50 MATHCOUNTS LECTURES
(34) AREA METHOD
BASIC KNOWLEDGE Theorem 1: The ratio of the areas of any two triangles is: 1
S ABC ABC S A B C 1
1
2 1
1
2
AB H
A1 B1 H 1
AB H A1 B1 H 1
Theorem 2: If two triangles have the same base, the ratio of the areas is the ratio of the heights.
S ABD ABD S ABC ABC
H h
Theorem 3: If two triangles have the height, the ratio of the areas is the ratio of the bases.
S ABC ABC S ADC
AB AD
;
S ABC ABC S DBC
AB DB
;
S ADC S BDC
AD DB
.
Theorem 4(a): If AB//CD If AB//CD,, then S ABC = S ABD and S AED = S BEC
(Same base and same height). 4(b): If S AED = S BEC , then then AB//CD. 227
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Theorem 5:
S AEC
Theorem 6:
S AED
S BEC
S BED
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AD DB
AD DB
Theorem 7: If AE = n and EC = m, then
Theorem 8: If AE = n and EC = m, then
S AED S EDC
S ABD S BDC
n m
n m
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.
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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 S AB AD DB DB 3 7 have: ACB 1 1 S ACD AD AD AD 4 4
S ACD
4 7
S ABC
4 7
175
4 25 100
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. 229
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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 cm2. Find the area of quadrilateral AEFD. Solution: 11. 2 Connect AF . We know that the area of AEF is 6 cm (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
2 3
6 9 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, S ABC
4, AE ED, AF
shaded area. Solution: 1. Method 1: Since E is the midpoint of AD, AEB and EBD 230
1 4
AC . Find the
50 MATHCOUNTS LECTURES
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have the same area. The shaded area given is the same as the area of triangle AFB. S AFB S FCB
AF FC
S AFB
1
S AFB S ABC S AFB
3
1
S AFB
3
4 S AFB
1 3
1
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 3 x. Since
S AFB
S FCB
AF FC
1 3
, 3( x + y) = 3 x + y + S CDE
S CDE = 2 y.
Since S ABC
4 , ( x + y) + 3 x + y + 2 y = 4
4 x +4 y = 4
x + y = 1
The answer is 1.
Example 7: In the obtuse triangle ABC, AM = MB, MD ABC is 24, find the area of BED. (1984 AMC).
BC,
EC BC . If the area of
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.
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50 MATHCOUNTS LECTURES
Figure 1
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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. 2 CE 2 AD DB 2 3 CE
BE
S ABE
3 3 2
CE BE
3
BE
3
S ACE S ABE
10 6 .
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
1 2
the altitude of BAC , and both
triangles share the same base, the area of BEC = BAC .
Area of EFC =
1 2
1 2
area of
area of BEC , and area of EGF =
232
1 2
area of
50 MATHCOUNTS LECTURES
EFC ;
therefore area of EGF =
(34) AREA METHOD 1 4
area of BEC . Thus, since area of BEC =
1 2
area
1
of ABC , area of EGF = area of ABC . 8
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. S ABO S ACO
S BOD
S COD
70 y
S ABO
Similarly, we have Or
70 y 70
3 70 4 y
84 x
S BCO
40 30
S AEO
84 x
S CEO
40 30
70
y
.
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 . S ABF S BFC
S AFD
5 x
S FDC
Similarly, we have
10
S ACF S BCF
S AEF
y 8
8 y 10
S EFB
x 5
Solving the equations, we get x =10, and y =12. The area of AEFD is 22.
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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)
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50 MATHCOUNTS LECTURES
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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 BC , AB, and AC, 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)
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Problem 10. Square ABCD has an area of 36 m . DE = 2 EC . 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).
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50 MATHCOUNTS LECTURES
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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)
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50 MATHCOUNTS LECTURES
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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 = 3 AD， BE =
2 BC . Find the area of △CDE .
Problem 21. As shown in the figure, in △ ABC, AB= 6 AD, and AC = 3 AE . If the area of 2
△ ADE is 1 cm , find the area of △ ABC .
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50 MATHCOUNTS LECTURES
Problem 22. As shown in the figure, S ABC
(34) AREA METHOD
5, AE ED, BD
2 3
BC . 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 ME AC . 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).
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50 MATHCOUNTS LECTURES
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ANSWER KEYS:
Problem 1. 50 (percent)
Problem 2. 16 sq units
Problem 3. 36
Problem 4.
1 3
2
Problem 5. Solution: 16 (units ).
Since triangles ABD and ACD have the same height and the same base, their areas are the same. A ABC
A ABD
AADC 8 8 16
Problem 6. 1/4
Problem 7. 60
Problem 8. Sqrt 3/4
Problem 9.
Problem 10.
1
Problem 11.
3
45
1 3
2 3
2
3 (cm )
(units )
2
Problem 13. 24 (units )
Problem 14. 12 sq units
Problem 15. 24 (units )
Problem 16. 12 sq units
Problem 17. 75 (%)
Problem 12.
2
2
2
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.
240
