AoPS

Introduction to

Geometry

AoPS

Chapter 4

Perimeter

and

Area

Problem 4.2

Find the perimeter of ΔDEF. D

FE30° 30°

8√3

8

Problem 4.2

Find the perimeter of ΔDEF.

Since <E = <F, we have DE = DF = 8. Therefore

the perimeter is DE + DF + EF = 16 + 8√3.

D

FE30° 30°

8√3

8

Problem 4.3

Given that each little square inthe grid is a 1 x 1 square, findthe perimeter of ABCDEFGHIJ.

A

B

J

IH

F G

ED

C

Problem 4.3

Given that each little square inthe grid is a 1 x 1 square, findthe perimeter of ABCDEFGHIJ.

A

B

J

IH

F G

ED

C

Let’s take the slick approach: The horizontal side on thebottom has length 8; therefore, the sum of the horizontalsides on top, which together cover the same horizontaldistance without bactracking, must be 8. Similarly,AB + CD = 5, so the vertical sides on the left have totallength 5. The ones on the right have the same total length,so they contribute 5 to the perimeter. P = 2(8 + 5) = 26.

Problem 4.4The length of each leg of an isosceles triangle is

three times the length of the base of the triangle.

The perimeter of the triangle is 91. What is the

length of the base of the triangle?

Problem 4.4The length of each leg of an isosceles triangle is

three times the length of the base of the triangle.

The perimeter of the triangle is 91. What is the

length of the base of the triangle?

Let the base of the triangle have length x. Then the length of eachlength is 3x. Since the perimeter of the triangle is 91, we musthave x + 3x + 3x = 91. Solving this equation for x, we find thatx = 13. The length of the base of the triangle is 13.

Exercise 4.1.5A triangle with perimeter 45 has one side that is

twice as long as the shortest side and another side

that is 50% longer than the shortest side. Find the

length of the shortest side of the triangle.

Let the shortest side have length x. Then the other two sides havelengths 1.5x and 2x. The perimeter of the triangle is 45, so45 = x + 1.5x + 2x, or 45 = 4.5x. Solving the equation, x = 10,which means the shortest side has length of 10.

Problem 4.6

The length of one side of a rectangle is 4 less than

3 times an adjacent side. The perimeter of the

rectangle is 64. Find the area of the rectangle.

Problem 4.6

The length of one side of a rectangle is 4 less than

3 times an adjacent side. The perimeter of the

rectangle is 64. Find the area of the rectangle.

Let the lengths of the two sides be x and y. We are given that x = 3y – 4. Since the opposite sides of a rectangle are equal, the perimeter of the rectangle is 2x + 2y = 2(3y – 4) + 2y =8y – 8. We are given that the perimeter is 64, so we have 8y – 8 = 64. Therefore, y = 9 and x = 3(9) – 4 = 23, so the area of the rectangle is 9(23) = 207.

Exercise 4.2.5A square poster is replaced by a rectangular

poster that is 2 inches wider and 2 inches shorter.

What is the difference in the number of square

inches between the area of the larger poster and

the smaller poster? (MATHCOUNTS)

Exercise 4.2.5A square poster is replaced by a rectangular

poster that is 2 inches wider and 2 inches shorter.

What is the difference in the number of square

inches between the area of the larger poster and

the smaller poster? (MATHCOUNTS)

Let the side of the square poster be x, in inches. Then the area ofthe square poster is x2. The area of the rectangular poster is(x + 2) (x – 2) = x2 – 4, so their positive difference is 4 in2.

Exercise 4.2.6The perimeter of a square garden is 64 meters. The path

surrounding the garden has a uniform width and has an

area of 228 m2. How many meters of fencing are needed

to surround the outer edge of the path? (MATHCOUNTS)

PPGardenGarden

Path

Exercise 4.2.6The perimeter of a square garden is 64 meters. The path

surrounding the garden has a uniform width and has an

area of 228 m2. How many meters of fencing are needed

to surround the outer edge of the path? (MATHCOUNTS)

PPGardenGarden

PathLet x and y be the side length of the garden andthe width of the path, respectively, in meters.The perimeter of the garden is 4x = 64, so x=16.

Problem 4.10Given that BC = 40 and CD = 20, what are the ratios of

the area of triangle ABC to area of triangle ACD and

area of triangle ABC to area of triangle ABD?




The altitude from A is the same forboth ACD and ABC. Let h be the length of this altitude. Hence, the area of ABC = ½ (BC)(h) = 20h andthe area of ACD = ½ (CD)(h) = 10h.




The altitude from A is the same forboth ACD and ABC. Let h be the length of this altitude. Hence, the area of ABC = ½ (BC)(h) = 20h andthe area of ACD = ½ (CD)(h) = 10h.

Therefore, area ABC/area ACD = 20h/10h = 2. Notice thatarea ABC/area ACD = (BC)/(CD).




