AREA AND PERIMETER WORD PROBLEMS
Problem 1 :
The length of a rectangle is 4 less than 3 times its width. If its perimeter is 32 cm, then find the area of the rectangle.
Solution :
Let x be the width of the rectangle.
Then, its length is (3x - 4).
Perimeter of the rectangle is 32 cm
2(l + w) = 32
Divide each side by 2.
l + w = 16
Substitute (3x - 4) for l and x for w.
3x - 4 + x = 16
4x - 4 = 16
Add 4 top each side.
4x = 20
Divide each side by 4.
x = 5
Therefore, width of the rectangle is 5 cm.
And length of the rectangle is
= 3(5) - 4
= 15 - 4
= 11 cm
Formula for area of a rectangle :
= l ⋅ w
Substitute 11 for l and 5 for w.
= 11 ⋅ 5
= 55
So, the area of the rectangle is 55 square cm.
Problem 2 :
Theresa wants new carpeting for her family room. Her family room is a 12 ft by 21 ft
rectangle. How much carpeting does she need to buy to cover her entire family room ?
Solution :
To find the amount of carpet required, we have to know its area. Because the family room is in rectangle shape, we can use the formula for area of a rectangle to find the area of the family room.
Formula for area of a rectangle :
= l ⋅ w
Substitute 12 for l and 21 for w.
= 12 ⋅ 21
= 252
So, Theresa needs to buy 252 square ft carpeting to cover her entire family room.
Problem 3 :
Lily to wants to do fencing around a circular garden that has a radius of 70 m. If the cost of fencing is $12 per meter, find the total cost of fencing for the entire garden (Use π = 22/7).
Solution :
Fencing is done around the circular garden. To find the total cost of fencing, we have to know the perimeter of the garden, Because the garden is in the shape of circle, we can use the formula for perimeter of a circle to find the perimeter of the garden.
Formula for perimeter of a circle :
= 2πr
Substitute 22/7 for π and 70 for r.
= 2 ⋅ 22/7 ⋅ 70
= 440
So, the perimeter of the garden is 440 meters.
The cost of fencing is $12 per meter.
Then, the total cost of fencing for 440 meters :
= 440 ⋅ 12
= 5280
So, the total cost of fencing for the entire garden is $5280.
Problem 4 :
If the cost of the carpeting is $15 per square meter, find the total cost of carpeting the floor of the room whose shape is a regular pentagon shown below.
Solution :
To find the total cost of carpeting the floor of the room, we have to know its area. Because the floor of the room is in the shape of regular pentagon (regular polygon), we can use the formula for area of a regular polygon to find the area of the floor.
Formula for area of a regular polygon :
= 1/2 ⋅ apothem ⋅ perimeter of polygon ----(1)
In the above regular pentagon, apothem is 4 m
The perimeter of a regular polygon is
= No. of sides ⋅ Length of each side
Therefore, the perimeter of the regular pentagon shown above is
= 5 ⋅ 6
= 30 m
To find the area of the regular pentagon, substitute 4 for apothem and 30 for perimeter in the formula (1).
(1) ----> = = 1/2 ⋅ 4 ⋅ 30
= 60 m2
So, the area of the floor is 60 square meters.
The cost of carpeting is $15 per square meter.
Then, the total cost of carpeting for 60 square meters :
= 60 ⋅ 15
= 900
So, the total cost of carpeting the floor of the room is $900.
Problem 5 :
The poster has a border whose width is 2 cm on each side. If the printed material has the length of 6 cm and width of 8 cm, then find the area of the border.
Solution :
Draw a sketch
In the above diagram, to find the area of the border, we have to subtract the area of the printed material from the complete area of the poster.
Complete area of the poster :
= l ⋅ w
Substitute 10 for l and 12 for w.
= 10 ⋅ 12
= 120 cm2
Area of the poster is 120 square cm.
Area of the printed material :
= l ⋅ w
Substitute 6 for l and 8 for w.
= 6 ⋅ 8
= 48 cm2
Area of the printed material is 48 square cm.
Area of the border :
= Area of the poster - Area of the printed material
= 120 - 48
= 72 cm2
So, the area of the border is 72 square cm.
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Word problems on quadratic equations
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on comparing rates
Converting customary units word problems
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Word problems on simple interest
Word problems on compound interest
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Complementary and supplementary angles word problems
Trigonometry word problems
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One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
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