Hard sat questions !!!!

fogcity
Thu, 17 Nov 2011 17:05:46 +0000

For 2, draw a perpendicular from R to the base PT. Assume it intersects the line PT at point X. For the purpose of calculating the areas think of PT as the base of the large enclosing triangle RQT and RX as its height. So the area of the large enclosing triangle RQT is (1/2) PT*RX.

You need to recall a bit of geometry to take the next step. The triangles RPT and RQS are “similar”. They have the same angles, and therefore their sides (including their height) have the same relative proportions – i.e. the small triangle’s sides are 1/2 of those of the large triangle, and the same for the heights.

If this isn’t obvious, you need to recall a simple theorem regarding the angles cut by the lines through the midpoints. Look this up.

Then the area of the small triangle (above the trapezoid) is (1/2) (PT/2)*(RX/2) or 1/4 of the area of the large triangle. That means that the trapezoid’s area is 3/4 that of the large enclosing triangle.

If you forgot all about similar triangles then you can still get this problem. Note that no where are the lengths of the sides of the triangle specified. That suggests that the ratio is most likely independent of the lengths! So pick a simple 3-4-5 right triangle for RPT (and hence area 6), and use the Pythagoras’s theorem to figure out the sides of the small triangle above. You’ll quickly see that The sides of the small triangle are 1.5, 2, and 2.5 (and hence area 1.5). The trapezoid’s area is the difference, 4.5.