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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.
Emily is helping her mom put some tiles on the backsplash. Some of them have an interesting shape.
The tile looks like what is left when some figures are removed from the corners of a rectangular tile.
Let's take a look at how tiles are formed.
The area of the tile is, therefore, equal to the area of the rectangle minus the total area of the four removed pieces. A = A_(rec) - (A_1 + A_2 + A_3 + A_4) First, let's find the dimensions of the rectangle.
Dimensions of The Rectangle | |
---|---|
Length, l | Width, w |
2+2+ 1=5 | 1+ π4+ 2=3+ π4 |
The area of a rectangle is the product of the length and width.
Next, we will find the area of the removed parts. Remember the area of a quarter of a circle is one-fourth the product of π and the radius squared. Let's use this formula to find the area of the pieces removed from the left-hand side of the tile.
Top Left Figure | Bottom Left Figure | |
---|---|---|
Formula | A = π r^2/4 | A = π r^2/4 |
Substitute radius | A_1 = π * 1^2/4 | A_2 = π * 2^2/4 |
Calculate | A_1 = π/4 | A_2 = π |
The area of a triangle is half the product of the base and height. Let's use this to find the area of the pieces removed from the right-hand side of the tile.
Big Triangle | Small Triangle | |
---|---|---|
Formula | A = bh/2 | A = bh/2 |
Substitute values | A_3 = 2* 2/2 | A_4 = 1* 1/2 |
Calculate | A_3 = 2 | A_4 = 1/2 |
Let's substitute the corresponding values into the equation we wrote for the area of a tile.
The tiles of this type have an area of 12.5 square inches. Since Emily will put 16 of these tiles, we multiply that value by 16. 16* 12.5in^2 = 200in^2 Emily will cover 200 square inches of the wall with those tiles.
Find the area of the following figure.
The U-shaped figure is what is left when a small rectangle is removed from a large rectangle.
Therefore, the area of the given figure is the area of the large rectangle minus the area of the small rectangle. A = A_(large)-A_(small) The large rectangle is 5 centimeters long and 4 centimeters wide. A_(large) = 5* 4 ⇒ A_(large) = 20cm^2 The small rectangle is 3.5 centimeters long and its width is 4-1.5-1.5 = 1 centimeter. A_(small) = 3.5* 1 ⇒ A_(small) = 3.5cm^2 Finally we subtract A_(small) from A_(large) to find the area of the given figure. A = 20-3.5 ⇒ A = 16.5cm^2 The U-shaped figure has an area of 16.5 square centimeters.
Dylan is having breakfast with his dad. He puts two placemats on the table.
The table space covered by the placemats is the sum of the areas of the placemats minus the area of the overlapping region. Each placemat is a rectangle 18 inches long and 12 inches wide. Let's multiply the length and width to find the area of one placemat. A_(placemat) &= 18* 12 &⇓ A_(placemat) &= 216in^2 Each placemat covers an area of 216 square inches. Together, they cover 2* 216= 432 square inches of the table. However, since they are overlapping each other, we need to subtract the area of the overlapping region. Otherwise, we would count this common area twice. Let's find its dimensions by focusing on the placemat on the right.
Since the placemat is 12 inches wide and 6.3 inches are uncovered, the covered side measures 12-6.3 = 5.7 inches. Similarly, since the placemat is 18 inches long and 14.1 inches are uncovered, the covered side measures 18-14.1 = 3.9 inches. Therefore, the overlapping region is a rectangle, 5.7 inches long and 3.9 inches wide. A_(overlapping) &= 5.7* 3.9 &⇓ A_(overlapping) &= 22.23in^2 Finally, let's subtract the area of the overlapping region from the area covered by the two placemats. 432 - 22.23 = 409.77in^2 The table space taken up by the two placemats is 409.77 square inches.
Consider the following figure.
We want to find the perimeter of the given figure. Recall that the perimeter of a figure is the distance around it.
We can see that the figure is composed of three semicircles surrounding a right triangle at the center. For this figure, we need to find the length of each semicircle. Recall that the circumference of a circle is 2π r. Since we only have semicircles, we need to divide this expression by 2. Length of a Semicircle 2π r/2 = π r We need the radius of each semicircle. The three semicircles have diameters 3, 4, and 5 units long. We will first divide each of these diameters by 2 because the radius is half the diameter. Then, we will find the length of each semicircle.
d | r = d/2 | π r | |
---|---|---|---|
Semicircle 1 | 3 | 3/2=1.5 | 1.5π |
Semicircle 2 | 4 | 4/2=2 | 2π |
Semicircle 3 | 5 | 5/2=2.5 | 2.5π |
Finally, we add the three lengths to find the perimeter of the figure. P = 1.5π+2π+2.5π ⇓ P = 6π The given figure has a perimeter of 6π.
