2 = a**2 + b**2; since a=b this can be rewritten so\n     c**2 = 2(a)**2\n     c**2 = 2(14)**2 -- Remember 14 is the length of a leg.\n     c**2 = 2(196)\n     c**2 = 392\n     c    = sq. rt. (392)\n     c = 19.7989\n     c is approx. 20 inches long.\n\n\n4.   The researcher will climb approx. 10 meters \n     to reach the crack.\n\n   Explanation:\n      \n      Using the Pythagorean Theorem: c**2 = a**2 + b**2\n      If we set a = 24, then c**2 = (24)**2 + b**2,\n      but we also know that c + b = 36 \n      which means c = 36 - b.\n\n      So to solve for this we need to set\n      (36-b)**2 = (24)**2 + b**2\n      1296-72b + b**2 = (24)**2 + b**2\n      1296 - 72b = (24)**2\n      1296 - 72b = 576\n      -72b = 576 - 1296\n      -72b = -720\n         b = -720/-72\n         b = 10 \n\n      And now we have our 10 meters the researcher \n      must climb to reach the crack in the tree.\n\n\n6.    Approximately a 42 inch stick is the longest stick\n      that will fit in a box that is 24 in X 30 in X 18 in.\n\n  Explanation:\n\n      First use Pythagorean Theorem to find the diagonal\n      inside the rectangle of 24 in X 30 in.\n\n      c**2 = (24)**2 + (30)**2\n      c**2 = 576 + 900\n      c**2 = 1476\n      c = approx. 38.418 in. -- we will call it 38 in.\n\n      Now, use Pythagorean Theorem to find the diagonal\n      inside the "new and invisible" rectangle that is \n      38 in. X 18 in.\n\n      c**2 = (38)**2 + (18)**2\n      c**2 = 1476 + 324\n      c**2 = 1800\n      c = sq. rt. (1800)\n      c is approx. 42.43 in.  \n      \n      We will call "c" 42 inches long.
331507
sorry hunny you need at least 2 side lengths in order for us to help you figure it out... plus once you have that the formula for figuring it out is using the pathagorean theorem! a2+b2=c2 (s = squared) say for instance... the 2 sides of your triangle are 7 and 3inches. 3>2+7>2=c>2\nwhich would turn into 9+49=c>2~~~~~~~ then 58=c>2 then to get c by itself(which is needed) you would take the square root of 58! your hypotenuse would equal 7.6 or rounded, 8!\nnow if you use my steps you should be able to do your problem! I'm a 10th grader in geometry and it's the most lovely subject!
420090
If the x and y axis are forming two of the sides of the triangle, the truangle is right angled so the area of the triangle will be 1/2 * base * height\n\nThe line through (3,2) therefore has to meet the x axis and the y axis to complete the triangle shape. \n\nConsider the gradient of the line = -1. The line would cut the x axis at (5,0) and the y at (0,5), making the triangle 0.5 * 5 * 5 = 12.5 units squared\n\nConsider gradient = -a. We'd have intersects at (3-2/a,0) and (0,2-3/a). The area of the triangle would be:\n\n0.5 * (3-2/a) * (2-3/a)\n= 0.5 * (6 - 4/a - 9/a + 6/a²)\n= 0.5 * (6 - 13/a + 6/a²)\n= 0.5 * (6a² - 13a + 6) ---- (multiply everything by a²)\n\nNow solve for minimum a by differentiating:\n\nd/da = 0.5 * (12a - 13)\n\nReaches a minimum when d/da = 0 :\n\n0 = 6a - 6.5\n6.5 = 6a\na = 1.086666\n\nNow put that back into our area equation, we get:\n\n= 0.5 * (3-2/a) * (2-3/a)\n= 11.556\n-------------
548292
Assuming that an area may be dissected into triangles, then the area of each triangle is given by \n\nArea = (s(s-a)(s-b)(s-c))^(1/2)\n\nwhere s = (1/2)(a + b + c)\n\nand a, b, and c are the lengths of sides of each triangle -\n\nThe sought for area is (finally) the sum of all the areas of the triangles.\n\nGiven sufficient patience, any arbitrary area will be found.\n\n(Drawing a diagram while slogging through the exercise and looping around the separate triangles in the same direction can yield a shortcut whereby all of the internal perimeters eliminate each other, leaving only the perimeter of the original figure.)
23271
The area of a triangle can be found by multiplying the base times the one-half the height.\n\nIf a triangle has a base of length 6 inches and a height of 4 inches, its area is 6*2=12 square inches
802049
The answer is (D).\n\nThe perimeter is 9x (just add up the 4 sides 2x+3x+3x+x to determine this) -- which eliminates (B) and (C) as a possible answer.\n\nThe area is 6x^2 -- which eliminates (A) as a possible answer, leaving only (D).\n\nLooks like you have a trapezoid, where the base is 2x, the sides are 3x each, and the top is x.  Draw straight lines down from your top to your base, so that your trapezoid is in 3 pieces:  a right triangle with base = x/2 and hypotenuse 3x, a rectangle x by 2x, and another right triangle with base = x/2 and hypotenuse 3x.  The area of the inner rectangle is 3x^2, and you can put the two triangles together to get another x by 2x rectangle for another 3x^2 area, or a total area of 6x^2.\n\nI hope that helped.\n\n~ ♥ ~
953741
Just break the pentagon up into 5 triangles.  You can then add together the area of the 5 triangles to get the total area.
778352
.............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.