CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 387
Concept Review 5
Your diagram should look something like the one
above, although the rectangle can also have dimen-
sions of 1 and 32 or 2 and 16.
- Think of the area of ∆FDBas the sum of the areas
of∆DCB,∆FDC,and∆FCB.Each of these trian-
gles has an area that is half of the rectangle, be-
cause each one has the same base and height as
the rectangle. (If you have a tough time seeing
this, think of BCas the base of ∆FCBandCD
as the base of ∆FDC.Since these triangles are ob-
tuse, their heights are “outside” the triangles.)
Therefore, the area of ∆FDBis 16 + 16 + 16 =48. - We just found that ∆FCBhas an area of 16. The
area of ∆FADis the area of ∆FCD+the area of
∆ADC,which is 16 + 16 =32. Therefore, the ratio
is 1:2.- Draw the extra segment as shown and determine
its length from the Pythagorean theorem. (It’s a
3-4-5 right triangle times 2!) The area of the semi-
circle is 16π/2= 8 π, the area of the rectangle is 40,
and the area of the triangle is 24, for a total area
of 64 + 8 π.
Answer Key 5: Areas and Perimeters
D
CA B848 F4P
5
11
10
5684 4SAT Practice 5
- E Move the shaded pieces around to see that
they make up half of the square. The area of the
square is 12 × 12 =144, so the shaded region has
area 144/2 =72.
2. C Find the area indirectly by subtracting the
three right triangles from the rectangle. The rec-
tanglehas area 8 × 24 =192, so the triangle has area
192 − 48 − 48 − 24 =72.
A
B
C
D
AB
D C
R
P
24
8
(^1212)
4
4
24
48
48
- The perimeter of the semicircle is 4π, so the
perimeter of the whole figure is 26 + 4 π. - E Each semicircle has a perimeter of 4π, which
means the circumference of a “whole” circle
would be 8πand therefore the diameter of each
circle is 8. Therefore, the height of the rectangle
is 16 and the length is 24. 24 × 16 = 384