The perimeter of square S is 40. Square T is inscribed in square S. What is the least possible area of square T?
The answer choices are 45, 48, 49, 50, 52. The answer is 50.
I don”t see why it can”t be 45 so that each side of Square T has length 45^(1/2). Please help. Thanks,
Sai Prasad Gochhayat
Why is the least possible value and why 49 cannot be the answer?
Feb 24, 2017•Comment
Cydney Seigerman, jovemaprendiz2019.org Tutor
Happy to help! We”re looking for the least possible value of the area, since that”s what the prompt asks for: “What is the least possible area of square T?”
Since square T is inscribed in square S, square T must touch square S on all four sides. In an earlier response, I explain how T will have the smallest area when the four vertices touch the center of each of the four sides of square S. If you have any specific questions my explanation, please let me know!
Also, this page uses figures to explain this question, that can be very useful to visualize the situation: http://math.stackexchange.com/questions/1623865/minimum-area-of-inscribed-square
I hope this helps!
Feb 24, 2017•Reply
I ajovemaprendiz2019.orge with Arvind. I don”t see any other possibility of inscribing the square. in all other cases the square wont be inscribed as it will fail to touch two sides
Sep 10, 2016•Comment
Cydney Seigerman, jovemaprendiz2019.org Tutor
Hi Bhavook 🙂
First, let”s look at the extreme case: the area of square T = 100. In this case, T would touch S in the maximum number of points: all of them!
Now, between the maximum area of 100 and the minimum of 50, T can be of any area. As we decrease the area of T, we need to rotate the square so that it maintains contact with the four sides of S. If T has any area smaller than 100, T will touch S at one point on each of the four sides and four congruent right triangles are formed from the area of S that is not shared by T. The length of the hypotenuse of each triangle is the side length of T.
When we increase the side length of T from 5sqrt(2), we must rotate T so that it remains inscribed in S. This changes the four congruent triangles formed. The hypotenuse is still the side length of T, while the two legs are not of equal length since the vertices of T are not at the midpoint of the sides of S. If we were to calculate the area of the triangles, we would see that this area is smaller than when T has side lengths of 5sqrt(2). And that makes sense, since T is larger and takes up a jovemaprendiz2019.orgater portion of the area of S.
Let”s look at an example 🙂 Say that the area of T = 52. So, the length of one side of T = sqrt(52) = 2sqrt(13). Since the sum of the length of the two legs must equal 10 (the length of a side of S), we can determine that one leg will have a length of 6 and the other a length of 4:
a^2 + b^2 = c^24^2 + 6^2 = 16 + 36 = 52c = sqrt(52) = 2sqrt(13)
So, the area of T is c^2 = 52. And the area of each of the four right triangles is 1/2*4*6 = 12.
The sum of the area of the 4 triangles (4*12) and the area of T should equal the area of S:
52 + 4*12 = 100 😀
This is also true for the smallest possible area of T (area = 50). The four triangles have areas of
1/2*5*5 = 12.5
So, the sum of the areas of the four triangles and T is
12.5*4 + 50 = 100.
You are watching: The perimeter of square s is 40
I hope this helps you understand a little bit better why 50 is the minimum possible area and not the only possible area of T 🙂
Sep 11, 2016•Reply
This is a bit tricky to explain without a visual, but bear with me.
Draw square S, first. We know that each side of the square is 10, because the perimeter is 40.
Now, we want to draw square T inside square S. But it”s not just inside—it”s inscribed. That means that the four corners of square T all fall on the lines of square S. In other words the two squares touch in four places (the corners of S).
We”ll try a few different ways to draw this to see how the placement of those four corners affect the size of square T. If the four points are very, very close to the four corners of square S, then the sizes would be very near equal. They would be almost the exact same square.
But if we draw the corners of square T further from the corners of S, the square shrinks. The further away, the smaller square T becomes. What”s the furthest each point can be from the corners of S? The center of each side of S. At this point, square T looks like a diamond—the corners point straight up and down, and the sides are at even 45 dejovemaprendiz2019.orge angles.
To find the area, we square that, and we get 25sqrt(4), or 25 * 2 = 50.
If it were any smaller, the corners would not reach all the way to fall on the sides of square S. It would be inside, but it would not be inscribed.
Jun 6, 2014•Comment
As Lucas mentioned above, there are multiple ways to draw square T so that it”s inside square S. And the way we draw it can change the area of square T.
Square T is inscribed in square S. So the four corners of square T all have to fall on the lines of square S. In other words the two squares touch in four places (the corners of S).
As Lucas explained above, the placement of those four corners affect the size of square T. If the four points are very, very close to the four corners of square S, then the sizes of squares S and T would be almost the same.
But if we draw the corners of square T further away from the corners of S, square T shrinks. The further away, the smaller square T becomes. If you find this hard to visualize, you can draw it yourself – and you”ll see that the farther away from the corners of square T are from the corners of square S, the smaller the area of square T! So therefore, there are many different possible values for the area of square T. 🙂