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| 8 Theory slides |
| 14 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Tearrik began training at a newly opened gym in his neighborhood called Little Muscles. In the swimming area there are three pools. One is a circular pool with an area of 28.2 square meters. Another is a rectangular pool with an area of 54 square meters.
The shape of the third pool surprised Tearrik a little. It looks like the combination of half the circular pool with the rectangular pool.
What is the area of the third pool?
The wrestling room of the gym where Tearrik works out has a circular carpet with the gym's logo inside.
The area of the logo is the sum of the areas of the involved figures. Then, start by recalling the formulas for finding the area of the corresponding figures.
Figure | Area's Formula |
---|---|
Rectangle | A=ℓ⋅w |
Trapezoid | A=2(b1+b2)h |
Semicircle | A=2πr2 |
Next, find the area of each of the pieces that make up the logo.
Figure | Dimensions (cm) | Area (cm2) |
---|---|---|
Green rectangle | ℓ=40 w=20 |
A1=40⋅20=800 |
Purple trapezoid | b1=100 b2=20 h=30 |
A2=2(100+20)30=1800 |
Gray trapezoid | b1=180 b2=100 h=30 |
A3=2(180+100)30=4200 |
Gray rectangle | ℓ=100 w=10 |
A4=100⋅10=1000 |
Red semicircle | r=20 | A5=2π(20)2≈628.32 |
Substitute values
Multiply
Add terms
a⋅cb=ca⋅b
Cross out common units
Cancel out common units
a⋅1=a
Calculate quotient
Round to 1 decimal place(s)
Substitute values
Calculate power
Add terms
LHS=RHS
a2=a
Calculate root
Rearrange equation
On the first floor of the gym, the weight room and the spinning room are connected by the locker room. The gym owner received many complaints about the current floor quality. The entire first floor, instead, will be covered using recycled yoga mats.
Add the areas of the two big rectangles and subtract the area of the locker room.
Notice that the three rooms together form a composite figure that is shaped as two overlapping rectangles.
The area of the entire floor is then the sum of the areas of the big rectangles minus the area of the locker room. The area of the locker room must be subtracted because it is common for both rectangles.
Room | Figure | Dimensions (ft) | Area (ft2) |
---|---|---|---|
Spinning | Upper Rectangle | ℓ=50 w=45 |
A1=50⋅45=2250
|
Weight | Lower Rectangle | ℓ=55 w=40 |
A2=55⋅40=2200
|
Locker | Small Rectangle | ℓ=55−35=20 w=40−25=15 |
A3=20⋅15=300
|
Substitute values
Add and subtract terms
The area of a regular polygon is half the perimeter times the apothem. The apothem is the distance from the center of the polygon to any of its sides. The handle is what is left when a small semicircle is removed from a large semicircle.
The area of a regular polygon is half the perimeter times the apothem.
p=108, a=16.3
Multiply
a⋅b1=ba
Calculate quotient
Semicircle | Radius (cm) | A=21πr2 | Area (cm2) |
---|---|---|---|
Outer | 6.75 | A1=21π(6.75)2 | A1=71.57 |
Inner | 6 | A2=21π(6)2 | A2=56.55 |
Tearrik noticed a poster on the wall saying that the gym would soon open its second floor with a boxing room and a multipurpose room. The blueprint for the second floor is shown next.
Use the coordinate plane to find the side lengths of the composite figure. The perimeter is the sum of the lengths of the exterior sides. Break down the figure into known figures and find the sum of their areas.
The first step involves finding the perimeter of the entire floor. After that, the area of the floor will be calculated.
Start by identifying the coordinates of the vertices of the blueprint and the lengths of the horizontal and vertical sides. Let m be the length of the slant side of the entrance.
Notice that the entire floor can be broken down into some known figures.
The area of the floor is the sum of the areas of the involved figures.
Place | Figure | Dimensions | Area |
---|---|---|---|
Entrance and locker room | Trapezoid | b1=6 b2=3 h=4 |
A1=2(6+3)4=18 |
Seating area, hall, and boxing room | Rectangle | ℓ=8 w=6 |
A2=8⋅6=48 |
Multipurpose room without the stage | Rectangle | ℓ=6 w=3 |
A3=6⋅3=18 |
Stage | Semicircle | r=3 | A4=21π(3)2≈14.14 |
Composite figures are everywhere, it is a matter of looking closely. For instance, a running track combines rectangles and semicircles.
Many traffic signs can also be decomposed into known figures. The traffic sign indicating the direction of a street is formed by a rectangle and a triangle. The traffic sign corresponding to no U-turn
combines two rectangles, a semicircle, and a triangle.
Kevin is putting together a pre-made desk that his parents bought him. One of the washers has the following shape.
The inner part of the washer is made of four semicircles and four straight sides.
Since all the semicircles have the same diameter, they all have the same length. The four straight sides have also the same length. With this in mind, we can write an expression for the inner perimeter. P_(inner) = 4* semicircles + 4* straight sides The length of a semicircle is half the circumference of a circle. Length of a Semicircle 2π r/2 = π r The diameter of each semicircle is 1 centimeter, this means that they all have a radius of 0.5 centimeters. The straight sides are all 1 centimeter long. We have all what we need to find the inner perimeter of the washer.
The inner part of the washer has a perimeter of 4+2π centimeters, about 10.3 centimeters.
We can see the washer as what is left when the inner shape is removed from a square.
The area of the washer is then the area of the square minus the area of the inner shape. A = A_(square) - A_(inner) The area of a square is the side length squared. In our case, the square has sides of 5 centimeters. A_(square) = 5^2 ⇓ A_(square) = 25cm^2 We can divide the inner shape into four semicircles and a regular octagon.
Since all the semicircles have the same diameter, each pair of them forms a full circle. Thus, area of the inner shape is the area of 2 circles plus the area of a regular octagon. A_(inner) = 2* A_(⊙) + A_(reg. octagon) The circles have a diameter of 1 centimeter. This means the radius is equal to 0.5 centimeters. The area of a circle is π times the radius squared.
The area of a regular polygon is half the product of the perimeter and the apothem. In our case, the apothem is 1.2 centimeters. Since all the sides have a length of 1 centimeter, the perimeter is 8* 1 = 8 centimeters.
We are ready to find the area of the inner shape.
Finally, we subtract the area of the inner shape from the area of the square.
The area of the washer is 20.2-0.5π square centimeters, or about 18.6 square centimeters.
Three small equilateral triangles of the same dimensions are removed from a larger equilateral triangle and the following figure is obtained.
All the heights of the triangle are rounded to the nearest integer.
The perimeter of a figure is the distance around it. We know the larger sides of the resulting figure are 32 centimeters long each.
Since the small triangles are equilateral, all the sides are 14 centimeters long. Therefore, the shorter sides of the resulting figure are 14 centimeters long each.
We have all what we need to find the perimeter of the resulting figure. P = 32+14+32+14+32+14 ⇓ P = 138 The perimeter of the resulting figure is 138 centimeters.
We can find the area of the resulting figure by subtracting the area of the three small triangles from the area of the large triangle.
Since the three equilateral triangles have the same dimensions, we can find the area of the resulting figure with the following expression. A = A_(large△) - 3 A_(small△) Recall that the area of a triangle is half the product of the base and height. A = b_1* h_1/2 - 3* b_2* h_2/2 The side length of the large equilateral triangle is 14+32+14 = 60 centimeters. This means that the base of the large triangle is 60 centimeters. The height of the large triangle is 52 centimeters. The base of the small triangle is 14 centimeters and the height is 12 centimeters.
The area of the resulting figure is 1308 square centimeters.