+569 ok finally got this working again, so going to post multiple questions again.
This is review for a test, which is why I am posting more than usual.
I am also not very good at proof problems (last two, so I would appreciate if someone took the time to show me how.
1. AD and BC are the two bases of trapezoid ABCD, whose side lengths are labelled in the diagram. The extensions of AB and DC intersect at P. Find [PAD].
2. Chords AC and BD intersect at P. We know BA = 10, AP = 4, and PD = 6. We can then conclude DC = ? . Prove this conclusion.
In parallelogram ABCD, M and N are the midpoints of AB and CD, respectively. P is the intersection of AC and DM. Q is the intersection of AC and NB. We can then conclude that PQ/AC = ?. Prove this conclusion.
+2407 Hi iamhappy again,
Good luck on your test. :))))))))
I'm gonna attempt problem 1.
Let's try to find PA.
(27-15)/20*27 = 22.5
We can find the height by using pythagreon theorum, sqrt(22.5^2-13.5^2 ) = 18.
18*13.5*2/2 = 243.
I hope this helped.
=^._.^=
+592 1.
we see that half of the original triangle is a 3-4-5 triangle, so $\frac{4\cdot\frac{13.5}3\cdot27}2=\boxed{243}$
2.
we know that two triangles are the same if they share the same intercepted arc and that one of the angles are similar, so the triangles are similar. this means $\frac46=\frac{10}{CD}$
$4CD=60$
$CD=15$
(I'll do third after class)
+592 1.
we see that half of the original triangle is a 3-4-5 triangle, so $\frac{4\cdot\frac{13.5}3\cdot27}2=\boxed{243}$
2.
we know that two triangles are the same if they share the same intercepted arc and that one of the angles are similar, so the triangles are similar. this means $\frac46=\frac{10}{CD}$
$4CD=60$
$CD=\boxed{15}$
3.
we can draw a line MN, that intersects at point R, which is also the midpoint of AC and NM. PQ/AC=PR/AR.
we see that PRM is similar to DPA because of the angles, and we know R is the midpoint of NM which is parallel to AD, so triangle DPA's lengths are 2 times the lengths of PRM, so PR/AP=PQ/AC=$\boxed{\frac12}$
