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=$\boxed{\frac12}$
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