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50 MATHCOUNTS LECTURES (34) AREA METHOD
227
BASIC KNOWLEDGE
Theorem 1: The ratio of the areas of any two triangles is:
111111
2
12
1
111HBA
HAB
HBA
HAB
S
S
CBA
ABC
Theorem 2: If two triangles have the same base, the ratio of the areas is the ratio of the
heights.
h
H
S
S
ABC
ABD
Theorem 3: If two triangles have the height, the ratio of the areas is the ratio of the bases.
AD
AB
S
S
ADC
ABC
; DB
AB
S
S
DBC
ABC
; DB
AD
S
S
BDC
ADC
.
Theorem 4(a): If AB//CD, then SABC = SABD and SAED = SBEC
(Same base and same height).
4(b): If SAED = SBEC, then AB//CD.
50 MATHCOUNTS LECTURES (34) AREA METHOD
228
Theorem 5: DB
AD
S
S
BEC
AEC
Theorem 6: DB
AD
S
S
BED
AED
Theorem 7: If AE = n and EC = m, thenm
n
S
S
EDC
AED
Theorem 8: If AE = n and EC = m, thenm
n
S
S
BDC
ABD
Theorem 9: For triangle ABC, extend AB to F such that AB = BF, extend BC to D such
that BC = CD, and extend CA to E such that CA = AE. The ratio of the area of triangle
ABC to the area of triangle DEF is 1/7.
Proof:
Connect FC, DA, and EB as shown in the figure. All the seven triangles
have the same areas.
50 MATHCOUNTS LECTURES (34) AREA METHOD
229
Therefore, the ratio of the area of triangle ABC to the area of
triangle DEF is 1/7.
Example 1: Triangle ABC has an area of 175 square units. Point D lies on side AB such
that AD : DB = 4 : 3. What is the area, in square units, of triangle ACD? (2010
Mathcopunts Handbook).
Solution: 100.
Since triangle ABC and triangle ACD share the same height, we
have:4
7
4
311
AD
DB
AD
DBAD
AD
AB
S
S
ACD
ACB
1002541757
4
7
4 ABCACD SS
Example 2: As shown in the figure, BC = CE,AD = CD. Find the ratio
of the area triangle ABC to the area of triangle CDE.
Solution: 2:1.
Connect BD. ABD and BDC have the same area.
BDC and CDE have the same area.
The ratio is then 2:1.
Example 3: Two squares with the side lengths of 10 cm and 12 cm, respectively. Find the
shaded area.
Solution: 50.
50 MATHCOUNTS LECTURES (34) AREA METHOD
230
We connect BF and we know that AC//BF. So the area of triangle ABH is the same as the
area of triangle CHF (Theorem 4(a)).
The shaded area is the area of triangle ACF, which is the same as the
area of triangle ABC, which is 10 10 2 = 50.
Example 4: The larger one of the two squares in the figure below has the side length of 10
cm. Find the shaded area.
Solution: 50.
We connect BD and we know that DB//GE. So the area of triangle
GHD is the same as the area of triangle BEH (Theorem 4(a)).
The area of the shaded region, triangle EDG, is the same as the area of
triangle BEG, which is 10 10 2 = 50.
Example 5: As shown in the figure, BD and CF cut the rectangle ABCD into 4 regions.
The area of ECF is 4 cm2 and the area of CEB is 6 cm
2. Find the
area of quadrilateral AEFD.
Solution: 11.
Connect AF. We know that the area of AEF is 6 cm2 (Theorem 4(a))
and the ratio of the area of AEF to the area of AEB is 4/6 = 2/3.
So the area of AEB is 963
2 and the area of ABC =ADC is
9 + 6 = 15.
The area of quadrilateral AEFD = 15 – 4 = 11.
Example 6: As shown in the figure, ACAFEDAES ABC4
1,,4 . Find the
shaded area.
Solution: 1.
Method 1: Since E is the midpoint of AD, AEB and EBD
50 MATHCOUNTS LECTURES (34) AREA METHOD
231
have the same area.
The shaded area given is the same as the area of triangle AFB.
3
1
FC
AF
S
S
FCB
AFB 3
1
AFBABC
AFB
SS
S
3
1
4
AFB
AFB
S
S
1AFBS
The answer is 1.
Method 2:
Connect CE. Let x denote the area of AEF and y denote the area of
AEB.
Then the area of EBD is y and the area of CEF is 3x.
Since 3
1
FC
AF
S
S
FCB
AFB , 3(x + y) = 3x + y + SCDE
SCDE = 2y.
