Answer:
Step-by-step explanation:
B(2,10); D(6,2)
Midpoint(x1+x2/2, y1+y2/2) = M ( 2+6/2, 10+2/2) = M(8/2, 12/2) = M(4,6)
Rhombus all sides are equal.
AB = BC = CD =AD
distance = √(x2-x1)² + (y2- y1)²
As A lies on x-axis, it y-co ordinate = 0; Let its x-co ordinate be x
A(X,0)
AB = AD
√(2-x)² + (10-0)² = √(6-x)² + (2-0)²
√(2-x)² + (10)² = √(6-x)² + (2)²
√x² -4x +4 + 100 = √x²-12x+36 + 4
√x² -4x + 104 = √x²-12x+40
square both sides,
x² -4x + 104 = x²-12x+40
x² -4x - x²+ 12x = 40 - 104
8x = -64
x = -64/8
x = -8
A(-8,0)
Let C(a,b)
M is AC midpoint
(-8+a/2, 0 + b/2) = M(4,6)
(-8+a/2, b/2) = M(4,6)
Comparing;
-8+a/2 = 4 ; b/2 = 6
-8+a = 4*2 ; b = 6*2
-8+a = 8 ; b = 12
a = 8 +8
a = 16
Hence, C(16,12)
A quadrilateral with only 2 right angles could be a trapezoid so Balavan is wrong. Denna is right because if 3 angles are right angles the the fourth angle must also be a right angle and has to be a square or rectangle.
4y.
Is the answer
Hope this helps
Yes compund interest is always greater than simple interest
simple interest = principal
compound interest = principal + accumulated interest
