How do you figure out the height of a triangle if you know the length of the sides?
411926
The height of a triangle is defined as the perpendicular distance from an angle to a side.  Now, since this is a right triangle, and depending on how it is situated, the height could be 10" or 12" if you observe from the two acute angles.  But from the right angle, the height goes from the angle to AC (your hypotenuse).  Finding this one will be tricky but can be done using the Pythagorean Theorem.\n\nIf you drop a height from the right angle to the hypotenuse, it breaks it into two parts.  The length of one part is x (since we don't know and since the height doesn't perfectly bisect the hypotenuse), and the length of the other part is 15.62-x (since the total length is 15.62").  It really doesn't matter which side is x, but for my explanation, I'll use x as the side closer to the leg that is 10".  This height creates two new right triangles, and you can use the Pythagorean Theorem to find out both the height and the missing side x.\n\nOne formula will end up being 10^2 = h^2 + x^2, and the other formula will end up being 12^2 = h^2 + (15.62-x)^2.  You can solve the first equation to find that h^2 = 100 - x^2 and substitute that in to the second equation:\n\n12^2 = h^2 + (15.62-x)^2\n144 = 100 - x^2 + 243.9844 - 31.24x + x^2\n144 = 343.9844 - 31.24x\n-199.9844 = -31.24x\nx = 6.4 (rounded)\n\nKnowing this, we can find that missing height:\nh^2 = 100 - x^2\nh^2 = 100 - (6.4)^2\nh^2 = 100 - 40.98\nh^2 = 59.02\nh = 7.68"\n\nSo it all depends on the specific height you're looking for - it's either 10", 12", or 7.68".\n\nAs for your trough, surface area is the sum of all the areas.  Break your trough into parts that you can find areas of.  If you're just looking for the outer surface area, for example, you have two triangles for the ends and two rectangles for the sides (assuming that it's an open trough).  The area of the triangles is A = 0.5bh, and the area of the rectangle sides is A = bh (b in this case will be a side of the trianlge, 2, and h will be the length of the trough, 16).  So just add up all the surfaces that will comprise your trough and you should have your surface area.  (I'm not sure if you need to include the inside as well, but if so you need 2 triangles and 2 rectangles for the outside, and 2 triangles and 2 rectangles for the inside as well.)
847510
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))
599733
I believe it would be one-half times the length of each diagonal.\n\nIn other words, if the diagonals are 4 & 6 the area would be:\n\nA = 1/2 * 4 * 6 = 12\n\nIf you don't know the diagonals, but you know the length of the sides, you could construct the diagonals and calculate their dimensions using the Pythagorean theorem:  c^2 = a^2 + b^2, and then solve from there.  Keep in mind that you'll have to do some extra geometric analysis to find the length of each leg of the right triangles that are created by the intersection of the diagonals before you can use the theorem.
324878
I'm assuming you are asking for a formula to calculate the area when you just know the base (x) and not the height.  But, you need to know the height, even if the triangle is a right triangle.  To see this, imagine you had a right triangle standing on it's shortest side.  That shortest side would be your base, x.  Now stretch the triangle taller and taller.  The base length still stays the same, but the area inside the triangle grows and grows.  (And the triangle is still a right triange.) So, unless you know the height, you can't know the area.\n\nNow, what I'm guessing, is that your triangle might actually be not only a right triangle, but an isosceles triangle too, meaning that 2 of the sides are the same length.  And because it's a right triangle too, that means the two shorter sides (that connect at the right angle) are the same length.  So, your base and your height are the same length.\n\nIf this is true...\n\nlet x = length of base\nsince triangle is isosceles and right, height also = x\n\nSo the traditional formula becomes:\n1/2 BH = 1/2 x*x\n(sorry, can't figure out how to write "x squared" right now)\n\nHope that helps...
483588
Trignometry could be made interestingly easy.\nUnderstand the basics:\nThe longest side of triangle is hypotenuse.\nThe side of the triangle opposite the angle(your angle) is the perpendicular.\nthe remaining side is called the base.\n\nsine theta=perpendicular/hypotenuse\ncosine theta=base/hypotenuse\ntangent theta=sine theta/cos theta=perpendicular/base\n\nAll other formulas are based on these relationships.\nlike sine square theta+cosine square theta=1 etc.
400701
what you want to do is draw an isosoles triangle with the 2 congruent angles the size you want. \n\nStart with a base with a certain length (use the compas to measure, say 4 inches).  Next determine the midpoint by marking arcs nboth above and below your line from each endpoint using the same compass diameter, then drawing a line through the two intersections.  You now have a right abngle at the midpoint.  You can now determine how tall your triangle must be using the sin table and use your compass to mark the point on the vertical line to complete the triangle with the angle you want.
443635
You can use Heron's formula if you know the length of all three sides. \nHeron's Formula:\nLet s be the semiperimeter (which is half the perimeter) and a,b,c be the length of the sides, then the area is\nA = sqrt[s*(s-a)*(s-b)*(s-c)]\nperimeter = 4+4+6 = 14\ns = 14/2 = 7 (because s is one-half the perimeter)\nA = sqrt[7(7-4)(7-4)(7-6)]\n= sqrt(7*3*3*1)\n= sqrt(63)\n= 3*sqrt(7) which is approximately 7.93725\n\nYou can also use the Pythagorean Theorem to find the height.
