How do you import a rounded rectangle from Adobe Photoshop into Adobe Flash?
Q. I am dragging it out of photoshop and dropping it into Flash, but the "rounded" rectangle is now a square rectangle. It automatically adds white corners... Is there a way around this?
Asked by Andrew A - Wed Jul 1 18:22:13 2009 - - 4 Answers - 0 Comments
A. Why don't you just create the rounded rectangle actually in flash as it will be a vector; whereas I do believe imported shapes are bitmaps unless you import a specific vector format such as eps. 1. Choose the rectangle tool. 2. In the properties you will see some curved corners, if they are locked (constrained) you need only change one corner using the slider or input box. You can unlock them and have individual values for each corner also. 3. Draw your rectangle. I don't think you can change the rounded corners after you have drawn the shape; at least not with the properties menu, you can however use the sub-selection tool (white arrow) to manipulate the vector path of any shape, holding down alt whilst moving a point will change the… [cont.]
Answered by unknown - Wed Jul 1 18:37:00 2009
Q. I am dragging it out of photoshop and dropping it into Flash, but the "rounded" rectangle is now a square rectangle. It automatically adds white corners... Is there a way around this?
Asked by Andrew A - Wed Jul 1 18:22:13 2009 - - 4 Answers - 0 Comments
A. Why don't you just create the rounded rectangle actually in flash as it will be a vector; whereas I do believe imported shapes are bitmaps unless you import a specific vector format such as eps. 1. Choose the rectangle tool. 2. In the properties you will see some curved corners, if they are locked (constrained) you need only change one corner using the slider or input box. You can unlock them and have individual values for each corner also. 3. Draw your rectangle. I don't think you can change the rounded corners after you have drawn the shape; at least not with the properties menu, you can however use the sub-selection tool (white arrow) to manipulate the vector path of any shape, holding down alt whilst moving a point will change the… [cont.]
Answered by unknown - Wed Jul 1 18:37:00 2009
What are the dimensions of such a rectangle with the greatest possible area?
Q. A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=2 x^2. What are the dimensions of such a rectangle with the greatest possible area?
Asked by S. - Fri May 1 09:47:11 2009 - - 1 Answers - 0 Comments
A. length of base: 2x height: y = 2 - x^2 A = 2xy = 2x(2 - x^2) get the derivative and set to zero find the desired x
Answered by Alam Ko Iyan - Fri May 1 11:18:09 2009
Q. A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=2 x^2. What are the dimensions of such a rectangle with the greatest possible area?
Asked by S. - Fri May 1 09:47:11 2009 - - 1 Answers - 0 Comments
A. length of base: 2x height: y = 2 - x^2 A = 2xy = 2x(2 - x^2) get the derivative and set to zero find the desired x
Answered by Alam Ko Iyan - Fri May 1 11:18:09 2009
How do you find the dimensions of a rectangle when perimeter is given?
Q. So the length of a rectangle is 1 inch less than 3 times its width. And the perimeter of the rectangle is 30 inches, what are its dimensions? How do you solve this step by step?
Asked by Cali Girl - Sat Sep 20 21:36:55 2008 - - 7 Answers - 0 Comments
A. Given: Rectangle = 4-sided, 2-dimensional figure P (Perimeter) = 30" L = Length 2x W = Width 2y x = (1/2P)-y y = (1/2P divided by3/1)-1 NOTE:(dividing 15 by 3/1 = multiplying 15 by 1/3) y =15 times 1/3, -1, = 5-1, which = 4" (= WIDTH of 1 side) x = 15-4 =11" (=LENGTH of 1side) 2x = 2 times 11" =22" 2y = 2 times 4" =8" proof: 8" (2x4) + 22" (2x11") = 30"
Answered by Ima D. Qwitta fwum Wasiwwa - Sat Sep 20 23:30:00 2008
Q. So the length of a rectangle is 1 inch less than 3 times its width. And the perimeter of the rectangle is 30 inches, what are its dimensions? How do you solve this step by step?
