Examples with solutions for Area of a Rectangle: Subtraction or addition to a larger shape

Exercise #1

Look at the two rectangles in the figure:

222444555222EEEDDDGGGFFFCCCBBBAAA

What is the area of the white area?

Video Solution

Step-by-Step Solution

As we know that EFGD is a rectangle, we also know that DE is equal to 2 and DG is equal to 4

In a rectangle, each pair of opposite sides are equal and parallel, therefore:

ED=FG=2 ED=FG=2

DG=EF=4 DG=EF=4

Now we calculate the area of the orange rectangle EFGD by multiplying the length by the width:

2×4=8 2\times4=8

Now we calculate the total area of the white rectangle ABCD:

AD=AE+ED=2+2=4 AD=AE+ED=2+2=4

DC=DG+GC=4+5=9 DC=DG+GC=4+5=9

The area of the entire rectangle ABCD is:

4×9=36 4\times9=36

Now to find the area of the white part that is not covered by the area of the orange rectangle, we will subtract the area of the rectangle EFGD from the rectangle ABCD:

368=28 36-8=28

Answer

28 cm²

Exercise #2

The quadrilateral ABCD is a rectangle.

Points E and F are located on sides DC and BC respectively.

FC = 1.5 cm
EC = 5 cm
DE = 3 cm

AD = 4 cm

Calculate the area of the quadrilateral ABFE.

4443335551.51.51.5AAABBBCCCDDDEEEFFF

Video Solution

Step-by-Step Solution

Let's calculate the side DC:

3+5=8 3+5=8

Now we can calculate the area of square ABCD:

4×8=32 4\times8=32

Let's calculate the area of triangle ADE:

4×32=122=6 \frac{4\times3}{2}=\frac{12}{2}=6

Let's calculate the area of triangle FCE:

5×1.52=7.52=3.75 \frac{5\times1.5}{2}=\frac{7.5}{2}=3.75

Now let's calculate the area of AEFB by subtracting the other areas we found:

AEFB=3263.75=22.25 AEFB=32-6-3.75=22.25

Answer

22.25

Exercise #3

Look at the rectangle in the figure.

A semicircle is added to each side of the rectangle.

What is the area of the entire shape?

444888

Video Solution

Step-by-Step Solution

The area of the entire shape equals the area of the rectangle plus the area of each of the semicircles.

Let's label each semicircle with a number:

4448881234Therefore, we can claim that:

The area of the entire shape equals the area of the rectangle plus 2A1+2A3

Let's calculate the area of semicircle A1:

12πr2 \frac{1}{2}\pi r^2

12π42=8π \frac{1}{2}\pi4^2=8\pi

Let's calculate the area of semicircle A3:

12πr2 \frac{1}{2}\pi r^2

12π22=2π \frac{1}{2}\pi2^2=2\pi

The area of the rectangle equals:

4×8=32 4\times8=32

Now we can calculate the total area of the shape:

32+2×8π+2×2π=32+16π+4π=32+20π 32+2\times8\pi+2\times2\pi=32+16\pi+4\pi=32+20\pi

Answer

32+20π 32+20\pi cm².

Exercise #4

The shape below consists of a rectangle from which the line segment BH has been erased.

AB = 6 cm

AH = 7 cm

EF = 3 cm

HG = 12 cm

BC = 3 cm


Calculate the area of the shape shaded orange.

666777333121212333333AAABBBCCCDDDHHHGGGFFFEEE

Video Solution

Step-by-Step Solution

Let's first calculate the area of triangle ABC:

6×72=422=21 \frac{6\times7}{2}=\frac{42}{2}=21

Since the shape before us is a rectangle, we can claim that:

AC=GD=7

Now let's calculate the area of triangle HGD:

7×32=212=10.5 \frac{7\times3}{2}=\frac{21}{2}=10.5

Let's draw an imaginary line between B and H to get square BEFH where each side equals 3 cm.

Let's calculate the area of BEFH:

3×3=9 3\times3=9

Let's calculate the area of rectangle ACDG:

7×12=84 7\times12=84

Now we can calculate the area of the brown shape by subtracting the other areas we found:

Answer

43.5 cm