Find the area of a triangle if b = 8 in; h= 12 in.?
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-
616025
Pretend that:\nx= length of base\n\nUsing the information given:\n4x-8= length of one of the legs\n\n\nProcedure:\n2(4x-8) + x = 29 ---> (there are 2 legs- so you x2)\n8x-16 + x= 29\n9x-16= 29\n9x= 29+16\n9x= 45\nx=5\n\nResult:\nSince x=5 and 4x-8 is one of legs of the triangle, by simplying applying the x=5 into 4x-8, you can get your length of one of the equal legs.\n\n4x-8= 4(5)-8= 20-8= 12\n\nANSWER: D) 12 cm
928672
let the base of the isosceles triangle be x\nheight of the triangle=h\ngiven the radius of the circle is 4\n(4-h)^2+(x/2)^2=4^2\n16-8h+h^2+x^2/4=16\n4h^2-32h+x^2=0\nx=rt(32-4h^2)\narea=1/2*x*h\n=1/2*h[2rt(8-h^2)]\ndA/dh=0 for max\nafter finding h sub to get x
155419
the base of a triangle times the height times 2\n\nb*h*2
503124
first split the triangle in half. \nfind the height of the triangle (with the 12cm side as base)\n\n7^2 - 6^2 = (height)^2\nheight = 3.61\n\nUse formula\n\nArea = (0.5) (height)(base)\n     = (0.5) (3.61)(12)\n     = 21.66 cm squared
876382
Area = (1/2) Base x Height.\n\nBase is x, because you can choose any side as base.  All are equal.\n\nThe Height (h), half of Base (0.5x), and one Side(x) form a right angled triangle.\n\nh/x = sin 60 = sqrt (3)/2\n\nh = (sqrt(3)/2)x.\n\nArea = (1/2)x (sqrt(3)/2)x\n=sqrt(3)*x^2/4
227708
There are two ways to do it...\n\n1.> If you know the base and the height of the triangle:\n\n(Base*Height)/2\n\nso if one of the side was 12 cm., and the perpendicular on it from the point where the other two sides meet is 6 cm., the area is = (12*6)/2 = 36 sq.cm.\n\n2.> If you know all the sides:\n\nHero's formula:\n\nFirst find s (semi-perimeter) as:\n\ns=0.5*(side1+side2+side3)\n\nnow the area= root (s * (s-side1) * (s-side2) * (s-side3))\n\nso if a triangle's sides are 3,4 and 5 cm.:\n\nsemi-perimeter = (3+4+5)/2 = 6\n\narea= root (6(6-3)*(6-4)*(6-5)) = root (6*3*2*1) = root (36) = 6 sq. cm.
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.)
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...
1086020
(Click on the link below for the diagram. Expand to get a clearer view)\n\nLet the point of intersection of altitude and base be P\n\nThe length BP and PC is x.\nThe length AB = AC = (40-x-x)/2 = 20-x\n\nThen we use pythagoras theorem to find x.\nx^2 + 8^2 = (20 - x)^ 2\n40x = 336\nx = 336/40 = 8.4\n\nHence the area of the triangle\n= 1/2(base*height)\n= 1/2(16.8*8)\n= 67.2
71096
let area be A\nwidth be W\nlength be L\nnow area of rectangle is = length * breadth \n    A = L * B\n  L = A/B\n L = 3x (squared) +14x+8 / x+4\n   = (3x+2)(x+4) / (x+4)\n   = 3x+2
68795
this looks fairly accurate to me\n\n x = b*h/(b + h)\n\nx=side of the square\nwhere b is the base of the triangle\nh is the height of the triangle
1091583
The altitude of the triangle is 5 times the square root of  3, or 5 times 1.7.\nArea of a triangle = 1/2 bh, so Area = 1/2 times (10)  times (5 times 1.7) = 42.5 or more accurately 43.30127019 using the correct value of the square root of 3.
816720
area of rhombus=(1/2)d1d2\n(1/2)*16*d2=120\n8d2=120\ndividing by 8\nd2=120/8=15\nthe other diagonal is 15'
606374
Let h = height, then \nb = (1/2)h\nA = (1/2)bh\n64 = (1/2)(1/2)h*h\n64*4 = h^2\n256 = h^2\nh = 16
1084262
use pythagoras theorm,if left hand side is equal to right hand side then it is a right triangle.\n\nsince 8 is the biggest no here that means the length of hypotenues is 8. \n\nlet side 1 be a=4\nlet side 2 be b=4 sqrt(3)\nlet hypotenuese be c=8\n\naccording to pythagoras theorm,\n\n(a)*(a)+(b)*(b)=(c)*(c)\n\nL.H.S\n\n=4*4+4sqrt(3)*4sqrt(3)\n=16+48\n=64\n\nR.H.S\n=8*8\n=64\n\nsince L.H.S=R.H.S\n\ntherefore, it is a right triangle\n\nhope u understand it\nif u hv any prob u can contact me
891617
The triangle area formula is\n\nA = 1/2 b h\n\na = 1/2(5)(2)\n\nA = 5/2(2)\n\nA = 10/2\n\nA = 5\n\nThe area of the triangle is 5 feet\n\nThe answer is " a "\n\n- - - - - - - - - - - -s-