Would you accept this answer to this geometry question?
Thanks for the A2A.The original answer is creative, and admirably simple and direct, but probably contrary to the spirit of the original question (though I might add that if the student had said 'a reflection about the line y=1' rather than a rotation, it might have been accepted).The transformation I would choose would be:x' = xy' = 1 - yThis can be achieved algebraically with the following matrix transformation:Associate each point (x,y) in triangle A with a vector [x,y] (drawn from (0,0) to (x,y));Multiply each vector [x,y] by the matrix [math][a_{11}, a_{12}, a_{21}, a_{22}] = [1,0,0,-1][/math], which reflects the vector in the x-axis (the rule for (2x2) matrix multiplication is [math][x',y'] = [(a_{11}x + a_{12}y), (a_{21}x + a_{22}y)][/math], where the element [math]a_{ij}[/math] denotes the element in the ith row and the jth column of the matrix);Add the constant vector [0,1] to the resulting vector, which translates all the points in the reflected triangle 1 unit in the positive y-direction,where the 2D domain of the transformation is restricted to the set of points comprising the interior and boundary of triangle A, i.e.1 <= x <= 3, and5 - x <= y <=4 for each value of x (derived from the equation for the hypotenuse of triangle A: y = mx + c where m (the slope) = -1 and c (the intercept of the extended hypotenuse with the y-axis) = 5).Job done!(If anyone wants to render this in proper LaTeX typesetting, please feel free :o) ).
Can you answer these geometry questions? 10 points to whoever gives the right answers! X-D?
Here are the links for the figures: http://www.learningbygrace.org/geom/images/geom_34.2.a.jpg http://www.learningbygrace.org/geom/images/geom_34.2.b.jpg http://www.learningbygrace.org/geom/images/geom_34.2.c.jpg Which Figure is the set of all points in space a fixed distance away from a given point? True or False. The solids in Figures A - I are all convex. True or False. The bases of the solid in Figure G are circular. What is the length of the edge of a cube with surface area 36 cm2? (BLANK) cm What is the volume of a pyramid of height 10 cm when the base is an equilateral triangle with sides of length 12 cm? (BLANK) cm3
Can you answer this question on geometry?
First notice that the side lengths of the squares are also 1, and so they are identical. Now notice that, since the line AQ is a diagonal of square ABQR, the angle BAQ is 45 degrees. Same is true for line AP and square ASPF, so angle FAP is 45 degrees as well. This leaves you with angle PAQ being 30, because ABCDEF is a regular hexagon. (An interior angle is 120 degrees)Now, another big revelation. You know the squares are identical and that AQ and AP are diagonals, so they must be equal too! In this case they are all sqrt(2). That’s not all, the triangle APQ is now an isosceles triangle! You know how to find the area of this triangle using trigonometry! So, area of APQ is 1/2 * sqrt(2) * sqrt(2) * sin(30) = 1/2.We know that angle APQ is 75 degrees (because the triangle APQ is isosceles), and AP is a diagonal, so angle APS is 45 degrees. AP = sqrt(2) and SP = 1. We can find AS. But, let’s hold onto that for a bit. Notice the same case also applies for line AR (look at lines RQ and AQ and angle RQA). It will yield exactly the same result. So mark the triangle ARS as isosceles for now. It’d be great if this was equilateral, right? Actually, it is! Look at what we said at the beginning, an interior angle of a regular hexagon is 120 degrees. A square takes 90 of it, so angle FAR is 30 degrees. Since FAP was 45 degrees, this leaves 15 degrees. Since we know ARS is isosceles and APQ is also isosceles, we can use triangle similarity to deduce that QAS is also 15. This makes 60 degrees on RAS. So, this is actually an equilateral triangle!Luckily for us we know AR = 1. So all sides of ARS are 1. So SR = 1. Due to symmetry, if we connected SQ, the area of the new formed triangle RQS would be the same as SRP. The reason is very simple => Observe the parallelity of RS to PQ, due to the fact that angles APQ and the intersection of AP with RS are same, which is 75. Same for the other side. So if we find area of RQS we are done. But we can easily find it:1/2 * sin30 * RS * RQ => 1/2 * 1/2 * 1 * 1 => 1 / 4. Why sin30? Because of the parallelity, the square takes 45 degrees, and we already know the angle APQ is 75. So 30 is left. And RQ is a side of the square, so 1.Ratio => 1/2 / 1/4 = 2.