Asked by Cali Girl - Sat Sep 20 21:36:55 2008 - - 7 Answers - 0 Comments
A. Given: Rectangle = 4-sided, 2-dimensional figure P (Perimeter) = 30" L = Length 2x W = Width 2y x = (1/2P)-y y = (1/2P divided by3/1)-1 NOTE:(dividing 15 by 3/1 = multiplying 15 by 1/3) y =15 times 1/3, -1, = 5-1, which = 4" (= WIDTH of 1 side) x = 15-4 =11" (=LENGTH of 1side) 2x = 2 times 11" =22" 2y = 2 times 4" =8" proof: 8" (2x4) + 22" (2x11") = 30"
Answered by Ima D. Qwitta fwum Wasiwwa - Sat Sep 20 23:30:00 2008
What bangs go with a rectangle shaped face?
Q. I have a rectangle/long shaped face and a long forehead,and have tried a couple different styles of bangs that would look good with my face shape. What kinds of bangs would work, if any?
Asked by alex_scx - Sun Jun 1 14:18:52 2008 - - 1 Answers - 0 Comments
A. Have you tried the side-swept bangs? They're perfect for a long forehead and create the illusion of a smaller one. I should know, my nick-name in high school was 5 head. lol Try that.
Answered by Kristin - Wed Jun 4 12:51:44 2008
Q. I have a rectangle/long shaped face and a long forehead,and have tried a couple different styles of bangs that would look good with my face shape. What kinds of bangs would work, if any?
Asked by alex_scx - Sun Jun 1 14:18:52 2008 - - 1 Answers - 0 Comments
A. Have you tried the side-swept bangs? They're perfect for a long forehead and create the illusion of a smaller one. I should know, my nick-name in high school was 5 head. lol Try that.
Answered by Kristin - Wed Jun 4 12:51:44 2008
What is the area of a rectangle with only knowing the perimeter?
Q. I have a rectangle with the perimeter of 64, but I need to find the area. How do I do that with only knowing the perimeter and nothing else. I'm in 7th grade. So please make it easy for me to understand.
Asked by niarafresh - Mon Nov 24 16:54:57 2008 - - 5 Answers - 0 Comments
A. Well, you know that a rectangle has 4 sides, 2 longer than the other two. And you also may know that to find the area of a rectangle you multiply the width by the length. If we divide 64 into eighths, we get 8. So if we make the two longer sides 3/8 and the shorter sides only 1/8, we can know the widths and lengths of the rectangle. 8x3=24, so now we know that the length of the shape is 24 units. 8x1=8, so now we know that the width of the rectangle is 8 units. To find the area then you would multiply 8 by 24 which equals 192 units. Hope this helps!
Answered by Kasey - Mon Nov 24 17:05:59 2008
Q. I have a rectangle with the perimeter of 64, but I need to find the area. How do I do that with only knowing the perimeter and nothing else. I'm in 7th grade. So please make it easy for me to understand.
Asked by niarafresh - Mon Nov 24 16:54:57 2008 - - 5 Answers - 0 Comments
A. Well, you know that a rectangle has 4 sides, 2 longer than the other two. And you also may know that to find the area of a rectangle you multiply the width by the length. If we divide 64 into eighths, we get 8. So if we make the two longer sides 3/8 and the shorter sides only 1/8, we can know the widths and lengths of the rectangle. 8x3=24, so now we know that the length of the shape is 24 units. 8x1=8, so now we know that the width of the rectangle is 8 units. To find the area then you would multiply 8 by 24 which equals 192 units. Hope this helps!
Answered by Kasey - Mon Nov 24 17:05:59 2008
What were the dimensions of the original rectangle?
Q. A rectangle is 4 times as long as it is wide. If the length is increased by 4 inches and the width is decreased by 1 inch, the area will be 60 square inches.
Asked by twilightgirl22 - Mon Feb 2 21:54:17 2009 - - 3 Answers - 0 Comments
A. let x be the one side then other side will be 4x after altrecation width will be x-1 and length will be 4x+4 so (x-1)(4x-4)=60 => 4x^2 -4x + 4x - 4 = 60 => 4x^2 - 4 = 60 => 4(x^2 - 1) = 60 => (x^2 - 1) = 15 => x^2 - 16 = 0 => x^2 - 4^2 = 0 now using a^2 - b^2 = (a + b) (a - b) => (x - 4) (x + 4) = 0 => x - 4 = 0 & x + 4 = 0 => x = 4 & - 4 ( as x can not be -ve) so x = 4 so original dimensiones will be 4 and 16.
