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I have this problem with special triangles. I have read the lesson over and over again...
I have this problem with special triangles. I have read the lesson over and over again but I still can't seem to grasp it.
Here are some sample questions that I don't understand:
Find the length of each leg.
Find the exact perimeter of quadrilateral ABCD. AE=12, CE=20, AF=EC, AE=FC,.
Triangle PQR is a 30-60-90 triangle with right angle Q and longer leg PQ. Find the possible coordinates of R if P(-4,1) and Q(6,1). Hint: There are two solutions.
You don't have to give me the answers. I just want to know how to go about solving it.
2 Answers | add yours
High School Teacher
The given quadrilateral is AECF , as per the measurement,
AE=12, EC=20, given. CF=AE=12 and FA=EC =20
Therefore, the perimeter= AE+EC+CF+FA=12+20+12+20=64 units.
To find coordinates of R:
The coordinates P qnd Q is known. By distance formula find PQ. Since PQ is known right angled triangle PQR, with angles P=30 and Q=90, QR must be half the PR.
See the Y coordinates of P and Q . They are same. Therefore, PQ || to X axis.
Q being a right angle, QR must be || to Y axis . Therefore the the X cordinate of R remains the same as that of Q , Y cordinate of R is either Y cordinate of Q + (1/2)PR or Cordinate of Q - (1/2) PR.
Posted by neela on June 18, 2009 at 4:42 AM (Answer #1)
Special right triangles are special because their components have specific relationships. For example, a 30-60-90 degree triangle has a specific relationship between its length of sides. That means if you know one leg of the triangle, you can always find the other legs.
If you have the coordinates of R & Q (as you do), you can figure out how long RQ is, which is 10. You'd then have the length of the longer leg, and since the longer leg is twice the shorter leg, you'd know the shorter leg is 10 divided by the square root of 3. You'd calculate that, then count that distance up or down from Q to get the two possible points.
Posted by gbeatty on February 26, 2009 at 4:09 AM (Answer #2)
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