Let's consider each of the geometric shapes that forms the shape. We can decompose the given figure into three semicircles and a right triangle.
The area of the figure is equal to the sum of the areas of the involved shapes. A = A_(△) + A_(sc_1) + A_(sc_2) + A_(sc_3) The triangle has a base of 4 units and it is 3 units high. Its area is half the product of the base and height. A_(△) = 4* 3/2 ⇒ A_(△) = 6 The area of a semicircle is half the area of the corresponding circle. Area of a Semicircle = π r^2/2 From the previous part, we already know the radius of each semicircle. We will substitute each radius in to the formula to find the corresponding area.
r | π r^2/2 | |
---|---|---|
Semicircle 1 | 1.5 | π (1.5)^2/2 = 2.25π/2 |
Semicircle 2 | 2 | π (2)^2/2 = 4π/2 |
Semicircle 3 | 2.5 | π (2.5)^2/2 = 6.25π/2 |
Let's substitute all the areas into the equation we wrote for the area of the given figure.
The given figure has an area of 6+6.25π square units.
Consider the figure shown on the coordinate plane.
The perimeter of the given figure is the distance around it. We can directly identify the lengths of the vertical and horizontal sides.
We can find the length of the slanted sides by using the Distance Formula. d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) First, we will find the length of the side with endpoints (1,5) and (3,3).
Next, we find the length of the side with endpoints (0,2) and (1,5).
Finally, we add all the side lengths to determine the perimeter of the given figure.
The given figure has a perimeter of about 22 units.
If we draw the segment connecting the vertices (3,2) and (3,3), we will get a rectangle on the right-hand side. But the other figure does not have a shape for which we know how to find its area.
However, if we extend the slant and the horizontal sides of the figure on the left, we will convert the unknown shape into a triangle. This allows us to see the given figure as a rectangle and a triangle with a triangular overlapping region.
This means that the area of the given figure is the sum of the areas of the triangle and rectangle minus the area of the overlapping region. A = A_(big triangle) + A_(rec) - A_(overlap) The base of the triangle is 4 and the height is 3. The length of the rectangle is 4 and the width is 3. The base and height of the overlapping triangle are both equal to 1.
The area of a triangle is half the product of the base and height. The area of a rectangle is the product of the length and width.
Figure | Formula | Substitute Values | Area |
---|---|---|---|
Big Triangle | A = b h/2 | A = 4* 3/2 | A=6 |
Overlapping Triangle | A = b h/2 | A = 1* 1/2 | A=1/2 |
Rectangle | A = w l | A = 4* 3 | A=12 |
We are ready to find the area of the given figure.
The given figure has an area of 17.5 square units.
Jordan has a heart-shaped mirror on her dresser. The mirror frame is 1 inch wide.
We can find the area of the mirror frame by determining the area of the mirror and frame together and then subtracting the area of the mirror. A_(frame) = A_(mirror and frame) - A_(mirror) Let's find those areas one by one.
The mirror and frame together can be divided into a square of side 10 inches and two semicircles of radius 5 inches.
The area of a square is the side length squared. Since the two semicircles have the same radius, together they form a full circle, whose area is equal to π times the radius squared.
Square | Circle | |
---|---|---|
Formula | A = l^2 | A = π r^2 |
Substitute values | A = 10^2 | A = π (5)^2 |
Calculate | A = 100 | A = 25π |
The mirror and frame together have an area of 100+25π square inches.
We are told that the frame is 1 inch wide. Therefore, the semicircles at the top have a diameter of 10-2* 1 = 8 inches. We can divide the remaining part of the mirror into a square of side 8 inches and two rectangles 8 inches long and 1 inch wide as shown in the diagram.
Again, since the two semicircles have the same radius, together they form a full circle of radius 4 inches. Also, both rectangles have the same dimensions so they have the same area. Recall that the area of a rectangle is the product of its dimensions.
Square | Circle | Rectangles | |
---|---|---|---|
Formula | A = l^2 | A = π r^2 | A= 2* l * w |
Substitute values | A = 8^2 | A = π (4)^2 | A = 2* 8* 1 |
Calculate | A = 64 | A = 16π | A = 16 |
The mirror has an area of 64+16π+16, which is the same as 80+16π square inches.
Finally, we subtract the area of the mirror from the area of the mirror and frame together to determine the area of just the frame.
The frame has an area of 20+9π square inches.