Answered by champtyagi - Tue Feb 3 10:46:30 2009
Q. A rectangle is 4 times as long as it is wide. If the length is increased by 4 inches and the width is decreased by 1 inch, the area will be 60 square inches.
Asked by twilightgirl22 - Mon Feb 2 21:54:17 2009 - - 3 Answers - 0 Comments
A. let x be the one side then other side will be 4x after altrecation width will be x-1 and length will be 4x+4 so (x-1)(4x-4)=60 => 4x^2 -4x + 4x - 4 = 60 => 4x^2 - 4 = 60 => 4(x^2 - 1) = 60 => (x^2 - 1) = 15 => x^2 - 16 = 0 => x^2 - 4^2 = 0 now using a^2 - b^2 = (a + b) (a - b) => (x - 4) (x + 4) = 0 => x - 4 = 0 & x + 4 = 0 => x = 4 & - 4 ( as x can not be -ve) so x = 4 so original dimensiones will be 4 and 16.
Answered by champtyagi - Tue Feb 3 10:46:30 2009
What happens to the perimeter of a rectangle if you double the length and keep the width the same?
Q. What happens to the perimeter of a rectangle if you double the length and keep the width the same? What happens to the perimeter of a regular hexagon if you double the length of each side? What happens to the circumference of a circle if you multiply the diameter by 3? Can someone help me please?
Asked by Sam - Tue May 5 15:56:43 2009 - - 2 Answers - 0 Comments
A. What happens to the perimeter of a rectangle if you double the length and keep the width the same? Let l = length Let w = width Perimeter_1 = l + w + l + w = 2l + 2w Double length Perimeter_2 = 2l + w + 2l + w = 4l + 2w Perimeter_2 - Perimeter_1 = 4l + 2w - (2l + 2w) = 2l. The rectangle will be 2l bigger. --- What happens to the perimeter of a regular hexagon if you double the length of each side? l = length of one side. Perimeter_1 = 6l Double length of each side. Perimeter_ 2 = 6(2l) Perimeter_ 2 / Perimeter_ 1 = 6(2l) / 6l = 2 The perimeter wil double. --- What happens to the circumference of a circle if you multiply the diameter by 3? Let d = diameter Circumference_1 = pi * d If multiply the diameter by 3 Circumference_2 =… [cont.]
Answered by Charles D - Tue May 5 16:30:00 2009
Q. What happens to the perimeter of a rectangle if you double the length and keep the width the same? What happens to the perimeter of a regular hexagon if you double the length of each side? What happens to the circumference of a circle if you multiply the diameter by 3? Can someone help me please?
Asked by Sam - Tue May 5 15:56:43 2009 - - 2 Answers - 0 Comments
A. What happens to the perimeter of a rectangle if you double the length and keep the width the same? Let l = length Let w = width Perimeter_1 = l + w + l + w = 2l + 2w Double length Perimeter_2 = 2l + w + 2l + w = 4l + 2w Perimeter_2 - Perimeter_1 = 4l + 2w - (2l + 2w) = 2l. The rectangle will be 2l bigger. --- What happens to the perimeter of a regular hexagon if you double the length of each side? l = length of one side. Perimeter_1 = 6l Double length of each side. Perimeter_ 2 = 6(2l) Perimeter_ 2 / Perimeter_ 1 = 6(2l) / 6l = 2 The perimeter wil double. --- What happens to the circumference of a circle if you multiply the diameter by 3? Let d = diameter Circumference_1 = pi * d If multiply the diameter by 3 Circumference_2 =… [cont.]
Answered by Charles D - Tue May 5 16:30:00 2009
What is the relationship between area of a rectangle and right triangle?
Q. It says discuss the relationship between the area of a rectangle with dimensions a and b and the area of a right triangle with legs of a and b. Anybody know what the relationship is?