Similarly, altitude from A is the same for both ABC and ABD, h. So, the area of ABC = ½ (BC)(h) andthe area of ABD = ½ (BD)(h).

Therefore, area ABC/area ABD = 20h/30h = 2/3. Notice thatarea ABC/area ABD = (BC)/(BD).




A key step in our solution is noting that the three triangles have an altitude in common. Notice that in each case the ratio of the areas of the triangles equals the ratios of the lengths of the sides to which this common altitude is drawn.

Problem 4.11Suba and Sam have been hired to paint a triangle with

base AB on the weirdly shaped wall in the diagram.

They are told to choose either C or D as the third

vertex of the triangle. Suba thinks using D will make

the triangle look cooler, but Sam thinks using C will

result in a smaller triangle to paint. How can Suba

convince Sam to choose D?

Problem 4.11Suba and Sam have been hired to paint a triangle with

base AB on the weirdly shaped wall in the diagram.

They are told to choose either C or D as the third

vertex of the triangle. Suba thinks using D will make

the triangle look cooler, but Sam thinks using C will

result in a smaller triangle to paint. How can Suba

convince Sam to choose D?

To figure out which of ABC or ABD will use more paint, we need to consider the areas of the two triangles. Sincearea of ABD = ½ (AB)(DE) andarea of ABC = ½ (AB)(BC) we see that DE/BC = ¾.

Problem 4.11So Suba should point that triangle ABD will require less

paint because its area is less than the area of triangle ABC.

To figure out which of ABC or ABD will use more paint, we need to consider the areas of the two triangles. Sincearea of ABD = ½ (AB)(DE) andarea of ABC = ½ (AB)(BC) we see that DE/BC = ¾.

!! IMPORTANT !!

1. If 2 triangles share an altitude, then the ratio

of their areas = the ratio of the bases to which

that altitude is drawn. This is particularly useful

in problems in which 2 triangles have bases

along the same line.

2. If 2 triangles share a base, then the ratio of

their areas = the ratio of the altitudes to that base.

Problem 4.12

AC and BD meet at X. Given [ABX] = 24,

[BCX] = 15, and [CDX] = 10, find [ADX].

Note: [ADX] means the area of triangle ADX.

Problem 4.12




Since triangles ABX and CBX share an altitude from B, we have

AX [ABX] 8 CX [CBX] 5

= =

Problem 4.12




Turning to triangles ADX and CDX, we have

AX [ADX] 8 CX [CDX] 5

= =

Problem 4.12




Turning to triangles ADX and CDX, we have

AX [ADX] 8 CX [CDX] 5

Therefore, [ADX] = (8/5)[CDX] = 16.

= =

Exercise 4.3.3Euclid Apartment Building has a wall that is 30 ft long and

10 ft high. Jean is supposed to paint a triangle with one vertex

at the top of the wall, and a base that runs 8 ft along the

bottom of the wall. Jean wants to put the vertex at one corner

and the base at the other side of the wall as shown on the left.

Exercise 4.3.3Euclid Apartment Building has a wall that is 30 ft long and

10 ft high. Jean is supposed to paint a triangle with one vertex

at the top of the wall, and a base that runs 8 ft along the

bottom of the wall. Jean wants to put the vertex at one corner

and the base at the other side of the wall as shown on the left.

George the building owner wants to save money onpaint. He insists that the top vertex be above the middle of the base as shown on the bottom left so thetriangle won’t be so big. However, to show he’s a nice guy, he says that she can make the base 10 ftwide instead. Jean argues that George’s design willuse more paint than hers. Is she right?

Exercise 4.3.3Jean’s triangle has a base of 8 ft and height of 10 ft, so the

area is ½ (8)(10) = 40 ft2. George’s triangle has base 10 ft and

height 10 ft, so the area is ½ (10)2 = 50 ft2. We could have seen this

without computing the areas by noting that the heights of the 2

triangles are the same, but George’s triangle has a longer base than

Jean.

s. George the building owner wants to save money onpaint. He insists that the top vertex be above the middle of the base as shown on the bottom left so thetriangle won’t be so big. However, to show he’s a nice guy, he says that she can make the base 10 ftwide instead. Jean argues that George’s design willuse more paint than hers. Is she right?

Exercise 4.3.4By what factor is the area of triangle multiplied if the

length of its base is doubled and the height is tripled?

(Source: MATHCOUNTS)




Solution 1: Let the triangle have base b and height h, so it has area bh/2.If the base is doubled to 2b and the height is tripled to 3h, then the areaof the new triangle is (2b)(3h)/2 = 3bh. Therefore, the area increases by a factor of 3bh

½ bh

= 6.




Solution 2: Doubling the base multiplies the area by 2 and tripling theheight multiplies the area by 2. Doing both of these will multiply thearea by 2 x 3 = 6.

Review Problem 4.15Find the perimeter of a square that has as area of 75.

