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
Since 34 only goes into 7 four times the answer is 28 with a remainder of 6:) so basically you would could by your 7s so 7, 14, 21, 28, 35 and since 35 is too much we would have to go with 28. so then we subtract 28 from 34 and that gives us 6 left over. hoped this helped:))