Asked by Xuxi - Tue Mar 31 07:08:29 2009 - - 13 Answers - 0 Comments
A. 1/2 a b = area of triangle a b = area of rectangle it is half of a b. HALF of a rectangle with the same lengths on either side of the 90 degrees. A rectangle has length a and width b. The area will be a b. Now, suppose you have a right angle triangle with one of the side a and another side b. As this is a triangle with a right angle, a and b is lies on the same place as the a and b in the rectangle. And the area will be halved as the triangle will have a third line which goes from the end of line a to the end of line b.
Answered by Cyber Electrons - Wed Apr 1 10:48:14 2009
Q. It says discuss the relationship between the area of a rectangle with dimensions a and b and the area of a right triangle with legs of a and b. Anybody know what the relationship is?
Asked by Xuxi - Tue Mar 31 07:08:29 2009 - - 13 Answers - 0 Comments
A. 1/2 a b = area of triangle a b = area of rectangle it is half of a b. HALF of a rectangle with the same lengths on either side of the 90 degrees. A rectangle has length a and width b. The area will be a b. Now, suppose you have a right angle triangle with one of the side a and another side b. As this is a triangle with a right angle, a and b is lies on the same place as the a and b in the rectangle. And the area will be halved as the triangle will have a third line which goes from the end of line a to the end of line b.
Answered by Cyber Electrons - Wed Apr 1 10:48:14 2009
How many ways can a rectangle just an ordinary rectangle be cut in half?
Q. There are 19 ways to cut a rectangle in half according to a hint i got. The dimensions are length 4 the height are 5. It's supposed to be geometry.
Asked by jackiewnstn - Tue Jan 3 16:54:40 2006 - - 6 Answers - 0 Comments
A. There's more to symmetry even than just mirror-plane symmetry. Any line bisecting the rectangle through the center will give two rotationally symmetric pieces.
Answered by Professor Beatz - Tue Jan 3 17:04:04 2006
Q. There are 19 ways to cut a rectangle in half according to a hint i got. The dimensions are length 4 the height are 5. It's supposed to be geometry.
Asked by jackiewnstn - Tue Jan 3 16:54:40 2006 - - 6 Answers - 0 Comments
A. There's more to symmetry even than just mirror-plane symmetry. Any line bisecting the rectangle through the center will give two rotationally symmetric pieces.
Answered by Professor Beatz - Tue Jan 3 17:04:04 2006
How do I find the dimensions of a rectangle with the area?
Q. The length of a rectangle is 10 feet less than twice the width. If the area of the rectangle is 168 square feet, how to do I find the dimension of the rectangle?? Thanks
Asked by k_artiaco - Sat Sep 15 14:48:21 2007 - - 1 Answers - 0 Comments
A. w=width l=length Then l=2*w - 10 Area = l*w So 168 = (2*w - 10) * w Then solve for w
Answered by Red Ibanez - Sat Sep 15 14:56:19 2007
Q. The length of a rectangle is 10 feet less than twice the width. If the area of the rectangle is 168 square feet, how to do I find the dimension of the rectangle?? Thanks
Asked by k_artiaco - Sat Sep 15 14:48:21 2007 - - 1 Answers - 0 Comments
A. w=width l=length Then l=2*w - 10 Area = l*w So 168 = (2*w - 10) * w Then solve for w
Answered by Red Ibanez - Sat Sep 15 14:56:19 2007
How do i calculate the length of a rectangle dissected with a 10mm diagonal line and 6mm wide?
Q. How long should i make the rectangle if i have a rectangle bisected with a 10mm long diagonal line and the rectangle 6mm wide?
Asked by ugot8too - Mon Jun 1 09:24:43 2009 - - 3 Answers - 0 Comments
A. Using pythagoras: a^2 = b^2 + c^2 Where a is the lenght of the hypotenuse and b and c are the other sides at right angles to each other. Using the information: 10^2 = 6^2 + x^2 Where x is the length of your rectangle Solve for x x^2 = 100 - 36 = 64 x = sq root (64) = 8 So you should make your rectangle 8mm long.
