How to make triangles in the Geometer's Sketchpad? Move ABC to a'b'c ', and then move ABC back from a'b'c'. Control the two movements separately

1. Make any triangle ABC, and translate point a by 0 degree to 7 cm to get point D; 2. Connect ad, take any point a 'in the line segment, select two points a', a in turn, click menu edit "move", change the label to "initial"; 3. Select two points a ', D in turn, click menu edit to move, change the label to "move"; 4. Select a, a', point

How to use Geometer's Sketchpad to animate the position relationship between line and circle

Make a straight line, and then make a circle. Find the center point when tangent. Find a center point when separated. Find a center point when intersecting. Make a button to move the center of the circle to the three points. This is the basic principle. When the circle can be dragged, you can also click the button to move the circle

How to make straight line x = a with Geometer's Sketchpad
I measured an angle (radian system), how to make a straight line x = the degree of this angle. I'm afraid it's inaccurate to draw points. Is there any way?

Draw a line, select the line, double-click the end of the line,
Click to open the transformation in the toolbar, then click to rotate, and enter the angle,
Click OK

How to write fractions in the Geometer's Sketchpad?

In calculator and function editor, you can use slash or △ sign directly. In text editing, you can use symbol panel to input more mathematical symbols. If you are in label, you need to use text and point combination to achieve label score

Can the results of Geometer's Sketchpad be expressed in fractions?

At present, only Casio's calculator can express the calculation results with fractions or irrational numbers with roots, which can't be done by the system itself in the Geometer's Sketchpad. But if you know the denominator of calculation, you can use the sketchpad function to convert it to "fractions". In 5.04, the system supports the parameter π of N, but the calculation results can't be automatically converted to fractions

Use Geometer's Sketchpad 5.0 to make a quadrilateral with the same perimeter but variable area, which is represented by animation,

Finally know how to do it. Press the following method. (a little difficult) (1) "data" new parameter, change to "perimeter = 1.00", unit: distance, OK. (for convenience, it's better to change the value of length to 8). (2) select the pop-up parameter box, right-click "mark distance", use the tool, on the screen

What if the Geometer's Sketchpad measures an angle greater than 180 degrees

First mark the angle with marker pen, then select the marker angle, right-click the mouse, properties → angle marker → marker pen → clockwise or counterclockwise

Use Geometer's Sketchpad 5.0 to make a quadrilateral with the same area and animation,
How to control constant area and movable vertex
I want to use animation to show the range of motion of the two vertices and keep the quadrilateral area constant

What do you want to embody? Simply, draw a trapezoid, and the upper bottom moves in a straight line parallel to the lower bottom

Find a point P (x1, Y1) on the parabola y = - x2 + 1 to minimize the area between the tangent of the parabola passing through the point P and the figure surrounded by the parabola and two coordinate axes
0

Let the tangent equation of P (a, b) be y-b = K (x-a)
Derivation of parabola
　　y'=-2x
　　y-b=-2a(x-a)
When x = 0, y = 2A ^ 2 + B
When y = 0, x = a + B / (2 * a)
The area bounded by tangent and XY axis s = (2a ^ 2 + b) * (a + B / (2a)) / 2
　　b=-a^2+1
Substituting
　　S＝（a^2+1)^2/4a
The intersection of parabola and X axis at (1,0)
So it's an integral with an area of 0 to 1
The size is 2 / 3
So PS = (a ^ 2 + 1) ^ 2 / 2A / 2-2 / 3
Find the extremum
Let PS' = 0, then a = radical 3 / 3
Then B = 2 / 3
So the minimum area of P (√ 3 / 3,2 / 3) is (4 √ 3-6) / 9

There is a straight line intersecting with the parabola y = x2 at two points a and B. the area enclosed by the line AB and the parabola is equal to 43. The trajectory equation of the midpoint P of the line AB is obtained

Let a (a, A2), B (B, B2) & nbsp; (a < b) then the area of the figure enclosed by the line AB and the parabola is s = ∫ Ba [(a + b) x − ab − x2] DX = (a + b2x2 − ABX − x33). Ba = 16 (B − a) 3 ∫ 16 (B − a) 3 = 43, ∫ B-A = 2 & nbsp; & nbsp; (6 points) let P (x, y) be the middle point of the line ab

How to make triangles in the Geometer's Sketchpad? Move ABC to a'b'c ', and then move ABC back from a'b'c'. Control the two movements separately