Geometry problem (simple)?
832160
The angles formed by crossing lines are supplementary (add to 180º).\n\nSo it is a simple matter of subtracting:\n180º - 145º = 35º\n\nThe acute angles will each be 35º.
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
586493
Quadrilateral..........D is the correct answer.....
809433
1.rectangle\n\n2.trapezoid
1090901
In this problem, you simply substitute "a" and "b" into the equation. The answer is 110 mm. Work shown:\n\n4a + 2b = perimeter\na = 10 mm\nb = 35 mm\n4(10) + 2(35) = perimeter\nPerimeter = 110 mm\n\n:)
572318
1) Let x:the length , y:the width of the picture.Note:x,y>0\nThe first equation we have is : x=y+4 (1)\nThe combined area has the length is x+4,the width is y+4       \nso the second equation we have is: (x+4)*(y+4)=140 (2)\nReplace (1) in (2)-->(x+4)x=140--> x^2+4x-140=0 -->x=10 or x=-14 .since x>0 then x=10(cm)-->y=6(cm).\n2)  Let x: the length,y:the width.Note:x,y>0\n    The first equation we have is x=y+3 (1)\n    The second is: the area=x*y=550 (2)\n    Replace (1) in (2)--> (y+3)y=550--> y^2+3Y-550=0 --> y=22 or y=-25.Since y>0 then y=22(cm)-->x=25(cm)\n3)  Let x,y:two positive integers.Supposed that x>y>0 \n    we have two equation:x*y=240(1) and x=y+8(2)\n    Replace (2) in (1)-->y(y+8)=240--> y^2+8y-240=0 -->y=12 or y=-20.Since y>0 then y=12-->x=20.
1155045
You are dealing with a triangle.  It is not lying flat on the ground, but all three lines are in the same plane.  One side is 350 ft, another is 400 ft, and the angle between them is 50 degrees.  This sounds like a Law of Cosines problem to me.\n\nCalling the unknown side x,\nx^2 = 350^2 + 400^2 - (2)(350)(400)(cos 50)\nLooks like 320.2 ft to me.
1099275
Oh c'mon! Smiddy hasn't had geometry for 20 years! but this is basic. Pythagorean theorem states that on a triangle A-squared plus B-squared = C-squared where C is the hypotenuse. The hypotenuse of a triangle is the diagonal bit if it's two halves of a square and we know by definition that on a square A and B are equal. Therefore 2X squared = 8-squared or 2x^2 = 64\n\nx^2 = 32\nx = sq.rt 32\n\nthe area of the original square is x times x so you're back at 32 square centimeters
906875
The trick is turning the word problem into equations.\n(Key: length = l; width = w)\n\nThey tell you the length is 3 feet more than twice the width, so:\nl = 2w + 3\n\nThen they tell you the perimeter (which is twice the width and twice the length) is 36, so:\n2l + 2w = 36\n\nYou should be able to take it from there.
711217
Assume the rectangle has dimensions a x b\nDraw a radius to one of the verticies of the rectangle.\nThen drop a perpendicular from the center of the circle to the midpoint of one of the sides of the rectangle.\nThis forms a right triangle with sides a/2, b/2 and hypotenuse 5.\n(see picture: http://mudandmuck.com/str2/optimization.JPG )\n\nSo (a/2)^2 + (b/2)^2 = 5^2 = 25\nor a^2/4 + b^2/4 = 25\na^2 + b^2 = 100\nand b = √(100 - a^2).\n\nNow we want to maximize the area of the rectangle..\nThe area y = ab = a√(100 - a^2)\n\nTake the derivative with respect to a and set = 0.\n\ndy/da =  √(100 - a^2) +  a * (1/2) * (100 - a^2)^(-1/2) * (-2a)\n\ndy/da = √(100 - a^2) - a^2/√(100 - a^2)  = 0\n\nmultiply both sides by √(100 - a^2) to get:\n\n100 - a^2 - a^2 = 0\n2a^2 = 100\na^2 = 50\na = √50 = 5√2\nb = √(100 - a^2) = √50 = 5√2\n\nSo the rectangle is indeed a square of side length  5√2
735406
Hint: the three side are equal long.\n\nTh
932841
This might help --> http://www.uwm.edu/~ericskey/TANOTES/Geometry/node15.html
87402
a: width\nb: length\nperimeter = 2 x (a+b)= 420 ---> a+b = 210\nwe need to find max (a x b)\n\na= 210-b ---> area = a x b = b x (210-b) = 210 - b2 = - b2 + 210b = - b2 + 2 x 105 x b - 105squared + 105squared\n= - (b2 - 2 x 105 x b + 105squared) + 105 squared\n= - (b - 105)squared + 11025\n\nSince (b - 105) squared is always >= 0,\n - (b - 105)squared is always <= 0\n  - (b - 105)squared + 11025 is always  <= 11025\n\nthe largest area will be 11025 squared feet. This happens when - (b - 105)squared = 0, which means b = 105.\n\na = 210 - b = 105\n\n\nthe shortest way is to say that a rectangle has the largest area when it's a square, so each side would be 420/4 = 105.
824775
The Area of a triangle is half the area of a square. So take one verticy and take a 90degree line from that up to the other corner.\n\nOh cr@p I'm starting to forget it all now. Too long ago.\n\nIt's something like that.
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).
1017007
Ive noticed you were asking quite alot of these geometry questions. I think you should pay more attention in class and you wouldnt need to ask home work questions, and plus, just knowing the answer wont help, youll have to know how to do it, especially if this is for a final exam...
323670
measure each diagonal, come up with the area of one triangle X it by 4, then work out the area of the base simple!