Since 4ABCS , (x + y) + 3x + y + 2y = 4 4x +4y = 4 x + y = 1
The answer is 1.
Example 7: In the obtuse triangle ABC, AM = MB, MD BC, EC BC. If the area of
ABC is 24, find the area of BED. (1984 AMC).
Solution: 12.
Connect MC (Figure 1).
Since MD and EC are parallel, the colored areas in Figure 2 are the same. The area of
BED is the same as the area of BMC (Figure 3), which is half of the area of ABC
(Figure 4). The answer is 24/2 = 12.
50 MATHCOUNTS LECTURES (34) AREA METHOD
232
Figure 1 Figure 2 Figure 3 Figure 4
Example 8: Triangle ABC in the figure has area 10. Points D, E, and F, all distinct from
A, B, and C, are on sides AB, BC and CA respectively, and AD = 2, DB = 3.
If triangle ABE and quadrilateral DBEF have equal areas, find that area.
(1983 AMC #28).
Solution: 6.
Since triangle ABE and quadrilateral DBEF have equal areas, we know that triangle ADG
has the same area as triangle EFG (Theorem 4(b)).
Therefore, AF//DE and ABC is similar to DBE.
BE
DB
CE
AD
BECE
32
BE
CE
3
2
ABE
ACE
S
S
3
2
61023
3
ABES .
Example 9: In ABC, D is the midpoint of side BC, E is the midpoint of AD, F is the
midpoint of BE, and G is the midpoint of FC. What part of the area of ABC
is the area of EFG?
Solution: 1/8.
Draw EC. Since the altitude of BEC is 2
1 the altitude of BAC, and both
triangles share the same base, the area of BEC = 2
1 area of
BAC. Area of EFC = 2
1 area of BEC, and area of EGF =
2
1area of
50 MATHCOUNTS LECTURES (34) AREA METHOD
233
EFC; therefore area of EGF = 4
1 area of BEC. Thus, since area of BEC =
2
1 area
of ABC, area of EGF =8
1area of ABC.
Example 10: As shown in the figure, ABC is divided into six smaller triangles by lines
drawn from the vertices through a common interior point. The areas of four of these
triangles are as indicated. Find the area ABC.
Solution: 315.
Let x and y be the areas for the small triangles as shown in the figure.
COD
BOD
ACO
ABO
S
S
S
S
30
40
70
84
y
x
Similarly, we have CEO
AEO
BCO
ABO
S
S
S
S
y
x 70
3040
84
Or y
y 70
4
3
70
70
.
x = 56 and y = 35. The total area is 84 + 70 + 40 + 30 + 35 + 56 = 315.
Example 11: Triangle ABC is divided into four parts with the areas of
three parts shown in the figure. Find the area of the quadrilateral
AEFD.
Solution: 22.
Connect AF.
FDC
AFD
BFC
ABF
S
S
S
S
810
5 yx
Similarly, we haveEFB
AEF
BCF
ACF
S
S
S
S
510
8 xy
Solving the equations, we get x =10, and y =12. The area of AEFD is
22.
50 MATHCOUNTS LECTURES (34) AREA METHOD
234
EXERCISES
Problem 1. In the rectangle shown, the ratio of width to length is 1: 4. What percent of the
rectangle is shaded? (Mathcounts Competitions)
Problem 2. In rectangle ACDE, B lies on AC , DC = 4, and DE = 8. Find the area of the
shaded region. (Mathcounts Handbooks)
Problem 3. In ∆ABC, B is a right angle. D is the midpoint of AC , F is the midpoint of
BC , and E is the midpoint of CF . AB = 12 cm. BC = 16 cm. What is the number of
square centimeters in the area of ∆ BDE? (Mathcounts Competitions)
Problem 4. Rectangle ABCD is shown, and E is the midpoint of AD . What is the ratio of
the area of the shaded region to the area of the unshaded region? (Mathcounts
Competitions)
50 MATHCOUNTS LECTURES (34) AREA METHOD
235
Problem 5. If BD = DC and the area of the triangle ABD is 8 square units, find the area of
triangle ABC. (Mathcounts Competitions).
Problem 6. If E is the midpoint of AD , what is the ratio of the area of triangle BED to
rectangle ABCD? (Mathcounts Handbooks)
Problem 7. What is the number of square centimeters in the area of a triangle whose sides
measure 8 cm, 15 cm, and 17 cm?
