Question 1178014: A trapezium in which AD is parallel to BC and ∠ADC = ∠BCD = 90 °, The points A,B and C are (a,18), (12,-2) and (2,-7) respectively. Given that AB = 2BC, Find the value of a. Find the equation of AD. Find the equation of CD.Find the coordinates of D.Find the area of trapezium.
Found 2 solutions by mananth, greenestamps:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website!
D(BC )
Distance Equation Solution:
d=sqrt((2−12)2+(−7−(−2)))
d=sqrt((−10)2+(−5)
d=sqrt((100+25))
d(BC)=√125
d(AB)
Distance Equation Solution:
d=sqrt((a−12)^2+(18−(−2)^2))
d=sqrt((a^2-24a+144 +400))
d=sqrt( a^2-24a+544)
AB = 2BC
sqrt( a^2-24a+544) =10sqrt(5)
a^2-24a +544 =100*5
a^2-24a+44=0
a^2-22a-2a+44=0
(a-2)(a-22)=0
a=2 Or 22
slope of BC
Slope =y2−y1/x2−x
of AD
−7−−2/2−12
−5/−10
1/2= slope of BC = slope of AD
equation of AD
use y = mx+c
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The response from the other tutor ignores the fact that two opposite angles of the trapezium are right angles, making the trapezium in fact a rectangle.
With the two right angles, solving the problem is far easier than the method she was trying to use.
From C(2,-7) to B(12,-2) is 10 units to the right and 5 units up; the slope of CB is 5/10 = 1/2.
Then, since the figure is a rectangle, the slope of AB is -2.
AB is twice BC; to get from B to A we need to go twice as far in each direction as we went from C to B. With a slope of -2, that means we need to go 10 units left and 20 units up; that puts us at A(2,18).
ANSWER: a is 2.
You now have the slopes of all the sides of the figure, and you have at least one point on each side. There are many ways to use that information to find the equations of each side of the figure.
For the area, since it is a rectangle, you can use the Pythagorean Theorem to find the length of CB; then you now the other dimension of the rectangle is twice that length. And lastly the area is the product of the length and width.
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