2 simple geometry questions?
1. Imagine the square cut by its diagonal lines, into four identical triangles. If you slide the left triangle to the right of the right triangle, and join them at the longest side, you will have a square with a side measure of 10. Move the top triangle below the lower triangle, and join them, and you have another square with a side of 10. So the combined area of these two squares is: 10 * 10 + 10 * 10 = 200. The area of the two smaller squares is equal to the area of the square inscribed in the circle. 2. The circumference of a circle is: C = 2 * Pi * R The area of a circle is: A = Pi * R * R The ratio of the circumference of the smaller circle to the larger circle is: (2 * Pi * r) / (2 * Pi * R) = 2 / 5 r / R = 2 / 5 R = 5 * r / 2 Now we know how large the radius of the larger circle is, in terms of the radius of the smaller circle. Next we need to find out the radius of the smaller circle. a = Pi * r * r = 10 * Pi r * r = 10 r = sqrt(10) R = 5 * r / 2 = (5 / 2) * sqrt(10) So the area of the larger circle must be: A = Pi * R * R = Pi * [(5 / 2) * sqrt(10)] * [(5 / 2) * sqrt(10)] A = Pi * [( 25 / 4 ) * 10] = Pi * 250 / 4 = Pi * 125 / 2 So, the final answer is C.
Can you help me with these analytic geometry questions?
The solution could be obtained defining circle center coordinates (Xc,Yc) and radius (R). Being an inscripted circle the center are where bisectors meet.One triangle vertex (1) is the intersection of x+3y=1 and x-3y=-7 which have symmetrical slopes then the visector have a constant ordinate (Y1=Yc=4/3) obtainedby getting common abcisa (X1) 1–3*Y1=3*Y1-7Another triangle vertex (2) is the intersection of x+3y=1 and 3x-y=5 which have orthogonal slopes (m and -1/m are orthogonal slopes) so the circle radius which are orthogonal to them close a square with points 2 and circle centre on one diagonal then the distance between 2 and C squared are equal to 2*R^2. By common ordinate (Y2) 3*X2–5=(1-X2)/3 then X2=8/5 and Y2=-1/5Last triangle vertex (3) is the intersection of x-3y=-7 and 3x-y=5 which have symmetry around a 45 degrees slope then the visector will have that slope andX3-Xc=Y3-Yc and we know that Yc=Y1=4/3 so from point 3 we get Xc. Lets get X3,Y3 from common ordinate (Y3) 3*X3–5=(X3+7)/3 then X3=11/4 Y3=13/4Now Y3-Yc=13/4–4/3=23/12=X3—Xc=11/4-Xc so Xc=(33-23)/12=10/12=5/6Squared distance between 2 and C is(Xc-X2)^2+(Yc-Y2)^2=(5/6–8/5)^2+(4/3+1/5)^2==(25–48)^2/900+4*(20+3)^2/900=5*23^2/900=23^2/180=2*R^2 orR^2=23^2/360 and the circle equation becomes (x-Xc)^2+(y-Yc)^2=R^2 or(x-5/6)^2+(y-4/3)^2=23^2/360
A couple of geometry questions?
The measures of the bases of a trapezoid are 22 and 28. What is the measure of the median of the trapezoid? 1. 50 2. 39 3. 36 4. 25 If QRST is an isosceles trapezoid with QR = ST, then angle Q is congruent to..... 1. Angle R 2. Angle S 3. Angle T 4. None of these
How do you answer this question?
In the figure ABD and BCD are right angled triangles.In the two triangles ABD and BCD
Can someone help me with these two geometry questions?
Can you help me find the answer and show me how you got it so i can understand? Given the volume of two cubes, find the scale factor. 6) V = 343 in^3 and V = 729 in^3 7) V = 1331 ft^3 and V = 125 ft^3
Need help with this Geometry question. will choose best answer!?
Name two methods for proving the two triangles congruent? Use one of your methods along with coordinate geometry to prove that the two triangles are congruent. Please answer the question fully and show your work :) Name 2 methods to prove congruent 2 points Use 1 method to correctly prove congruence 2 points Use correct calculations to prove congruence. 2 points ( I really need this) Thank you!!!!!!!!!!