Given that the radius of O is 10 cm and OA = 8 cm, then the position relationship between point a and O is: point a is at o

Given that the radius of O is 10 cm and OA = 8 cm, then the position relationship between point a and O is: point a is at o

Because the radius of OA is 8cm
We can see the radius of OA ＜⊙ o
So: point a is inside ⊙ o

100 Mathematical Olympiad questions in grade three of primary school

The playground of Renmin Road Primary School is 90 meters long and 45 meters wide. After the renovation, the length and width of the playground will be increased by 10 meters and 5 meters respectively?
The current area of the playground is (90 + 10) × (45 + 5) = 5000 (square meters), and the original area of the playground is: 90 × 45 = 4050 (square meters), so it is 5000-4050 = 950 square meters more than before
(90 + 10) × (45 + 5) - (90 × 45) = 950 (M2)
Exercise (1) there is a rectangular board, 22 decimeters long and 8 decimeters wide. If the length and width are reduced by 10 decimeters and 3 decimeters respectively, how many square decimeters is the area reduced?
Exercise (2) a rectangle is 80 meters long and 45 meters wide. If the width is increased by 5 meters, how many meters should the length be reduced to keep the area unchanged?
2. If the width of a rectangle remains unchanged and the length increases by 6 meters, its area will increase by 54 square meters. If the length remains unchanged and the width decreases by 3 meters, its area will decrease by 36 square meters. What is the original area of the rectangle?
[thought navigation] from "width unchanged, length increased by 6 meters, then its area increased by 54 square meters", we can see that its width is 54 △ 6 = 9 (meters); from "length unchanged, width decreased by 3 meters, then its area decreased by 36 square meters", we can see that its length is 36 △ 3 = 12 (meters), so the area of this rectangle is 12 × 9 = 108 (square meters). (36 △ 3) × (54 △ 9) = 108 (square meters)
Exercise (1) if the width of a rectangle is unchanged and the length is reduced by 3 meters, then its area will be reduced by 24 square meters. If the length is unchanged and the width is increased by 4 meters, then its area will be increased by 60 square meters. What is the original area of the rectangle?
Exercise (2) if the width of a rectangle remains unchanged and the length increases by 5 meters, then its area increases by 30 square meters. If the length remains unchanged and the width increases by 3 meters, then its area increases by 48 square meters. How many square meters is the original area of the rectangle?
Exercise (3) if the length of a rectangle is reduced by 3 meters, or its width is reduced by 2 meters, then its area is reduced by 36 square meters. Find the original area of the rectangle
3. The picture below is a rectangular chicken farm enclosed by a 16 meter fence
[thought navigation] according to the title, because one side uses a wall, the two lengths plus one width equals 16 meters, and the width is 4 meters, so the length is (16-4) △ 2 = 6 (meters). Therefore, the floor area is 6 × 4 = 24 (square meters)
(16-4) △ 2 × 4 = 24 (M2)
Exercise (1) the following picture shows a poultry farmer using a 13 meter long fence to form a rectangular chicken farm. How large is the area of the chicken farm?
Exercise (2) use a 56 meter long wooden fence to form a rectangle 20 meters long or wide. One side of the fence is used. How can we maximize the enclosed area?
4. For a square steel plate, first cut off the rectangle 5 decimeters wide and then cut off the rectangle 8 decimeters wide (as shown in the figure below). The area is 181 square decimeters less than the original square. What is the side length of the original square?
[thought navigation] cut out the shadow part, and put the two small squares together (as shown in the figure below), and then add a small rectangle with the length of 8 decimeters and the width of 5 decimeters. The area of the combined rectangle is 181 + 8 × 5 = 221 (square decimeters), the length is the side length of the original square, and the width is 8 + 5 = 13 (decimeters), The side length of the original square is 221 △ 13 = 17 (decimeter)
(181 + 8 × 5) / (8 + 5) = 17 (decimeter)