Answered by ChristopheraX - Mon Jun 1 09:33:16 2009
Q. How long should i make the rectangle if i have a rectangle bisected with a 10mm long diagonal line and the rectangle 6mm wide?
Asked by ugot8too - Mon Jun 1 09:24:43 2009 - - 3 Answers - 0 Comments
A. Using pythagoras: a^2 = b^2 + c^2 Where a is the lenght of the hypotenuse and b and c are the other sides at right angles to each other. Using the information: 10^2 = 6^2 + x^2 Where x is the length of your rectangle Solve for x x^2 = 100 - 36 = 64 x = sq root (64) = 8 So you should make your rectangle 8mm long.
Answered by ChristopheraX - Mon Jun 1 09:33:16 2009
How do you work out the diagonal measurements of a rectangle?
Q. If a rectangle measures 2.5 metres by 3.5 metres how do you work out the diagonal measurement from corner to corner?
Asked by Colbag - Fri Feb 29 14:41:04 2008 - - 4 Answers - 0 Comments
A. By using pythagoras' theory.
Answered by judge mark - Fri Feb 29 14:45:16 2008
Q. If a rectangle measures 2.5 metres by 3.5 metres how do you work out the diagonal measurement from corner to corner?
Asked by Colbag - Fri Feb 29 14:41:04 2008 - - 4 Answers - 0 Comments
A. By using pythagoras' theory.
Answered by judge mark - Fri Feb 29 14:45:16 2008
How do I find the dimensions of a rectangle?
Q. The length of a rectangle is ten inches longer then twice the width. If the perimeter of the rectangle is 344 inches, how do I find the dimensions?
Asked by k_artiaco - Fri Sep 14 11:26:36 2007 - - 9 Answers - 0 Comments
A. P = 2L + 2W L = 2W + 10 344 = 2(2W + 10) + 2W ==> distribute 344 = 4W + 20 + 2W ==> subtract 20 from both sides and combine like terms 324 = 6W ==> divide both sides by 6 54 = W ==> plug this into the equation for length: L = 2(54) + 10 = 108 + 10 = 118 The dimensions are 118 inches x 54 inches.
Answered by C-Wryte - Fri Sep 14 11:30:40 2007
Q. The length of a rectangle is ten inches longer then twice the width. If the perimeter of the rectangle is 344 inches, how do I find the dimensions?
Asked by k_artiaco - Fri Sep 14 11:26:36 2007 - - 9 Answers - 0 Comments
A. P = 2L + 2W L = 2W + 10 344 = 2(2W + 10) + 2W ==> distribute 344 = 4W + 20 + 2W ==> subtract 20 from both sides and combine like terms 324 = 6W ==> divide both sides by 6 54 = W ==> plug this into the equation for length: L = 2(54) + 10 = 108 + 10 = 118 The dimensions are 118 inches x 54 inches.
Answered by C-Wryte - Fri Sep 14 11:30:40 2007
What is the percentage of decrease in the area of the resulting rectangle?
Q. Two opposite sides of a square are increased by 25% and the other two are decreased by 40%. What is the percentage of decrease in the area of the resulting rectangle?
Asked by mhmm - Sun Nov 23 16:26:57 2008 - - 3 Answers - 0 Comments
A. well if u say the square had sides L then its area at first is = L now, the two sides are increased by 25% = L + 25% of L = L + 0.25L = 1.25L the other side is decreased by 40% = L - 40% of L = L - 0.4L = 0.6L so now the area is = 1.25L x 0.65L = 0.75L so percentage change = (L - 0.75L ) / L x 100% = 25% hope that helped:)
Answered by Sweetgurl - Sun Nov 23 16:32:42 2008
Q. Two opposite sides of a square are increased by 25% and the other two are decreased by 40%. What is the percentage of decrease in the area of the resulting rectangle?
