chapter 34 area method

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.


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.


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.


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


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.


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


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.


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)


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)


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).


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)


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.


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).


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.

chapter 34 area method

Documents