Given EF is the midsegment of Isosceles trapezoid ABCD. BC = 17x,EF = 22.5x + 9 and AD = 30x + 12, find AD.?
921148
AD is larger than BC, and the two sides are opposite one another, but the legs of the isosceles trapezoid are congruent. AD and BC must be the parallel sides. The midsegment of an isosceles trapezoid is equal to the average of the two parallel sides, so (AD + BC) / 2 = EF.\n\nAD + BC = 2EF\n(30x + 12) + (17x) = 2(22.5x + 9)\n47x + 12 = 45x + 18\n2x = 6\nx = 3\n\nAD = 30x + 12 = 30(3) + 12 = 102
966570
Is it isosceles (with the legs the same length)?\n\nThen you can solve using:\na² + b² = c²\n\nc² = sqrt(52)² = 52\n\nThus a² + b² = 52\n\nIf a = b, then a² = 26\na = b = sqrt(26)\n\nOtherwise, if you don't have some other information, you can't uniquely solve for the length of the legs.  The legs could have lengths from nearly 0 to nearly sqrt(52).
973910
17.\n5x+7y=-1  (1)\n-2y+3z=9  (2)\n7x-z=27    (3)\n(3)*3\n21x-3z=81\n-2y+3z=9\nadding\n21x-2y=90 *7\n147x-14y=630\n10x+14y=-2\n157x=628\nx=4\nsub in (1)\ny=-3\nsubin (2)\nz=1\n\n\n18.\nr-s+3t=-8    (1)     \n2s-t=15       (2)\n3r+2t=-7      (3)\n\n(2)*2  4s-2t=30\n          3r+2t=-7\nadding\n4s+3r=23\n(2)*3  6s-3t=45\n       r-s-3t=-8\nadding\n5s+r=37\n4s+3r=23\n15s+3r=111\nsubtracting\n11s=88\ns=8\nsub in (2)\n16-t=15\nt=1\nsub in (3)\n3r+2=-7\nr=-3
404470
a) AX = 2.25 cm.\n\nb) area AXY upon area ABC = 0.5625\n\nc) area AXY upon area trapezium XYCB = 0.4375
814748
The obtuse angles of the trapezoids will be (in degrees) 90+invtan((57.35-29.27)/(2*27.54)), which is approximately 117 degrees.\n360 degrees minus the sum of two of these angles (126 degrees) is the interior angle of whatever regular polygon you are trying to make.\nThe interior angle of a regular hexagon is 120 degrees.\nThe interior angle of a regular heptagon (7-sided polygon) is approximately 128.57 degrees.\n\nTherefore, you could put 6 of these tables into a circular arrangement and have some extra space between the front edges of the tables, or you could put 7 of these tables into a circular arrangement (a better fit) and have a little bit of extra space between the back edges of the tables.\n\nIt would also look more interesting to have heptagonal arrangements of tables in the room.
889089
hi\nthe details which you gave have the answer for this question.\n\nangle E= angle C and angle A= angle B and DA=DB\n\nfrom this given data i will not assume that the triangles are equilateral, I will prove they are.\n\ncalculating the area of the 2 triangles with the formula (1/2)*base*height we can prove these 2 triangles are congruent and AC=BE\n\nsince the angle E=angle C and DA=DB then DB and DA must be the sides opposite these angles E and C respectively. \nhence  the line drawn(at angle of 90 degrees to base DB) from angle E to DB and line drawn(at and angle of 90 degrees to DA) from angle C to DA must be of equal magnitude (otherwise the angles cant be equal and/or sides cant be equal) ,  \n\nhence the length of this line drawn which is called as the height of the triangle is proved to be equal\n\nnow calculating the area with above formula (1/2)*base*height\nheights of 2 triangles are proved to be equal equal and bases are \ngiven to be equal hence areas of these 2 triangles must be equal.\n\nwhen area of 2 trianlges are equal and 2 sets of angles are equal and one side is equal then the 2 triangles must be absolutely congruent. hence AC must be equal to BE
824963
Given:  BX/D = CE/F\n\nFind:  "X"\n\nSolution:\nBX/D = CE/F\n(B/D)(X) = CE/F\n(D/B)(B/D)(X) = (D/B)(CE/F)\n(1)(X) = CDE/BF\n\nAnswer:\nX = CDE over BF  or  X = CDE/BF
802843
Area of a trapezoid = 1/2 Height (Base1 + Base 2)\n224 sqm = 1/2 (16)(20 + x)\nmultiply both sides of the equation by 2 to remove fractions\n448 = 16 (20 + x)\n448 = 320 + 16x\ncombine\n16x = 128\nx = 8 m
275679
qw\n 12a-43b-22c = 3a-69b+22c\n   .........+22c........... .... -22c\n  12a-43b = 3a-69b
776798
Every side of this triangle has a pair of ends and all the three pairs are cyclic. ie, if AB,BC & CA are the three sides of the given triangle , then the side AB has the ends (4,5),(1,1), BC has the ends  (1,1)and (-2,5) & CA has the pair of ends (4,5),(-2,5).\nBy simple inspection of these pairs we can find that the points C and A are at the same height from the abcissa hence parallel to it. By this we can conclude easily that the length of CA=6 units.\nFurther, to be an isocelous triangle in this case the point B should be on the mid line of CA, and it is normal to CA. The mid point of CA is three units away from its ends C&A. As CA is parallel to absissa we can clearly conclude that any point between (1,0)&(1,5) will be the vertex of an isosceles triangle. Here the point B is (1,1) on the said line. Thus proved that the given points form an isosceles triangle.\nWe alredy know that CA = 6\nAnd the other sides are equal to sqrt(4-1)^2 +(5-1)^2 = 5 units\nNow, we can find the area of this triangle by Heron's Formula \nie sqrt[8(8-6)(8-5)^2] =12 sq units
797921
"A trapezoid is a 4-sided figure with one pair of parallel sides. To find the area of a trapezoid, take the sum of its bases, multiply the sum by the height of the trapezoid, and then divide the result by 2." The height is the perpendicular distance between the two parallel bases.\n\nArea (trapezoid) = 1/2 h (b1 + b2) , where b1 and b2 are the lengths of the two parallel sides, and h is the perpendicular distance between them.
