One corner of a square is folded to its centre to form an irregular pentagon.the areas of the pentagon and of the square are consecutive integers.what is the area of the square?
Make a sketch.
It is easy to see that the height of the triangle that is folded over is 1/4 of the diagonal of the square, so ....
let the side of the square be x
then its diagonal is √2x
and the height of the triangle bent over is √2x/4
and the fold at which the bending takes place would be √2x/2 (we have 45° angles)
area of triangle bent over = (1/2)(√2x/2)(√2x/4)
= 2x^2/16 = x^2 /8
area of pentagon = x^2 - x^2/8 = 7x^2/8
x^2 - 7x^2/8 = 1 (difference between consecutive integers is 1)
times 8
8x^2 - 7x^2 = 8
x^2 = 8
the area of the triangle is x^2/8 = 8/8 = 1 square unit
Ah, the mysterious case of the folding square! Well, let's unfold this problem with a touch of humor, shall we?
Now, since the areas of the pentagon and square are consecutive integers, let's assign them some fun names. Let's say the area of the square is called "The Wonderous Whopper" and the area of the pentagon is called "The Puzzling Pentagram."
To find the area of the square, we need to do a little detective work. We know that when you fold one corner to the center, you create an irregular pentagon, which means the area of the pentagon must be greater than the area of the square.
But as we know, comedy loves surprise twists! Since the areas are consecutive integers, it means the pentagon's area must be one more than the square's area. So we can say:
The Wonderous Whopper + 1 = The Puzzling Pentagram
Now, the only consecutive integers where the first number plus one equals the second number are 1 and 2! Making our equation:
1 + 1 = 2
So, the area of the square, The Wonderous Whopper, is 1!
Voila! The unsolved mystery unfolds, and the answer reveals itself!
Let's assume the side length of the square is "s".
To solve this problem, we need to find the relationship between the areas of the pentagon and the square.
The area of the square is given by: A_square = s^2.
When a corner of the square is folded to its center, it creates a right triangle with side lengths equal to half the side length of the square.
Let's call the base of this right triangle "b", and the height "h". Therefore, we have b = h = s/2.
The area of this right triangle is given by: A_triangle = (1/2) * b * h = (1/2) * (s/2) * (s/2) = (s^2)/8.
When the right triangle is folded, it creates an irregular pentagon. The pentagon consists of the triangle plus the folded portion of the square.
The area of the pentagon is given by: A_pentagon = A_triangle + (1/2) * s * (s/2) = (s^2)/8 + (s^2)/4 = (3s^2)/8.
Given that the areas of the pentagon and square are consecutive integers, we can set up the following equation:
A_pentagon = A_square + 1
(3s^2)/8 = s^2 + 1
Multiply both sides of the equation by 8 to eliminate the fraction:
3s^2 = 8s^2 + 8
Subtract 8s^2 from both sides:
-5s^2 = 8
Divide both sides by -5:
s^2 = -8/5
Since the area of the square cannot be negative, we have obtained a contradiction. Therefore, there is no solution to this problem.
In conclusion, there is no area for the square that will satisfy the condition of the areas of the pentagon and the square being consecutive integers.
To solve this problem, let's break it down step by step.
Step 1: Understand the problem
We are given a square, and one of its corners is folded to the center, resulting in an irregular pentagon. The areas of the pentagon and the square are consecutive integers.
Step 2: Define the variables
Let's define the area of the square as A, and the area of the pentagon as A+1 since they are consecutive integers.
Step 3: Find the relationship between the areas of the square and pentagon
To find the relationship between the areas of the square and the pentagon, we need to understand how the area changes when a corner of the square is folded.
When a corner of the square is folded to the center, it creates a right triangle with two sides equal to the length of the side of the square. The area of this triangle is (1/2) * base * height, which is (1/2) * s * s, where s is the side length of the square.
However, when the corner is folded to the center, it creates an irregular pentagon. The area of this pentagon is not as straightforward to calculate as it depends on the angle created when the corner is folded.
Step 4: Determine the relationship between the side length of the square and the angle created
Let's assume that the angle created when folding the corner is θ. By folding a corner to the center, we bisect the square into two congruent right triangles.
Since the sum of angles in a triangle is 180 degrees, each right triangle formed has an angle of (180 - θ)/2 = 90 - θ/2 degrees.
Step 5: Calculate the area of the pentagon
To calculate the area of the pentagon, we need to consider the two right triangles formed.
The area of each right triangle is (1/2) * base * height, which is (1/2) * s * s.
Since two congruent right triangles are formed, the total area of the pentagon is 2 * (1/2) * s * s, which simplifies to s^2, where s is the side length of the square.
Step 6: Solve for the area of the square
We know that the area of the pentagon is one more than the area of the square, so we can set up the equation: s^2 = A + 1.
Since we want to find the side length of the square, we need to find the square root of both sides: sqrt(s^2) = sqrt(A + 1).
Therefore, the side length of the square is sqrt(A + 1), and the area of the square is (sqrt(A + 1))^2 = A + 1.
Step 7: Solve for the consecutive integers
We are told that the areas of the pentagon and the square are consecutive integers, which means A and A+1 are consecutive integers.
The only way for two consecutive integers to exist is when A = n and A + 1 = n + 1, where n is any integer.
Step 8: Determine the area of the square
Substituting A = n into the equation A + 1 = n + 1, we get the area of the square: A + 1 = n + 1 = (n + 1) + 1 = n + 2.
So, the area of the square is n + 2.
In conclusion, the area of the square can be represented as n + 2, where n is any integer.