Area of Squares and Rectangles: Problems with Solutions

Solution:
Area of a rectangle is [tex]L \times W[/tex] where L is the length and W is the width
[tex]6 \times 4 = 24 \text{in}^2.[/tex]

Solution:
The sides of a square are equal size. The square's area is [tex]a \times a[/tex] where a is the side of the square.
[tex]5 \times 5 = 25[/tex]

Solution:
Area = [tex]a \times a[/tex] where a is the length of a side
[tex]a \times a = 16[/tex]
[tex]4 \times 4 = 16[/tex]
[tex]a = 4[/tex]

Solution:
Area = [tex]L \times W[/tex]
[tex]45 = 9 \times W[/tex]
[tex]W = 45 \div 9 = 5[/tex]

Solution:
Perimeter is [tex]4 \times a[/tex] where a is the side of the square
[tex]4 \times a =24[/tex]
[tex]a = 24 \div 4 = 6[/tex]
Area = [tex]a \times a = 6 \times 6 = 36 \text{in}^2[/tex]

Solution:
[tex]L = 12[/tex]
[tex]W = 12 - 5 = 7[/tex]
Area = [tex]L \times W = 12 \times 7 = 84 \text{in}^2[/tex]

Solution:
[tex]L = 12[/tex]
[tex]W = 12 \div 3 = 4[/tex]
Area = [tex]L \times W = 12 \times 4 = 48 \text{in}^2[/tex]

Solution:
The area of the first rectangle is [tex]L \times W = 6 \times 4 = 24 \text{in}^2[/tex]
The new length is [tex]6 + 2 = 8[/tex] in
[tex]8 \times W = 24[/tex] then [tex]W = 24 \div 8 = 3[/tex] in

Solution:
The area of the square is [tex]2 \times 2 = 4 \text{in}^2[/tex]
The area of the rectangle is [tex]L \times W = 24 \times 8 = 192 \text{in}^2[/tex]
The number of squares is [tex]192 \div 4 = 48[/tex]

Solution:
The area of the square is [tex]6 \times 6 =36 \text{in}^2[/tex]
The area of the rectangle is [tex]L \times W[/tex]. Because the areas are equal [tex]L \times W = 36[/tex]
[tex]L \times 4 = 36[/tex]
[tex]L = 36 \div 4 = 9 in[/tex]

Solution:
The area of the first square is [tex]5 \times 5 = 25 \text{in}^2[/tex]
The area of the new square is [tex]10 \times 10 = 100 \text{in}^2[/tex]
The area is 4 times bigger

Solution:
The field is a rectangle whose length is 25 feet and width is 12 feet. The area is [tex]L \times W = 25 \times 12 = 25 \times 4 \times 3 = 100 \times 3 = 300 \text{ft}^2[/tex]
The field covered by the house is a square with the side of 9 feet. The area is [tex]9 \times 9 = 81 \text{ft}^2[/tex]
The garden's area will be [tex]300 - 81 = 219 \text{ft}^2[/tex]

Solution:
The area of the blue zone is half of the rectangle's area.
Area = [tex](L \times W) \div 2 = (12 \times 9) \div 2 = 108 \div 2 = 54 \text{ft}^2[/tex]

Solution:
[tex][AF] = 24\text{ ft}[/tex]
[DE] = 15 ft
[tex][BC] = [AF] - [DE] = 9\text{ ft}[/tex]
[tex][FE] = 20\text{ ft}[/tex]
[tex][AB] = 9\text{ ft}[/tex]
[tex][CD] = [FE] - [AB] = 11\text{ ft}[/tex]
The area of the red zone is equal to the difference between the area of the rectangle with a length of 24 feet and a width of 20 feet and the area of the rectangle with a length of 11 feet and a width of 9 feet
Area = [tex](24 \times 20) - ( 11 \times 9) = 480 - 99 = 381 \text{ft}^2[/tex]

Solution:
The area of the park is [tex]30 \times 30 = 900 \text{ft}^2[/tex]
The alley is shaped like a rectangle with a length of 30 ft and a width of 2 ft. Area is [tex]L \times W = 30 \times 2 = 60 \text{ft}^2[/tex]
The green zone has an area of [tex]900 - 60 = 840 \text{ft}^2[/tex]

Solution:
The length of the screen is [tex]80 - (5 + 5) = 80 - 10 = 70[/tex] in, the width of the screen is [tex]50 - (5 + 5) = 50 - 10 = 40[/tex] in
The area of the screen is [tex]70 \times 40 = 2800 \text{in}^2[/tex]

Solution:
Because there are 5 squares on the width of the rectangle and 7 squares on its length, then the side of the square is 2 in. The area of a square is 4 [tex]\text{in}^2[/tex]. There are 18 blue squares, so their area will be [tex]4 \times 18 = 72 \text{in}^2[/tex]

Solution:
The area of the rectangle ABCD is 48 [tex]\text{in}^2[/tex]. Rectangle ABCD is split into 4 equal rectangles. Thus, the area of rectangle AMOQ is 12 [tex]\text{in}^2[/tex]. Rectangle AMOQ is split into 4 equal rectangles. The area of a blue rectangle is 3 [tex]\text{in}^2[/tex]. The total area of the 2 blue rectangles is 6 [tex]\text{in}^2[/tex].

Solution:
The area of the square is 4 [tex]\text{in}^2[/tex]. It's split in 4 equal squares. The area of a small square is 1 [tex]\text{in}^2[/tex], the area of the rectangle is 54 [tex]\text{in}^2[/tex], so the area of the red zone is 54 - 1 = 53 [tex]\text{in}^2[/tex].

Solution:
The sides of the box are equal in pairs. Two of the sides have edges of 10 in and 6 in.
Their area is [tex](10 \times 6) \times 2 = 60 \times 2 = 120 \text{in}^2[/tex]
Two of the sides have edges of 10 in and 8 in.
Their area is [tex](10 \times 8) \times 2 = 80 \times 2 = 160 \text{in}^2[/tex]
Two of the sides have edges of 8 in and 6 in.
Their area is [tex](8 \times 6) \times 2 = 48 \times 2 = 96 \text{in}^2[/tex]
The total area is [tex]120 + 160 + 96 = 376 \text{in}^2[/tex]