The figures in (a) and (b) below are made up of semicircles and quarter circles; the figure in (c) shows a quarter of a circle in a square. Find the area and the perimeter of each figure in terms of π.
since I can't put a URL to the pic I will try to describe it
1/ a square with the top right corner as A, The bottom right is B, and C is next to B and D is next to A
2/ side DC is 8 in
3/ a diagonal line from B to D
4/ and arc from B to D in triangle ABD
So first of all, I made the 'arc' as part of a circle, by construction, then I found the diameter, circumference, and area. But now I don't know what else to do.
I am only in seventh grade so I don't really understand the topic, my teacher also wants us to make a statement and reason chart. I have a lot of difficulty writing the reason so can you help by showing the chart and "explaining" it, it would be a real help. :)
(I know it's a lot, but if you want you can just show me an equation to find the answer)
Thanks, :)
I can definitely help you with that! Let's break down the problem step by step and create a statement and reason chart to solve it.
Figure (a):
- Given that the figure is a square with side DC measuring 8 inches.
- The area of a square is given by the formula A = s^2, where s is the length of a side.
Therefore, the area of the square in figure (a) is A = 8^2 = 64 square inches.
- The perimeter of a square is given by the formula P = 4s, where s is the length of a side.
Therefore, the perimeter of the square in figure (a) is P = 4(8) = 32 inches.
Figure (b):
- Given that the figure is the same square as in figure (a), but with a diagonal line from B to D and an arc from B to D within triangle ABD.
- Let's first find the length of the diagonal line BD using the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In figure (a), AD and AB are sides of a square with length 8, so they are equal in length.
Using the Pythagorean theorem: BD^2 = AD^2 + AB^2
BD^2 = 8^2 + 8^2
BD^2 = 64 + 64
BD^2 = 128
Taking the square root of both sides: BD = sqrt(128) = 8sqrt(2) inches (rounded to two decimal places).
- The arc from B to D within triangle ABD is actually a quarter of a circle.
- The circumference of a circle is given by the formula C = 2πr, where r is the radius.
Since the arc from B to D forms a quarter of a circle, the radius would be half the diagonal BD.
The radius of the circle is r = (1/2) * BD = (1/2) * 8sqrt(2) = 4sqrt(2) inches (rounded to two decimal places).
The circumference of the circle (the arc in this case) is C = 2πr = 2π * 4sqrt(2) = 8πsqrt(2) inches (rounded to two decimal places).
- To find the area of the sector (the shaded region formed by the arc), we need to find the angle of the sector.
We know that the whole circle has 360 degrees and the arc is a quarter of the circle.
Therefore, the angle of the sector is 1/4 * 360 degrees = 90 degrees.
The area of the sector is given by the formula A = (θ/360) * πr^2, where θ is the angle in degrees and r is the radius.
Substituting the values: A = (90/360) * π * (4sqrt(2))^2 = 4π square inches (rounded to two decimal places).
- The perimeter of figure (b) is the sum of the lengths of all the sides.
In this case, we have the perimeter P = AB + AD + BD + the quarter-circle arc from B to D.
The lengths of AB and AD are both 8 inches.
The length of BD is 8sqrt(2) inches.
The quarter-circle arc from B to D has a length equal to one-quarter of the circumference of the circle, which is 8πsqrt(2) inches (as calculated earlier).
Therefore, the perimeter of figure (b) is P = 8 + 8 + 8sqrt(2) + 8πsqrt(2) inches.
I hope this explanation and the statement and reason chart help you understand the problem better! Let me know if there's anything else I can assist you with.
Of course! I'll do my best to help you with this problem.
Let's break it down step by step.
First, let's calculate the area and perimeter of figure (a). Since figure (a) is made up of a square and a quarter circle, we'll calculate those separately and then add them together.
1. Calculate the area of the square:
The area of a square is found by multiplying the length of one of its sides by itself. In this case, the side length is given as 8 inches. Therefore, the area of the square is (8 in) * (8 in) = 64 square inches.
2. Calculate the area of the quarter circle:
To find the area of a quarter circle, we need to find the area of a full circle and then divide it by 4. The quarter circle has its center at point B, and the diameter is the length of side DC, which is also 8 inches. The radius of the circle is half of the diameter, so the radius is (8 in) / 2 = 4 inches. The area of a full circle is πr^2, so the area of the quarter circle is (1/4) * π * (4 in)^2 = 4π square inches.
3. Calculate the total area of figure (a):
To find the total area, we add the area of the square and the area of the quarter circle. Therefore, the total area is 64 square inches + 4π square inches.
Now let's move on to calculating the perimeter of figure (a):
4. Calculate the perimeter of the square:
The perimeter of a square is found by adding up the lengths of all four sides. In this case, all sides of the square are equal, so the perimeter is 4 times the length of one side. Therefore, the perimeter of the square is 4 * 8 inches = 32 inches.
5. Calculate the circumference of the quarter circle:
The circumference of a quarter circle is equal to one-quarter of the circumference of a full circle. Since the quarter circle has the same radius as mentioned earlier (4 inches), we can calculate its circumference as (1/4) * 2π * 4 inches.
6. Calculate the total perimeter of figure (a):
To find the total perimeter, we add the perimeter of the square and the circumference of the quarter circle. Therefore, the total perimeter is 32 inches + (1/4) * 2π * 4 inches.
Now that we've completed the calculations for figure (a), you can follow a similar process to find the area and perimeter of figure (b) and figure (c). Remember to apply the appropriate formulas for the shapes involved in each figure.
Regarding the statement and reason chart, it's a tool used to organize your thoughts and explanations when solving mathematical problems. It helps you structure your reasoning and provide logical justifications for each step of your solution. Below is an example of how you could set up the chart for this problem:
Statement | Reason
-------------------------
Area of square = 64 sq in | Formula for the area of a square is side^2
Area of quarter circle = 4π sq in | Formula for the area of a quarter circle is (1/4) * π * radius^2
Total area = 64 sq in + 4π sq in | The total area is the sum of the area of the square and the area of the quarter circle
Perimeter of square = 32 in | The perimeter of a square is 4 times the length of one side
Circumference of quarter circle = (1/4) * 2π * 4 in | The circumference of a quarter circle is one-quarter of the circumference of a full circle with the same radius
Total perimeter = 32 in + (1/4) * 2π * 4 in | The total perimeter is the sum of the perimeter of the square and the circumference of the quarter circle
I hope this explanation and the chart structure provided are helpful to you! If you have any further questions or need additional assistance, feel free to ask.
what should I do with the area of the triangle? should I subtract it from the area of the quarter of the whole circle? and to find the perimeter should I find the perimeter of the whole circle then divide it by 4?
thanks for your help Reiny! :)
from 4/
an arc from B to D suggests that the arc passes through A
If that is the case, BD is the diameter and your arc is a half-circle.
(Any angle subtended by a diameter is 90° )
I am going to guess that half the square lies within the half-circle.
If that is the case, we can find BD by Pythagoras:
BD^2 = 8^2 + 8^2
BD^2 = 128
BD = √128 or 8√2
The centre of your half-circle is the midpoint of BD, thus its radius is 4√2
The area of the semicircle is (1/2)π(4√2)^2
= (1/2)π(32)
= 16π
the triangle has area of (1/2)(8)(8) = 32
find any other stuff from here, such as shaded areas, perimeters of whatever