Unit 7:Polygons & Quadrilaterals Homework 4: Rhombi and squares
1 answer:
Based on the quadrilateral given, the values of the missing measures will be:
- VU = 15
- SU = 21.21
- TV = 21.21
- SW = 10.61
A quadrilateral simply means a polygon that has four sides, edges and vertices.
Based on the information given, SU = TV and this will be:
= ✓450 = 21.21
SW = 21.21/2 = 10.61
Also, for the second shape, the values will be:
OM = 46; PN = 23, ON = 46; MN = 46.
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Unfortunately the portion of the graph shown does not show the locations of points A and B, nor does it show sufficient parts of the curve.
Fortunately, to do part C, it is not necessary to use the graph of the function, nor the locations of points A & B, since we know that t=-1 at A.
The solution is as follows:
We know that the tangent line has the equation
l : 2x -5y -9 = 0 ...................(1)
and that the parametric equation is
x = t^3 -8t; y = t^2 ..............(2)
We can obtain the intersections of the tangent line (1) with the original curve (2) by substituting (2) in (1), giving
f(t) = 2*(t^3-8*t) -5*(t^2)-9 = 0
or on simplification,
f(t) = 2t^3 -5t^2 -16t -9 = 0 ....................(3) parametric equation of intersection points
Solution of which gives t=9/2 or t=-1.
Since t=-1 is the given tangent point, we conclude that t=9/2 is point B.
Translating into rectangular coordinates for t=9/2, we get
x=t^3-8t=(9/2)^3-8(9/2)= 441/8 = 55.125
y=t^2=(9/2)^2= 81/4 = 20.25
Therefore the coordinates of B:
t=9/2, or (55.125, 20.25)
Attached is the graphics of f(t), showing that t=-1 is a tangent line, and that the solutions are t=-1, or t=9/2.
Fortunately, to do part C, it is not necessary to use the graph of the function, nor the locations of points A & B, since we know that t=-1 at A.
The solution is as follows:
We know that the tangent line has the equation
l : 2x -5y -9 = 0 ...................(1)
and that the parametric equation is
x = t^3 -8t; y = t^2 ..............(2)
We can obtain the intersections of the tangent line (1) with the original curve (2) by substituting (2) in (1), giving
f(t) = 2*(t^3-8*t) -5*(t^2)-9 = 0
or on simplification,
f(t) = 2t^3 -5t^2 -16t -9 = 0 ....................(3) parametric equation of intersection points
Solution of which gives t=9/2 or t=-1.
Since t=-1 is the given tangent point, we conclude that t=9/2 is point B.
Translating into rectangular coordinates for t=9/2, we get
x=t^3-8t=(9/2)^3-8(9/2)= 441/8 = 55.125
y=t^2=(9/2)^2= 81/4 = 20.25
Therefore the coordinates of B:
t=9/2, or (55.125, 20.25)
Attached is the graphics of f(t), showing that t=-1 is a tangent line, and that the solutions are t=-1, or t=9/2.