Diagonal of a Rhombus are perpendicular & intersects in their middle point:
Assume the diagonals intersects in H
A(0,-8), B(1,-0), C(8,-4) & D(x, y) are the vertices of the rhombus and we have to calculate D(x, y)
Consider the diagonal AC. Find the coordinate (x₁, y₁) H, the middle of AC
Coordinate (x₁, y₁) of H, middle of A(0,-8), C(8,-4)
x₁ (0+8)/2 & y₁=(-8-4)/2 ==> H(4, -6)
Now let's calculate again the coordinate of H, middle of the diagonal BD
B(1,-0), D(x, y)
x value = (1+x)/2 & y value=(y+0)/2 ==> x= (1+x)/2 & y=y/2
(1+x)/2 & y/2 are the coordinate of the center H, already calculated, then:
H(4, -6) = [(1+x)/2 , y/2]==>(1+x)/2 =4 ==> x=7 & y/2 = -6 ==> y= -12
Hence the coordinates of the 4th vertex D(7, -12)
Answer:
(-2,-5)
Step-by-step explanation:
point A=(-8,5)
point M=(-5,0)
The difference between them is that point m is 3 to the right and 5 down. We have to do this again to find point b
Point B= (-2,-5)
Answer:
SA = 251.33 yd ^2
V = 301.59 yd ^3
Step-by-step explanation:
Right cylinder 2π = 6.28318530718
Solve for surface area
A =2πrh+2πr2 = 24 x 2π + 2π x 16
A =150.796447372 + 100.530964915
A ≈251.33 yd^2
Right cylinder π = 3.14159265359
Solve for volume
V = πr2h = 16 x 6 = 96π
V = 3.14159265359 x 96
V = 301.592894745
V= 301.59 yd ^3
<span>0.00168918918
this is a great answer for a calculator! </span>
If we are solving for x then it would be -1 :)
