Find the missing diagonal of a rhombus if the area is 120ft2 and the other diagonal is 16 feet.?
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'
549797
enclose the ireegular shape with a rectangle and find the empty spaces. then subtract that with the area of the enclosed rectangle to get ur answer.
765211
h = -16x² + 48x + 8\n\nNow clearly it is projected from an elevated position and that position is 8 ft above the horizontal ( when x = 0 h = 8)\n\nSo the NEXT time it reaches 8 ft will be the second time it is there (on the way down)\n\n8 = -16x² + 48x + 8\nie 0 = -16x² + 48x\n= 16x(3 - x)\nSo x = 0 or 3\n\nThus the next time it reaches 8 ft is in 3 second
504740
After several attempts, it seems that it is rather impossible.  If you start in the bottom corner, you can get all but 3 squares rather easily.  If you're careful, you can have 1 remaining if you start diagonal to the corner.\n\nWhat are you trying to do this for?  Is it a website?  I'd love to try it again.
150640
Since D= (L^2 + w^2 + H^2)^1/2 then\nD^2 = L^2 + W^2 + H^2\nW^2 =  D^2 - L^2 - H^2\nW = (D^2 - L^2 - H^2)^(1/2)\nOR W = SQRT (D^2 - L^2 - H^2)\nNOW FOR L = 12, D= 13 AND H=3 YOU HAVE\nW= SQRT (169 - 144 -9) = SQRT (16) = 4
103215
With this given data, it is not possible\n\nalthough the area is totalling to be 70 for both cases (7*10 and two pieces of 8*8 + 1*6) but with one division only, this is not possible\n\nPLEASE DO LET ME KNOW IF THERE IS SOME SOLUTION (i found it very interesting)
841450
s√3 where s is the length of an edge.\n\nFirst, the diagonal of a face is (df)^2 = s^2 + s^2\nor (df)^2 = 2s^2 and df = √2s\nThen, the diagonal of the cube is a hyponenuse of a right triangle with lengths s and √2s\nSo (dc)^2 = s^2 + (√2s)^2 = 3s^2\nand dc = s√3
893957
well darling ;\nimagine this like that\n\ny^2 + ( a+b)y + ab \nyou need to find " a+b " & " ab" \nin our function \na+b = 16 & ab = 64 \n( y + a)( y + b) \n( y + 8 ) ( y + 8)\n{ ' 8 + 8 = 16 '   &  ' 8 * 8 = 64 ' }\nso\n( y + 8)^2\n\nGood Luck sweetheart.
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.
801034
Here's the answer: (I'm assuming the top of the box is counted in the area)---------   x = side of square base, h = height, S = surface area, P = perimeter length\n\nS = 2x^2 + 4xh\n\nP = 8x + 4h ---->  h = (P-8x)/4  Substituting,\n\nS = 2x^2 + 4*x*(P-8x)/4 = Px - 6x^2\n\nFor your particular case where P = 120 cm,\n\nS = (120x - 6x^2) cm^2
815860
Hello Dear \n\n- Take " √ " of eachsides\n√16y^2 =  √ 25 \n± (4y = 5)\ny = ± 5/4\n\nGood Luck dear.
777066
(x+4)(2x+4)-2x^2=88\n2x^2+12x+16-2x^2=88\n12x+16=88\nadding -16\n12x=72\ndividing by 12\nx=6\nso the dimensions of the original rectangle are  6m*12m
117044
Area equals half the base times the height.  \nA=1/2bh.  Plug in and solve!
588642
a^2 + b^2 = c^2\n\nthis assumes right angles, a and b are the length and width, c is the diagonal
471423
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
733290
We first find the most obvious altitude (the altitude to the base 6).  That is h = √(8² - 3²) = √55. (Pythagorean Theorem)\n\nThus, the area is \nA = bh/2 = 6√55/2 = 3√55\n\nNow, any combination of base and altitude will yield the same area, thus,\nbh/2 = 3√55\n\nWe need to find the length of the altitude to the base 8, so\n8h/2 = 3√55\n4h = 3√55\nh = 3/4 √55\n\nSince we know that 3/4 √55 < √55, then the length of the shortest altitude of the triangle is 3/4 √55.\n\n^_^
1152688
Use the Pythagorean theorem.\n50^2+120^2=x^2\nx^2=2,500+14,400\nx^2=16,900\nx=130