Hello friends!IN composition of the Unified State Examination in mathematicsincludes tasks related to finding the area of a circle or its parts (sectors, ring elements). The figure is set on a sheet of paper in a checkered pattern. In some problems the scale of the cell is given as 1×1 centimeter, in others it is not specified - the area of the element of the circle or the circle itself is given.
The tasks are shallow, you need to remember the formula for the area of a circle, be able to visually (by cells) determine the radius of the circle, what proportion of the circle is the selected sector. By the way, on the blog about the area of the sector. Its content has nothing to do with solving the problems presented below, but for those who want to remember the formula for the area of a circle and the area of a sector it will be very useful. Consider the tasks (taken from the open task bank):
Find (in cm2) the area S of the figure shown in checkered paper with a cell size of 1 cm x 1 cm. Write down S/l in your answer.
In order to obtain the area of a figure (ring), it is necessary to subtract the area of a circle with radius 1 from the area of a circle with radius 2. The formula for the area of a circle is:
Means,
Divide the result by Pi and write down the answer.
Answer: 3
Two circles are drawn on checkered paper. The area of the inner circle is 51. Find the area of the shaded figure.
The area of the shaded figure can be found by calculating the difference between the area of the larger circle and the area of the smaller one. Let us determine how many times the area of the larger one differs from the area of the smaller one. Let the radius of the smaller one be equal to R, then its area is equal to:
The radius of the larger circle is twice as large (visible by the cells). So its area is equal to:
We found that its area is 4 times larger.
Therefore, it is equal to 51∙4 = 204 cm 2
Thus, the area of the shaded figure is 204 – 51 = 153 cm 2.
*Second method. It was possible to calculate the radius of the small circle, then determine the radius of the larger one. Next, find the area of the larger one and calculate the area of the desired figure.
Two circles are drawn on checkered paper. The area of the inner circle is 1. Find the area of the shaded figure.
This problem is practically no different from the previous one in its solution; the only difference is that the circles have different centers.
Despite the fact that it is clear that the radius of the larger circle is 2 times greater than radius smaller, I advise you to designate the size of the cell with the variable x (x).
Just as in the previous problem, let’s determine how many times the area of the larger one differs from the area of the smaller one. Let's express the area of the smaller circle, since its radius is 3x:
Let's express the area of the larger circle, since its radius is 6x:
As you can see, the area of the larger circle is 4 times larger.
Therefore, it is equal to 1∙4 = 4 cm 2
Thus, the area of the shaded figure is 4 – 1 = 3 cm 2.
Answer: 3
Two circles are drawn on checkered paper. The area of the inner circle is 9. Find the area of the shaded figure.
Let us denote the size of the cell by the variable x (x).
Let us determine how many times the area of the larger circle differs from the area of the smaller one. Let's express the area of the smaller circle. Since its radius is 3∙ x, then
Let's express the area of the larger circle. Since its radius is 4∙ x, then
Divide the area of the larger one by the area of the smaller one:
That is, the area of the larger circle is 16/9 times more area less, therefore it is equal to:
Thus, the area of the shaded figure is 16 – 9 = 7 cm 2.
*Second method.
Let's calculate the radius of the smaller circle. Its area is 9, which means
Let's find the size of the cell and then we can determine the radius of the larger circle. The cell size is:
Since the radius of the larger circle corresponds to 4 cells, its radius will be equal to:
Determine the area of the larger circle:
Find the difference: 16 – 9 = 7 cm 2
Answer: 7
A circle with an area of 48 is drawn on checkered paper. Find the area of the shaded sector.
In this problem, it is obvious that the shaded part is half the area of the entire circle, that is, equal to 24.
Answer: 24
A short summary.
In problems related to the area of a sector of a circle, it is necessary to be able to determine what proportion it makes up of the area of the circle. This is not difficult to do, since in such problems central angle sector is a multiple of 30 or 45.
In problems related to finding the areas of ring elements, there are different ways for solution, both are shown in the solved problems. The method in which the size of the cell is indicated through the variable x, and then the radii are determined, is more universal.
But the most important thing is not to memorize these methods. You can find a third and fourth solution. The main thing is to know the formula for the area of a circle and be able to reason logically.
That's all. Good luck to you!
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shaded figure.
188. Prototype of task B5 (No. 315124)
Two circles are drawn on checkered paper. The area of the inner circle is 9. Find the area of the shaded figure.
189. Prototype of task B5 (No. 315132)
A circle with an area of 48 is drawn on checkered paper. Find the area of the shaded sector.
190. Prototype of task B5 (No. 315133)
A circle is depicted on checkered paper. What is the area of the circle if the area of the shaded sector is 32?
193. Prototype of task B5 (No. 319057)
The area of parallelogram ABCD is 176. Point E is the midpoint of side CD. Find the area of triangle ADE.
194. Prototype of task B5 (No. 319058)
Square triangle ABC equal to 12.DE - the middle line parallel to side AB. Find the area of the trapezoidABDE.
195. Prototype of task B5 (No. 324460)
Points A and B are marked on checkered paper with a square size of 1 × 1. Find the length of segment AB.
196. Prototype of task B5 (No. 324461)
An angle is depicted on checkered paper with a square size of 1 × 1. Find its degree value.
197. Prototype of task B5 (No. 324462)
On checkered paper with a square size of 1 × 1, triangle ABC is depicted. Find its length midline, parallel to side AB.
