New York City SHSAT 2017

(Marvins-Underground-K-12) #1
fraction of the large triangle’s area is shaded.

All of the small triangles are congruent and have the same area. There are 10 unshaded and 6 shaded triangles. This means
that 6 out of the 16 small triangles are shaded and that of the area of triangle XYZ is shaded. This gives you a shaded area
of

C
The height of a shape can be drawn anywhere as long as it is perpendicular to the base. Be strategic about where you put it.

Area = b × h (the same as l × w for a rectangle, but the former formula applies to many more shapes). You are given the base
of this rectangle (PO = 6), but you need to find the height. Draw a perpendicular line from point Q to and label the point
R as shown below:

Now look at right triangle PQR. In an isosceles triangle, the altitude bisects the base, so bisects Therefore PR =

Hypotenuse has length 5, so this is a 3:4:5 right triangle. Therefore, segment must have length 4. This is equal to the
height of rectangle MNOP, so the area of the rectangle is b × h = 6 × 4 = 24, (C).

49.


H


This is a strange looking figure, but you should immediately notice two important parts of the figure. Intersecting lines tell you
that vertical angles will come into play, and triangles with two angles given tell you that you will use the fact that the sum of
the interior angles equals 180 degrees.

50.

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