Asked by mhmm - Sun Nov 23 16:26:57 2008 - - 3 Answers - 0 Comments
A. well if u say the square had sides L then its area at first is = L now, the two sides are increased by 25% = L + 25% of L = L + 0.25L = 1.25L the other side is decreased by 40% = L - 40% of L = L - 0.4L = 0.6L so now the area is = 1.25L x 0.65L = 0.75L so percentage change = (L - 0.75L ) / L x 100% = 25% hope that helped:)
Answered by Sweetgurl - Sun Nov 23 16:32:42 2008
How does one solve this golden rectangle problem?
Q. IF a rectangular billboard is to have a height of 18 feet, how long should it be if it is to form a golden rectangle? Round to the nearest tenth of a foot. Thank you so much for your help! I'm lost on how to solve this problem.
Asked by Rogue - Tue Jun 9 21:19:07 2009 - - 1 Answers - 0 Comments
A. A golden rectangle is a rectangle whose sides are in the golden ratio, which is approximately 1.618. (The exact value is (1+sqrt(5))/2.) So the width of the billboard is 18 * 1.618 = 29.1 feet.
Answered by fisdof4 - Tue Jun 9 21:47:38 2009
Q. IF a rectangular billboard is to have a height of 18 feet, how long should it be if it is to form a golden rectangle? Round to the nearest tenth of a foot. Thank you so much for your help! I'm lost on how to solve this problem.
Asked by Rogue - Tue Jun 9 21:19:07 2009 - - 1 Answers - 0 Comments
A. A golden rectangle is a rectangle whose sides are in the golden ratio, which is approximately 1.618. (The exact value is (1+sqrt(5))/2.) So the width of the billboard is 18 * 1.618 = 29.1 feet.
Answered by fisdof4 - Tue Jun 9 21:47:38 2009
What is the percentage change of area of rectangle, when length is increased by 10% and width is decreased by?
Q. What is the percentage change of area of rectangle, when length is increased by 10% and width is decreased by 10%?
Asked by aruna g - Sat Apr 19 16:07:39 2008 - - 4 Answers - 0 Comments
A. 1% Say the original length was L and the original width was W.So the original area was A = LW. The new length is L + 0.1L = 1.1L The new width is W - 0.1W = 0.9W The new area is then 1.1W*0.9W = 0.99WL = 0.99A A - 0.99A = 0.01A or 1% decrease.
Answered by Cappucino - Sat Apr 19 16:17:24 2008
Q. What is the percentage change of area of rectangle, when length is increased by 10% and width is decreased by 10%?
Asked by aruna g - Sat Apr 19 16:07:39 2008 - - 4 Answers - 0 Comments
A. 1% Say the original length was L and the original width was W.So the original area was A = LW. The new length is L + 0.1L = 1.1L The new width is W - 0.1W = 0.9W The new area is then 1.1W*0.9W = 0.99WL = 0.99A A - 0.99A = 0.01A or 1% decrease.
Answered by Cappucino - Sat Apr 19 16:17:24 2008
The length of a rectangle is fixed at 11cm. What widths will make the perimeter greater than 96 cm?
Q. The length of a rectangle is fixed at 11cm. What widths will make the perimeter greater than 96 cm? The width must be greater than ___ cm.
Asked by heavenlysunriseqt - Mon Jul 28 11:02:08 2008 - - 3 Answers - 0 Comments
A. Perimeter of a rectangle = 2(length+breadth) 96 cm =2(11cm+breadth) 96cm =22cm + 2 X breadth 96cm - 22cm = 2 X breadth 74 cm / 2 = breadth 37cm = breadth So the breadth or width of the rectangle is 37 cm. to make the perimeter greater that 96 cm the width should be greater than 37 cm --- perimeter is directly proportional to width. Good luck with maths.
Answered by firebolt sruti - Mon Jul 28 11:17:17 2008
Q. The length of a rectangle is fixed at 11cm. What widths will make the perimeter greater than 96 cm? The width must be greater than ___ cm.
Asked by heavenlysunriseqt - Mon Jul 28 11:02:08 2008 - - 3 Answers - 0 Comments
A. Perimeter of a rectangle = 2(length+breadth) 96 cm =2(11cm+breadth) 96cm =22cm + 2 X breadth 96cm - 22cm = 2 X breadth 74 cm / 2 = breadth 37cm = breadth So the breadth or width of the rectangle is 37 cm. to make the perimeter greater that 96 cm the width should be greater than 37 cm --- perimeter is directly proportional to width. Good luck with maths.
