# REAL WORLD PROBLEMS INVOLVING AREA AND PERIMETER

Problem 1 :

The diagram shows the shape and dimensions of Teresa’s rose garden.

(a) Find the area of the garden

(b) Teresa wants to buy mulch for her garden. One bag of mulch covers 12 square feet. How many bags will she need? Solution :

By drawing a horizontal line, we have divided the given shape as two parts. (1) ABCD is a rectangle

(2) CEFG is also a rectangle

Area of the garden

= Area of rectangle ABCD + Area of the rectangle CEFG

Area of rectangle ABCD :

length AB = 15 ft and width BD = 9 ft

= length x width

=  15 x 9

=  135 ft²  ----(1)

Area of rectangle CEFG :

length CE = 24 ft and width CF = AF - AC ==> 18 - 9 = 9 ft

= length x width

=  24 x 9

=  216 ft²  ----(1)

(1) + (2)

Area of the rose garden = 135 + 216 ==> 351 ft²

Number of bags that she needed = 351/12 ==> 29.25

So, she will need 30 bags of mulch

Problem 2 :

The length of a rectangle is 4 less than 3 times its width. If its length is 11 cm, then find the perimeter.

Solution :

Let w be the width of the rectangle.

Then, its length is (3w - 4).

Given : Length is 11 cm.

Then,

Length (l)  =  11

3w - 4  =  11

3w  =  15

w  =  5

So, the perimeter of the rectangle is

=  2(l + w)

=  2(11 + 5)

=  2(16)

=  32 cm

Problem 3 :

The diagram shows the floor plan of a hotel lobby. Carpet costs \$3 per square foot. How much will it cost to carpet the lobby? Solution :

By observing the above picture, we can find two trapeziums of same size. Since both are having same size. We can find area of one trapezium and multiply the area by 2.

Area of trapezium = (1/2) h (a +  b)

h = 15.5 ft  a = 30 ft  and b = 42 ft

=  (1/2) x 15.5 x (30 + 42)

=  (1/2) x 15.5 x 72 ==> 15.5 x 36==> 558 square feet

Area of floor of a hotel lobby = 2 x 558

=  1116 square feet

Cost of carper per square feet = \$3

=  3 x 1116 ==> \$ 3348

Amount spent for carpet =  \$ 3348.

Problem 4 :

The cost of fencing a circle shaped garden is \$20 per foot. If the radius of the garden is 14 feet, find the total cost of fencing the garden. (π  =  22/7).

Solution :

To know the length of fencing required, find the circumference of the circle shaped garden.

Circumference of the circle shaped garden is

=  2πr

Substitute 22/7 for π and 14 for r.

=  2(22/7)(14)

=  88 feet

Total cost of fencing is

=  88(20)

=  \$1760

Problem 5 :

Jess is painting a giant arrow on a playground. Find the area of the giant arrow. If one can of paint covers 100 square feet, how many cans should Jess buy? Solution :

Now we are going to divide this into three shapes. Two triangles and one rectangle. Area of rectangle = length x width

=  18 x 10 ==> 180 square feet

Area of one triangle = (1/2) x b x h

=  (1/2) x 6 x 10 ==> 30 square feet

Area of two triangles = 2 x 30 = 60 square feet

Total area of the given shape = 180 + 60

=  240 square feet

one can of paint covers 100 square feet

Number of cans needed = 240/100 = 2.4 approximately 3.

So, Jessy has to 3 cans of paint. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

If you have any feedback about our math content, please mail us :

v4formath@gmail.com

You can also visit the following web pages on different stuff in math.

WORD PROBLEMS

Word problems on simple equations

Word problems on linear equations

Algebra word problems

Word problems on trains

Area and perimeter word problems

Word problems on direct variation and inverse variation

Word problems on unit price

Word problems on unit rate

Word problems on comparing rates

Converting customary units word problems

Converting metric units word problems

Word problems on simple interest

Word problems on compound interest

Word problems on types of angles

Complementary and supplementary angles word problems

Double facts word problems

Trigonometry word problems

Percentage word problems

Profit and loss word problems

Markup and markdown word problems

Decimal word problems

Word problems on fractions

Word problems on mixed fractrions

One step equation word problems

Linear inequalities word problems

Ratio and proportion word problems

Time and work word problems

Word problems on sets and venn diagrams

Word problems on ages

Pythagorean theorem word problems

Percent of a number word problems

Word problems on constant speed

Word problems on average speed

Word problems on sum of the angles of a triangle is 180 degree

OTHER TOPICS

Profit and loss shortcuts

Percentage shortcuts

Times table shortcuts

Time, speed and distance shortcuts

Ratio and proportion shortcuts

Domain and range of rational functions

Domain and range of rational functions with holes

Graphing rational functions

Graphing rational functions with holes

Converting repeating decimals in to fractions

Decimal representation of rational numbers

Finding square root using long division

L.C.M method to solve time and work problems

Translating the word problems in to algebraic expressions

Remainder when 2 power 256 is divided by 17

Remainder when 17 power 23 is divided by 16

Sum of all three digit numbers divisible by 6

Sum of all three digit numbers divisible by 7

Sum of all three digit numbers divisible by 8

Sum of all three digit numbers formed using 1, 3, 4

Sum of all three four digit numbers formed with non zero digits

Sum of all three four digit numbers formed using 0, 1, 2, 3

Sum of all three four digit numbers formed using 1, 2, 5, 6

Featured Categories

Math Word Problems

SAT Math Worksheet

P-SAT Preparation

Math Calculators

Quantitative Aptitude

Transformations

Algebraic Identities

Trig. Identities

SOHCAHTOA

Multiplication Tricks

PEMDAS Rule

Types of Angles

Aptitude Test 