CHAPTER 10 / ESSENTIAL GEOMETRY SKILLS 393
Concept Review 6
- If two figures are similar, then their correspond-
ing sides are proportionaland their correspond-
ing angles are equal (orcongruent). - a. two pairs of corresponding angles are equal
b. two pairs of corresponding sides are propor-
tional and the included angles are equal
c. all three pairs of corresponding sides are
proportional - The ratio of the sides is 4:6 or 2:3. The ratio of the
areas is the square of the ratio of sides, which is
4:9. If xis the area of the smaller triangle, then
x/27=4/9. Solving for xgivesx=12.
4. If 1 ⎥⎥ 2 , then the two triangles must be similar.
Since corresponding sides are proportional,
AC/AE=BC/DE.
Substituting, this gives 4/10 =5/DE.
Cross-multiply: 4DE= 50
Divide by 4: DE=12.5
5. A 5- 8-inch rectangle has an area of 40 square
inches. The ratio of areas, then, is 40:1,000, or 1:25.
This is the square of the ratio of lengths, so the
ratio of lengths must be 1:5. If xis the length of the
larger tree image, then 3/x=1/5. Cross-multiplying
givesx=15, so the tree is 15 inches high in the
larger photograph.
Answer Key 6: Similar Figures
1 2A
BC
DE
4
5
6
SAT Practice 6
- D If the ratio of the areas is 4:1, then the ratio
of corresponding lengths is the square root: 2:1.
If the perimeter of the smaller square is 20, then
the perimeter of the larger one is twice as big. - B Find the width of the patio with a proportion:
5/7=x/21
Cross-multiply: 7 x= 105
Divide by 7: x= 15
So the patio is a 15- × 21-foot rectangle, which has
an area of 15 × 21 =315 square feet.- A The two regions are similar, because the central
angles are the same. The ratio of their correspond-
ing lengths is 3:2, so the ratio of their areas is 9:4.
Since the larger area is 9, the smaller area must be 4. - C IfEFhas length xand
EGhas length y,thenFG
must have length y−x,
as shown. Since
the two lines
are parallel,
∆ABCis sim-
ilar to ∆AEF
and∆ACDis
similar to
∆AFG.ThereforeAC/CF=BC/xand
AC/CF=CD/(y−x). So BC/x=CD/(y−x),
and therefore CD/BC=(y−x)/x=y/x−1.
5. B The height of the larger
cone can be found with the
Pythagorean theorem to be
12. (It’s the old 5-12-13 right
triangle!) Since the two trian-
gles are similar, x/12=8/13.
Multiplying by 12 gives
x=96/13.
6. 28 The two trian-
gles are similar be-
causetheir
corresponding
angles are equal.
Since they are
right triangles, the
missing sides can be found with the Pythagorean
theorem. Your diagram should look like the one
above. The perimeter is 3 + 8 + 5 + 12 =28.
7. A Since the lines are parallel, ∆ABCis similar to
∆AEFand∆ACDis similar to
∆AFG.Therefore,
AD/AG= AC/AF=6/10=3/5.
The ratio of areas between
∆ABCand∆AEFis the
square of the ratio
of sides, which is
(3/5)^2 =9/25. Since
∆AEFhas an area of 75, (the area of ∆ABC)/75=9/25.
So∆ABChas an area of 27.
85
512xA
E F G
B C D
6
4
6
3
10
8
12
5
B C D
EF G
A
1
2
x y-x
y