Problem 8. Given equilateral triangle ABC with sides of length 2. M, N and P are
midpoints of sides ,AC and ,, ABBC respectively. If A is folded to M, B is folded to P, and
C is folded to N, what would the new area be of the folded region? (Mathcounts
Handbooks)
Problem 9. Find the ratio of the area of triangle ACE to the area of rectangle ABCD.
Express your answer as a common fraction. (Mathcounts Competitions)
50 MATHCOUNTS LECTURES (34) AREA METHOD
236
Problem 10. Square ABCD has an area of 36 m2. DE = 2EC. What is the ratio of the area
of ∆ BED to the area of square ABCD? Express your answer as a common fraction.
(Mathcounts Competitions)
Problem 11. ABC is an equilateral triangle with sides equal to 2 cm. BC is extended its
own length to D, and E is the midpoint of AB . ED meets AC at F. Find the area of the
quadrilateral BEFC in square centimeters in simplest radical form. (Mathcounts
Competitions)
Problem 12. In the figure, AE = 6, EB = 7, and BC = 5. What is the area of quadrilateral
EBCD? Express your answer as a common fraction. (Mathcounts Competitions).
Problem 13. Rectangle CDEF is inscribed in ∆ ABC. In the triangle, AC = 8, CB = 12, and
D and F are the midpoints of sides BC and AC respectively. Find the number of square
units in the area of the shaded region. (Mathcounts Competitions).
50 MATHCOUNTS LECTURES (34) AREA METHOD
237
Problem 14. ABC is a right triangle. CDEF is a rectangle with D and F midpoints of sides
BC and AC. If AC = 6 and BC = 8, find the area of the shaded region. (Mathcounts
Handbooks).
Problem 15. Rectangle CDEF is inscribed in ∆ ABC. In the triangle, D and F are the
midpoints of sides BC and AC respectively. If the area of triangle ABC is 48 square
units, find the number of square units in the area of the shaded region.
Problem 16. ABC is a right triangle. CDEF is a rectangle with D and F midpoints of sides
BC and AC. If the area of triangle ABC is 24 square units, find the area of the shaded
region.
Problem 17. ABCD is a rectangle that is four times as long as it is wide. Point E is the
midpoint of BC . What percent of the rectangle is shaded? (Mathcounts Handbooks)
50 MATHCOUNTS LECTURES (34) AREA METHOD
238
Problem 18. In rectangle ABCD, AB = 16 cm and AD = 5 cm. FG , BD and AC are
concurrent at point O. What is the number of square centimeters in the area of the shaded
region? (Mathcounts Competitions)
c
Problem 19. As shown in the figure, square ABCD has the side length of 6 cm. The areas
of ABE, ADF and quadrilateral AECF are all the same. Find the area of AEF.
Problem 20. As shown in the figure, the area of triangle ABC is 1 and AC = 3AD,BE =
2BC. Find the area of △CDE.
Problem 21. As shown in the figure, in △ABC, AB= 6AD, and AC = 3AE. If the area of
△ADE is 1 cm2, find the area of △ABC.
50 MATHCOUNTS LECTURES (34) AREA METHOD
239
Problem 22. As shown in the figure, BCBDEDAES ABC3
2,,5 . Find the shaded
area.
Problem 23. Rectangle ABCD has AB = 8 and BC = 6. Point M is the midpoint of
diagonal AC, and E is on AB with MEAC. What is the area of AME? (2009 AMC 12 B).
Problem 24. As shown in the figure on the right, ∆ ABC is divided into six smaller triangles by
lines drawn from the vertices through a common interior point. The areas of four of these triangles
are as indicated. Find the area of ∆ ABC. (1985 AIME 6).
50 MATHCOUNTS LECTURES (34) AREA METHOD
240
ANSWER KEYS:
Problem 1. 50 (percent) Problem 2. 16 sq units
Problem 3. 36 Problem 4. 3
1
Problem 5. Solution: 16 (units2).
Since triangles ABD and ACD have the same height and the same base, their areas are the
same. 1688 ADCABDABC AAA
Problem 6. 1/4 Problem 7. 60
Problem 8. Sqrt 3/4 Problem 9. 3
1
Problem 10. 3
1 Problem 11. 3
3
2 (cm2)
Problem 12. 2
45(units
2) Problem 13. 24 (units
2)
Problem 14. 12 sq units Problem 15. 24 (units2)
Problem 16. 12 sq units Problem 17. 75 (%)
Problem 18. 20 (square centimeters)
Problem 19. 10. Problem 20. 2/3.
Problem 21. 18. Problem 22. 2.
Problem 23. 75/8 Problem 24. 315.