Prove that the quadrilateral whose vertices are I(-2,3), J(2,6), K(7,6), and L(3, 3) is a rhombus.
I think in these problems the first step is to express each side as a vector. A vector is the difference between points. When two sides have the same vector (or negatives) it means they're parallel and congruent. So in a rhombus IJKL the vectors IJ and LK should be the same, as should JK and IL. That much assures a parallelogram; we check IJ and JK are congruent to complete the crowing of the rhombus.
Let's calculate these vectors:
IJ = J - I = (2,6) - (-2,3) = (2 - -2, 6 - 3) = (4, 3)
LK = K - L = (7, 6) - (3, 3) = (4, 3)
IJ = LK, so far so good
(Note: If you haven't got to vectors yet you can just show the two sides are the same length, 5, and have the same slope, 3/4, both of which can be read off the vectors.)
JK = K - J = (7,6) - (2,6) = (5,0)
IL = L - I = (3, 3) - (-2, 3) = (5, 0)
Those are the same too.
Now we have to show IJ ≅ JK
The length of IJ is the cliche √4²+3² = 5, the same as JK, so IJ ≅ JK
We showed all four sides are congruent and we have two pair of parallel sides, so we have a rhombus.
Answer:
x = 2
Step-by-step explanation:
x is the height of these triangles. Thus, we know all three sides of either triangle: height = x; base = 8/2, or 4; and hypotenuse √20.
Using the Pythagorean Theorem, we find x as follows:
x^2 + 4^2 = 20, or
x^2 = 4. Then x must be 2.
x = 2
Answer:
35 idek
Step-by-step explanation:
35 tiles
B)
70 tiles
C)
100 tiles
D)
300 tiles
Number of tiles required is 300 tiles
Given that;
Dimensions of room = 15 ft × 20 ft
Area of tiles = 1 ft. square
Find:
Number of tiles required
Computation:
Area of room = 15 × 20
Area of room = 300 ft, square
Learn more:
Answer:
D) 300 tiles
Step-by-step explanation:
20 * 15 = 300