Answered by firebolt sruti - Mon Jul 28 11:17:17 2008
How to build a rectangle&a triangle using a tangram?
Q. I'm learning about tangrams right now, and this is a little challenge in my class. I need a little help with building a rectangle and a triangle.
Asked by S-MOMENT - Wed Jun 7 18:06:24 2006 - - 1 Answers - 0 Comments
A. Because the tangram is a set of physical shapes, it's difficult to demonstrate how to construct shapes with it in the limited capacity of a Yahoo! answer. See the wikipedia web page I list in Source(s) below as a start. This page does show a rectangle, comprising two squares, composed out of the tangram pieces. Tangrams are a great way to put your knowledge of geometry to use. If you understand the properties of isosceles trianges, squares and parallelograms, it's easier to put the pieces together in the shapes you want. But trial and error works, too. Good luck and have fun!
Answered by ~unfolding.fire~ - Sun Jun 11 17:01:21 2006
Q. I'm learning about tangrams right now, and this is a little challenge in my class. I need a little help with building a rectangle and a triangle.
Asked by S-MOMENT - Wed Jun 7 18:06:24 2006 - - 1 Answers - 0 Comments
A. Because the tangram is a set of physical shapes, it's difficult to demonstrate how to construct shapes with it in the limited capacity of a Yahoo! answer. See the wikipedia web page I list in Source(s) below as a start. This page does show a rectangle, comprising two squares, composed out of the tangram pieces. Tangrams are a great way to put your knowledge of geometry to use. If you understand the properties of isosceles trianges, squares and parallelograms, it's easier to put the pieces together in the shapes you want. But trial and error works, too. Good luck and have fun!
Answered by ~unfolding.fire~ - Sun Jun 11 17:01:21 2006
How to build a strong bridge out of 30 wood rectangle sticks?
Q. How to build a bridge that holds weight using wood coffee stirrers? Needs to be 8 inches long, have a 4 inch long roadway that holds weight on top and the road way must be 1 in. by 4 in. or larger. I only get 30 rectangle flat wood coffee stirrers and wood glue- nothing else. I need a plan/design and tips on how to make it strong/ design ideas or links.
Asked by zebraz25 - Tue May 19 01:24:14 2009 - - 1 Answers - 0 Comments
A. Build a truss bridge. Don't leave open rectangles - put a link across the diagonal to make triangles. Make the bridge as deep as possible in the direction in which the load is applied. Make the bridge as short as possible.
Answered by lithiumdeuteride - Tue May 19 01:28:47 2009
Q. How to build a bridge that holds weight using wood coffee stirrers? Needs to be 8 inches long, have a 4 inch long roadway that holds weight on top and the road way must be 1 in. by 4 in. or larger. I only get 30 rectangle flat wood coffee stirrers and wood glue- nothing else. I need a plan/design and tips on how to make it strong/ design ideas or links.
Asked by zebraz25 - Tue May 19 01:24:14 2009 - - 1 Answers - 0 Comments
A. Build a truss bridge. Don't leave open rectangles - put a link across the diagonal to make triangles. Make the bridge as deep as possible in the direction in which the load is applied. Make the bridge as short as possible.
Answered by lithiumdeuteride - Tue May 19 01:28:47 2009
What is the formula to get the area of a rectangle?
Q. One sid is 7 and the other side is 2, what is the area of this rectangle?
Asked by regie909 - Tue Feb 27 01:33:07 2007 - - 20 Answers - 0 Comments
A. Length x Width. 7 x 2 = 14
Answered by shakensunshine86 - Tue Feb 27 01:35:24 2007
Q. One sid is 7 and the other side is 2, what is the area of this rectangle?
Asked by regie909 - Tue Feb 27 01:33:07 2007 - - 20 Answers - 0 Comments
A. Length x Width. 7 x 2 = 14
Answered by shakensunshine86 - Tue Feb 27 01:35:24 2007
From Yahoo Answer Search: 'rectangle'
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