Answer:
Step-by-step explanation:
A
<span>Triangle ACD is similar to triangle ABD</span>
For line B to AC: y - 6 = (1/3)(x - 4); y - 6 = (x/3) - (4/3); 3y - 18 = x - 4, so 3y - x = 14
For line A to BC: y - 6 = (-1)(x - 0); y - 6 = -x, so y + x = 6
Since these lines intersect at one point (the orthocenter), we can use simultaneous equations to solve for x and/or y:
(3y - x = 14) + (y + x = 6) => 4y = 20, y = +5; Substitute this into y + x = 6: 5 + x = 6, x = +1
<span>So the orthocenter is at coordinates (1,5), and the slopes of all three orthocenter lines are above.</span>
Remark
A kite is constructed such that AB = BC and AD = DC. AB = sqrt( (1/2)AC + 18^2) see diagram. AD = sqrt(24^2 + 32^2)
Step One
Solve for AB
1/2 AC = 24 (AC is given as 48)
18 is a given length
AB = sqrt(24^2 + 18^2) = sqrt(576 + 324) = sqrt(900) = 30
Step Two
Find the length of AD
AD = sqrt(32^2 + 24^2) = sqrt(1024 + 576) = sqrt(1600) = 40
Step Three
Find the Perimeter.
P = 2 * 30 + 2*40 = 60 + 80 = 140
P = 140 <<<<< Answer