198. Prototype of task B5 (No. 324463)
On checkered paper with a square size of 1 × 1, triangle ABC is depicted. Find the length of its height lowered to side AB.
191. Prototype of task B5 (No. 317338)
The area of parallelogram ABCD is 189. Point E is the midpoint of side AD. Find the area of the trapezoid AECB.
192. Prototype of task B5 (No. 319056)
The area of parallelogram ABCD is 153. Find the area of parallelogram A " B " C " D " whose vertices are the midpoints of the sides of this parallelogram.
199. Prototype of task B5 (No. 324464)
An isosceles right triangle is depicted on checkered paper with a square size of 1 × 1. Find the length of its median drawn to the hypotenuse.
200. Prototype of task B5 (No. 324465)
Points A, B and C are marked on checkered paper with a square size of 1 × 1. Find the distance from point A to line BC.
201. Prototype of task B5 (No. 324466)
A triangle is depicted on checkered paper with a square size of 1 × 1. Find the radius of the circle described around it.
Circles require a more careful approach and are much less common in tasks B5. At the same time, general scheme the solutions are even simpler than in the case of polygons (see lesson “ Areas of polygons on a coordinate grid »).
All that is required in such tasks is to find the radius of the circle R. Then you can calculate the area of the circle using the formula S = πR 2. It also follows from this formula that to solve it it is enough to find R 2.
To find the indicated values, it is enough to indicate a point on the circle that lies at the intersection of the grid lines. And then use the Pythagorean theorem. Let's consider specific examples radius calculations:
Task. Find the radii of the three circles shown in the figure:
Let's do it additional constructions in each circle:
In each case, point B is chosen on the circle to lie at the intersection of the grid lines. Point C in circles 1 and 3 complete the figure to right triangle. It remains to find the radii:
Consider triangle ABC in the first circle. According to the Pythagorean theorem: R 2 = AB 2 = AC 2 + BC 2 = 2 2 + 2 2 = 8.
For the second circle everything is obvious: R = AB = 2.
The third case is similar to the first. From triangle ABC using the Pythagorean theorem: R 2 = AB 2 = AC 2 + BC 2 = 1 2 + 2 2 = 5.
Now we know how to find the radius of a circle (or at least its square). Therefore, we can find the area. There are problems where you need to find the area of a sector, and not the entire circle. In such cases, it is easy to find out what part of the circle this sector is, and thus find the area.
Task. Find the area S of the shaded sector. Please indicate S/π in your answer.
Obviously, the sector is one quarter of a circle. Therefore, S = 0.25 S circle.
It remains to find S of the circle - the area of the circle. To do this, we perform an additional construction:
Triangle ABC is a right triangle. According to the Pythagorean theorem we have: R 2 = AB 2 = AC 2 + BC 2 = 2 2 + 2 2 = 8.
Now we find the area of the circle and the sector: S circle = πR 2 = 8π ; S = 0.25 S circle = 2π.
Finally, the desired value is S /π = 2.
Sector area with unknown radius
This is absolutely new type tasks, there was nothing like this in 2010-2011. According to the condition, we are given a circle certain area(namely area, not radius!). Then, inside this circle, a sector is selected, the area of which needs to be found.
The good news is that similar tasks- the easiest of all the problems on the area that occur in the Unified State Examination in mathematics. In addition, the circle and sector are always placed on a coordinate grid. Therefore, to learn how to solve such problems, just look at the picture:
Let the original circle have an area S circle = 80. Then it can be divided into two sectors of area S = 40 each (see step 2). Similarly, each of these “halves” sectors can be divided in half again - we get four sectors with area S = 20 each (see step 3). Finally, we can divide each of these sectors into two more - we get 8 “scraps” sectors. The area of each of these “scraps” will be S = 10.
Please note: there is no smaller partition in any Unified State Exam task no in math! Thus, the algorithm for solving Problem B-3 is as follows:
- Cut the original circle into 8 “scraps” sectors. The area of each of them is exactly 1/8 of the area of the entire circle. For example, if according to the condition the circle has an area S of the circle = 240, then the “scraps” have an area S = 240: 8 = 30;
- Find out how many “scraps” fit in the original sector, the area of which needs to be found. For example, if our sector contains 3 “scraps” with an area of 30, then the area of the desired sector is S = 3 · 30 = 90. This will be the answer.
That's it! The problem is solved practically orally. If something is still not clear, buy a pizza and cut it into 8 pieces. Each such piece will be the same sector-“scraps” that can be combined into larger pieces.
Now let’s look at examples from the trial Unified State Exam:
Task. A circle is drawn on checkered paper with an area of 40. Find the area of the shaded figure.
So, the area of the circle is 40. Divide it into 8 sectors - each with area S = 40: 5 = 8. We get:
Obviously, the shaded sector consists of exactly two “scraps” sectors. Therefore, its area is 2 · 5 = 10. That's the whole solution!
Task. A circle is drawn on checkered paper with an area of 64. Find the area of the shaded figure.
Again, divide the entire circle into 8 equal sectors. Obviously, the area of one of them is exactly what needs to be found. Therefore, its area is S = 64: 8 = 8.
Task. A circle is drawn on checkered paper with an area of 48. Find the area of the shaded figure.
Again, divide the circle into 8 equal sectors. The area of each of them is equal to S = 48: 8 = 6. The required sector contains exactly three sectors - “scraps” (see figure). Therefore, the area of the required sector is 3 6 = 18.