746097
Simple algebra.\n\nLeft-Hand Side\n(x/2)*.57 = 0.285x\n\nRight-Hand Side\n(x+1)/2*50 = 25(x+1) = 25x + 25\n\nTogether\n0.285x = 25x + 25\n-25 = (25 - 0.285)x = 24.715x\nx = -25/24.715 = -1.0115
945908
Hi Dear Ruth ; \n\n{ as you know Sum ( angles of trapezoid) = 360\nif ' a ' & ' b ' & ' c ' & ' d ' are angles of trapezoid so \na + b + c + d = 360 } ....\n\n▪ Step 1 \na = 6x - 4   , b = 5x + 1 , c = 4x  , d = 3x + 3\nas , a + b + c + d = 360 so ;\n6x - 4 + 5x + 1 + 4x + 3x + 3 = 360\n[ 6x + 5x + 4x + 3x  ] + [ -4 + 1 + 3 ] = 360\n18x + 0 = 360 \nx = 360 / 14\nx = 20 \n\n▪ Step 2\nas x = 20 \na = 6x - 4  ; 6(20) - 4 = 116 \nb = 5x + 1 ; 5(20) + 1 = 101\nc = 4x ; 4(20) = 80 \nd = 3x + 3 ; 3(20) + 3 = 63\nso we have ;\na = 116 ; b = 101 , c = 80 , d = 63 \n\nGood Luck dear ♣
875314
Fisrt divide the regular octagon into 8 isosceles triangle. The resulting vertex angle would be 45 deg.\nDivide the isoscles into congruent right triangles with an angle of 22.5 deg, 67.5 deg and 90 deg.\nThe length of the sides opposite 22.5 deg would b equal to\nhypotenuse (radius of the circle) times sin 22.5 deg.l\nThat would be 6.046398231 cm.\nDoubling this value would result to the side of the octagon,\nand that is 12.09279646.\nMultiplying the side by 8, we get the final answer or the perimeter of the regular octagon inscribed in the circle.\nthe answer is\n96.7423717 cm.\n\nYou my use this formula for any regular octagon inscribed in a circle with radius x.\nP = 16x sine 22.5\nContact me if this is wrong.
18379
(XA+XB)/2, (YA+YB)/2\n\n(-7+(-1))/2 = -4 as the X coordinate\n(6+1)/2 = 3.5 as the Y coordinate\n\nThe midpoint is (-4, 3.5)
278653
15: First you cut the triangle in half so the hypotenuse is 17 and the bottom leg is 8 (half of 16=8) and then use Pythagorean theorem so it would be 17^2 - 8^2 (=225) and take the square root of that (=15).\nGood luck on your midterm!
850777
Area of figure = Area of equilateral triangle + Area of arc AB + Area of arc BC.\n------------------------------------------------------------------------------------\nNow, to find the area of an equilateral triangle, use the formula\n\nA = ((s^2)/4)*(sqrt3) where s is the length of the side. Here, S= 7, so now solve the equation.\n\n((s^2)/4)*(sqrt3)\n=((7^2)/4)*(sqrt3)\n=(49/4)*(sqrt3)\n=12.25*sqrt3\n------------------------------------------------------------------------------------\nTo find the area of the arc, you need to see angle ABC as a central angle for a circle. Basically, see ABC as a sector of a circle where AB is the radius. Each angle in an equilateral triangle is 60 degrees.\n\nIn order to find the area of a sector, use the formula\n\na = (area of circle)*(d/360) where d is the number of degrees. Here, d equals 60. The area of a circle is (pi)(r^2) where r is the radius of a circle. Here, r equals 7. Now solve.\n\na = ((pi)(r^2))*(d/360)\na = ((pi)(7^2))*(60/260)\na = ((pi)(49))*(1/6)\na = 49pi * (1/6)\na = (49(pi))/6\n\nThe area of the sector is (49pi)/6. This area, however, includes the equilateral triangle. So, in order to find the area of the arc, we need to subtract the area of the triangle from the area of the sector.\n\na = ((49pi)/6) - 12.25(sqrt3)\nSimplify this on paper to get\na = (98pi - 147sqrt3)/12\n\nThe area of arc AB equals the area of arc BC. So, to get the total area of the two arcs, multiply by two to get\n\narea of two arcs = (98pi - 147sqrt3)/6\n------------------------------------------------------------------------------------\nNow add the area of the triangle and the two sectors.\n\nA = (12.25pi * sqrt3) + (98pi - 147sqrt3)/6\nSimplify this to get a total area of the figure...\n\nA = (196pi - 147sqrt3) / 12\n------------------------------------------------------------------------------------