A)
To be similar triangles have to have equal angles
triangle ZDB'
1)angle Z=90 degrees
triangle B'CQ
1) angle C 90 degrees
angle A'B'Q=90
DB'Z+A'B'Q+CB'Q=180, straight angle
DB'Z+90+CB'Q=180
DB'Z+CB'Q=90
triangle ZDB'
DZB'+DB'Z=180-90=90
DB'Z+CB'Q=90
DZB'+DB'Z=90
DB'Z+CB'Q=DZB'+DB'Z
2)CB'Q=DZB' (these angles from two triangles ZDB' and B'CQ )
3)so,angles DB'Z and B'QC are going to be equal because of sum of three angles in triangles =180 degrees and 2 angles already equal.
so this triangles are similar by tree angles
b)
B'C:B'D=3:4
B'D:DZ=3:2
CQ-?
DC=AB=21
DC=B'C+B'D (3+4= 7 parts)
21/7=3
B'C=3*3=9
B'D=3*4=12
B'D:DZ=3:2
12:DZ=3:2
DZ=12*2/3=8
B'D:DZ=CQ:B'C
3:2=CQ:9
CQ=3*9/2=27/2
c)
BC=BQ+QC=B'Q+QC
BQ' can be found by pythagorean theorem
<span>The quadrilateral ABCD have vertices at points A(-6,4), B(-6,6), C(-2,6) and D(-4,4).
</span>
<span>Translating 10 units down you get points A''(-6,-6), B''(-6,-4), C''(-2,-4) and D''(-4,-6).
</span>
Translaitng <span>8 units to the right you get points A'(2,-6), B'(2,-4), C'(6,-4) and D'(4,-6) that are exactly vertices of quadrilateral A'B'C'D'.
</span><span>
</span><span>Answer: correct choice is B.
</span>
Answer:
d - 7
Step-by-step explanation:
The word "difference" tells us that we will be using subtraction. Also, subtraction is not the same if you reverse the terms like, x - 2 and 2 - x are not the same. In your question it say "d and 7" so that's the order we use in the algebraic expression.
Answer:
C
Step-by-step explanation:
B can immediately be eliminated because it has an area of 14.
A and C have an area of 10.
So add the sides.
C:
4 + 4 + 1 + 1 + 2 + 2 + 1 + 3
= 18
A: 3 + 3 + 2 + 2 + 1 + 1 + 1 + 1 =
10 + 4 =
14
Answer:
√3 is irrational
Step-by-step explanation:
The location of the third point of a triangle can be found using a rotation matrix to transform the coordinates of the given points.
<h3 /><h3>Location of point C</h3>
With reference to the attached figure, the slope of line AC is √3, an irrational number. This means the line AC <em>never passes through a point with integer coordinates</em>. (Any point with integer coordinates would be on a line with rational slope.)
<h3>Equilateral triangle</h3>
The line segments making up an equilateral triangle are separated by an angle of 60°. If two vertices are on grid squares, the third must be a rotation of one of them about the other through an angle of 60°. The rotation matrix is irrational, so the rotated point must have irrational coordinates.
The math of it is this. For rotation of (x, y) counterclockwise 60° about the origin, the transformation matrix is ...
![\left[\begin{array}{cc}\cos(60^\circ)&\sin(60^\circ)\\-\sin(60^\circ)&\cos(60^\circ)\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}x'\\y'\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Ccos%2860%5E%5Ccirc%29%26%5Csin%2860%5E%5Ccirc%29%5C%5C-%5Csin%2860%5E%5Ccirc%29%26%5Ccos%2860%5E%5Ccirc%29%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%27%5C%5Cy%27%5Cend%7Barray%7D%5Cright%5D)
Cos(60°) is rational, but sin(60°) is not. For any non-zero rational values of x and y, the sum ...
cos(60°)·x + sin(60°)·y
will be irrational.
As in the attached diagram, if one of the coordinates of the rotated point (B) is zero, then one of the coordinates of its image (C) will be rational. The other image point coordinate cannot be rational.