188413
Draw a picture.  You have 2 right triangles, with the satellite in the middle perpindicular to the ground.  Call this height "x"\n\nThen you have a right triangle coming off this height going off to the left with a lower-left angle of 37 degrees.  Another right triangle to the right with a lower-right angle of 57 degrees.  This will make one big triangle, with the height of the satellite coming down from the top vertex to somewhere along the bottom side.\n\nYou know that the entire length of this bottom side is 54, but you don't know how to divide it up.  Call the distance from the first observer (on the left) to the tower "y", and then the remaining distance will be "54-y"\n\nNow you have 2 variables, and enough information to create 2 equations:\n\ntan(31)= x/(54-y)\ntan(37)=x/y\n\nBy multiplying both sides of the second equation by "y" you get y*tan(37)=x; now you can substitute y*tan(37) back into the first equation\n\ntan(31)=(y*tan(37))/(54-y)  now it's just simple algebra.  But be careful, once you've solved it you have solved for y, so be sure to go back and find x.
303035
If the triangle is a right triangle, you can find the other side by Py's Th and then by the formula,(A=1/2bh).\nIf it is not, you should know the angle they make each other or an adjacent angle. Only if, you can find the area by sine laws or cosine laws.
155419
the base of a triangle times the height times 2\n\nb*h*2
990570
You could use the law of cosines to find the lengths of the diagonals.\n\nThe measure of interior angle B = 128 4/7 degrees.\nAC^2 = AB^2 + BC^2 - 2(AB)(BC)(cos B).\nSince the heptagon is regular, BC = AB, so you can substitute:\nAC^2 = 2AB^2 + - 2(AB)^2(cos B)\nAC^2 = 2AB^2(1 - cos B)\nAC^2 = 2AB^2(1 - cos (128 4/7))\nAC = sqrt [2AB^2(1 - cos (128 4/7))]\n\nThe measure of angle BCD = 128 4/7 degrees.\nThe measure of angle BCA = 25 5/7 degrees because that triangle is isosceles.\nTherefore, the measure of angle DCA is 102 6/7 degrees.\n\nAD^2 = AC^2 + CD^2 - 2(AC)(CD)(cos DCA)\n\nAnd so on...\n\nCheating with a calculator, it's this:\n\nAC = sqrt [2AB^2(1 - (-0.6234898))]\nAC = sqrt [2AB^2(1.6234898)]\nAC = sqrt [AB^2(3.2469796)]\nAC = sqrt [3.2469796](AB)\nAC = 1.8019377358(AB)\n\nThe measure of angle BCD = 128 4/7 degrees.\nThe measure of angle BCA = 25 5/7 degrees because that triangle is isosceles.\nTherefore, the measure of angle DCA is 102 6/7 degrees.\n\nAD^2 = AC^2 + CD^2 - 2(AC)(CD)(cos DCA)\nAgain, CD = AB and you've already found AC, so substitute:\nAD^2 = [1.8019377358(AB)]^2 + AB^2 - 2[1.8019377358(AB)](AB)(cos DCA)\nAD^2 = 3.2469796(AB^2) + AB^2 - 3.60838754716(AB)^2(cos DCA)\nAD^2 = 4.2469796(AB^2) - 3.60838754716(AB)^2(cos DCA)\nAD^2 = 4.2469796(AB^2) - 3.60838754716(AB)^2(cos 102 6/7)\nAD^2 = 4.2469796(AB^2) - 3.60838754716(AB)^2(-0.22252)\nAD^2 = 4.2469796(AB^2) + 0.8019377358(AB)^2\nAD^2 = 4.2469796(AB^2) + 0.8019377358(AB)^2\nAD^2 = 5.04891734(AB)^2\nAD = sqrt [5.04891734(AB)^2]\nAD = 2.2469796(AB)\n\n1/AC + 1/AD = 1/1.8019377358(AB) + 1/2.2469796(AB)\n1/AC + 1/AD = 2.2469796/4.04891734(AB) + 1.8019377358/4.04891734(AB)\n1/AC + 1/AD = 4.04891734/4.04891734(AB)\n1/AC + 1/AD = 1/AB\n\nI like this problem, and I'll give it a shot without a calculator later this afternoon.
766258
The length of the wire must be (2x5)+(2x7)=24cm.  So each side of the triangle must be 24/3=8cm. long.
605775
Are you talkin about a trapezium or a trapezoid.\nif its a trapezium,here's how : \nlet the height be "h".\ndrop two perpendiculars equal to h on the base.the lengths b&#92;w the perps.is 6 units.therefore the rest of the length is 10.each individual length to the left of the perp is 5 units.\nin the triangle with the diagonal,the side adjacent to the angle with the base and diagonal is 6+5=11\ntherefore the angle is cos^-1(11/12)=a\ntherefore tan a=h/11\ntherefore h=11tan a\nanswer is height,h=4.795831523
778088
Perimeter means the total length or distance around the sides of the triangle, so if you add all three measurements you get D - 18 inches.
748059
First use the Pythagorean Theorem.\n\nThe hypotenuse is 20. One of the legs has length x. Call the length of the other leg y.\n\nx^2 + y^2 = 20^2\nx^2 + y^2 = 400\n\nNow solve for y in terms of x.\n\ny^2 = 400 - x^2\ny = sqrt(400 - x^2)\n\nNow, since the triangle is right, the two legs are the altitude (or height) and the base.\n\nA = (1/2)*b*h\nA = (1/2)*x*y\nA = (1/2)*x*sqrt(400-x^2)\n\nThere's the area in terms of x!
385120
When you are trying to find out various measurements of a right triangle (angle measurements or lengths of sides). Try this mnemonic (memory device):\n\nSaddle Our Horses\nCanter Away Happily\nTo Other Adventures
833189
Area = 5 square meters\nBase = 2x\nArea of any triangle = (Base*Height)/2\n5 = (2x*Height)/2\n5 = x*Height (The 2's cancel each other out)\nx*Height = 5\nHeight = 5/x(Divide both sides by 'x')\n\nI hope that helped