Express your answer in radical form.

Review Problem 4.15Find the perimeter of a square that has as area of 75.

Express your answer in radical form.

Let the side of the square be s. Then the area is s2. Since thearea of the square is 75, we have s2. Solving s2 = 75, givesus √75 = √25 * 3 = 5√3 as the side length of the square.There are four sides, so the perimeter is 4(5√3) = 20√3.

Problem 4.17The length of a given rectangle is 2 less than 4

times the width of the rectangle. Its perimeter is

51. What is its area?

Problem 4.17The length of a given rectangle is 2 less than 4

times the width of the rectangle. Its perimeter is

51. What is its area?

Let the width of the rectangle be w, so the length is 4w – 2.In terms of w, the perimeter then is 2(w) + 2(4w – 2) = 10w – 4. Since the perimeter = 51, we have10w – 4 = 51, so w = 5.5. Therefore, the length is4w – 2 = 20, so the area is (5.5) (20) = 110.

Problem 4.20A rectangle has a perimeter of 28 cm and an area

of 48 cm2. Find the dimensions of the rectangle?



Let x and y be the dimension of the rectangle. Then the perimeter of the rectangle is 2(x + y) = 28, so x + y = 14.The area of the rectangle is xy = 48.



Let x and y be the dimension of the rectangle. Then the perimeter of the rectangle is 2(x + y) = 28, so x + y = 14.The area of the rectangle is xy = 48. Substituting y = 14 – x into xy = 48, we get x(14 – x) = 48.Multiplying we get x2 – 14x + 48 = 0 which factors to(x – 6)(x – 8) = 0. The solutions are x = 6 and x = 8,which lead to y = 8 and y = 6, respectively.



Let x and y be the dimension of the rectangle. Then the perimeter of the rectangle is 2(x + y) = 28, so x + y = 14.The area of the rectangle is xy = 48. Substituting y = 14 – x into xy = 48, we get x(14 – x) = 48.Multiplying we get x2 – 14x + 48 = 0 which factors to(x – 6)(x – 8) = 0. The solutions are x = 6 and x = 8,which lead to y = 8 and y = 6, respectively. Therefore, the rectangle is 6 cm by 8 cm. [We could also have used trial and error to solve x + y = 14and xy = 48. ]

Problem 4.25In the figure below, AB = 12 and BC = AD = 8.

BC | AB and DA | AB. How many square units

are in the area of the shaded region?


B

C

A

D

Problem 4.25In the figure below, AB = 12 and BC = AD = 8.

BC | AB and DA | AB. How many square units

are in the area of the shaded region?


B

C

A

DLet E be the point on AB where the 2 trianglesmeet. Then the area of the shaded region is area ΔADE + ΔBCE =½ AE * AD + ½ BE * BC =½AE * 8 + ½BE * 8 = 4AE + 4BE = 4AB = 48 sq units.

E

Problem 4.26In the figure below, [WOZ] = 12, [ZOY] = 18,

and [WXYZ] = 50. Find [WOX].


and [WXYZ] = 50. Find [WOX]. W

X

Z

YO



X

Z

YO ΔWOZ and ΔWYZ share altitude from Z,

so WO/WY = [WOZ]/[WZY] = 12/30 = 2/5.



X

Z

YO ΔWOZ and ΔWYZ share altitude from Z,

so WO/WY = [WOZ]/[WZY] = 12/30 = 2/5.

We also have [WXY] = [WXYZ] – [WZY] = 20. Since [WOZ]/[WZY] = WO/WY = 2/5, we have [WOX] = (2/5)[WXY] = 8.

Problem 4.29We begin with an equilateral triangle. We divide each side into

three segments of equal length, and add an equilateral triangle

to each side using the middle third as a base. We then repeat

this, to get

Given that the perimeter of the 1st figure is 12, what is the

perimeter of the 2nd figure? the 3rd figure?

Problem 4.29Given that the perimeter of the 1st figure is 12, what is the

perimeter of the 2nd figure? the 3rd figure?

The key to this problem is realizing that for each segment, wedivide it into 3 smaller segments and end up with 4 copies ofthe smaller segment.Hence, the perimeter increases by a factor of 4/3 with eachstep. Therefore, the 2nd figure has perimeter (4/3)(12) = 16and the 3rd has perimeter (4/3)(16) = 64/3.

Problem 4.30Suppose we continue the process described in

the previous problem forever. What is the

perimeter of the resulting figure?

Problem 4.30Suppose we continue the process described in

the previous problem forever. What is the

perimeter of the resulting figure?After each step, the perimeter is 4/3the perimeter of the previous step.Therefore, the perimeter grows andgrows without bound – the perimeteris infinite. However we can containthe whole figure on the page (play with it and see). How can this be?!?(If you’re unconvinced, look upKoch’s snowflake on the internet.)

Fini!

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