A square garden has fourteen posts along each side at equal intervals. Find how many posts are there in all four sides ? Please explain in detail.
Answer
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Hint: This particular question can be done in two cases. In the first case, we consider that the posts are between the corners. There are no posts which are attached to the corner. In the second case, we should consider that the posts are starting from the corner. If we consider this way, then there would be a post at each corner of the square. Here we have to be careful as there is a chance of re-counting the same post again.
Complete step by step solution:
Let us look at the first case. In this case , we consider that the posts are between the corners.
It is mentioned in the question, that there are $14$ posts spread at equal intervals on each side.
We all know that a square has $4$ side. And each side has $14$ posts.
So the number of posts in the first case would be $14\times 4=56$.
Now let us look at the second case. In this case, we will consider that the posts are starting from the corner. If we consider this way, then there would be a post at each corner of the square.
So on the first side there would be as usual $14$ posts with one post standing at corner $1$ and another post standing at corner $2$. Now let us come to the next side. This side would have $13$ posts.
It will not have $14$. Since the post at corner $2$ is common to the first side and second side. Now let us come to the third side. This side would have $13$ posts too since the post at corner $3$ is common to the second side and third side.
Now let us come to the next side. This side would have $12$ posts. Since the post at corner $1$ would be common to the first side and fourth side and the post at corner $4$ would be common to side three and side four.
We eliminate this way so as to avoid re-counting.
So the total number of posts in the second case would be $4+13+13+12=52$
Note: We should be very careful while solving these types of questions. It is always better to take two cases and solve for each case. We should have enough practice to have logical thinking. We should be able to think about the posts at the corner too and how they are common with the other side and how to avoid re-counting. Practice is needed to complete these kinds of questions quickly and accurately in the exam.
Complete step by step solution:
Let us look at the first case. In this case , we consider that the posts are between the corners.
It is mentioned in the question, that there are $14$ posts spread at equal intervals on each side.
We all know that a square has $4$ side. And each side has $14$ posts.
So the number of posts in the first case would be $14\times 4=56$.
Now let us look at the second case. In this case, we will consider that the posts are starting from the corner. If we consider this way, then there would be a post at each corner of the square.
So on the first side there would be as usual $14$ posts with one post standing at corner $1$ and another post standing at corner $2$. Now let us come to the next side. This side would have $13$ posts.
It will not have $14$. Since the post at corner $2$ is common to the first side and second side. Now let us come to the third side. This side would have $13$ posts too since the post at corner $3$ is common to the second side and third side.
Now let us come to the next side. This side would have $12$ posts. Since the post at corner $1$ would be common to the first side and fourth side and the post at corner $4$ would be common to side three and side four.
We eliminate this way so as to avoid re-counting.
So the total number of posts in the second case would be $4+13+13+12=52$
Note: We should be very careful while solving these types of questions. It is always better to take two cases and solve for each case. We should have enough practice to have logical thinking. We should be able to think about the posts at the corner too and how they are common with the other side and how to avoid re-counting. Practice is needed to complete these kinds of questions quickly and accurately in the exam.
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