A circle with an inscribed square - can you find the area of the square?
Вставка
- Опубліковано 8 жов 2024
- Geometry area problem square inscribed in circle. Learn more math at TCMathAcademy.....
TabletClass Math Academy - TCMathAcademy....
Help with Middle and High School Math
Test Prep for High School Math, College Math, Teacher Certification Math and More!
Popular Math Courses:
Math Foundations
tabletclass-ac...
Math Skills Rebuilder Course:
tabletclass-ac...
Pre-Algebra
tabletclass-ac...
Algebra
tabletclass-ac...
Geometry
tabletclass-ac...
Algebra 2
tabletclass-ac...
Pre-Calculus
tabletclass-ac...
Math Notes: tcmathshop.com/
If you’re looking for a math course for any of the following, check out my full Course Catalog at: TCMathAcademy....
• MIDDLE & HIGH SCHOOL MATH
• HOMESCHOOL MATH
• COLLEGE MATH
• TEST PREP MATH
• TEACHER CERTIFICATION TEST MATH
Pssst .... too much talk and too much explanation causes confusion. You only needed about 2 minutes to adequately explain the method and solution.
For real, as soon as I looked at the thumbnail, I knew that since the radius is 4, the diameter is 8, then turn the square into two triangles, and the hypotenuse is 8. Then use Pythagrian Theorum to find the sides of the square. Side times side equals area of the square
I did this in my head in about 30 seconds
A couple of ways...
The diagonal of the square = the diameter of the circle = 8
From a bit of Pythagoras (or from knowing that the sine and cosine of 45 degrees is 1/√2), the diagonal of a square is √2 × the side length, so the side length is 8/√2
So area = (8/√2)² = 64/2 = 32
Or...
Drawing the diagonal of the square makes two triangles, each with base length = diameter of the circle = 8 and triangle height = radius = 4.
Triangle area = ½ × base × height = ½×8×4 = 16. But we have two such triangles so total area = 16+16 = 32.
Your first way of solving is the simplest, most straightforward way of solving the problem. I essentially solved the problem that way, except I converted 8 to "2 cubed," divided by sort. 2, then squared what was left.
@@robertstuart6645 Interesting. I think I found the second way simpler. In my head I didn't actually do ½×base×height and then double it. I just did base×height.
@@gavindeane3670how about this solution.
An easy way to calculate it is as follows.
1. Draw two diagonals creating 4 small triangles.
2. Label the top edge of the square C and the two legs of the small triangle, A and B.
3. We know that A=4. B=4
4. Using a^2 + b^2 = c^2
5. 4^2 + 4^2 = c^2 = 32
6. We also know that c^2 is the area of the square. Area = 32.
Area of triangle = .5 x base x height = 4 x4 x2 = 32
Thank you for the videos. You are awesome at explaining everything.
This one is really easy again... rule of Pythagoras a² + b² = c² with for the square a=b
so the area of the square A = a² and diagonal of the square c = 2r = 8
a² + a² = 8² so 2a² = 64 and a² = 64/2 = 32 and that is the area of the square.
Very abstract. Good one.
Yes! Don’t even need to know the sides.
32 squ. in about 10 seconds in my head. It's a circumscribed square, so finding the area of one the right triangles made by drawing lines from the center to a corner of the square is easy enough. Actually, the area is 2r², where r is the radius of the circle.
2 x radius = the diameter of the circle = 8
The diameter of the circle = one of the 2 diagonals of the square = 8
The diagonal split the square in 2 congruent right angle triangles with angles 45° 45° 90° --->
1 : 1 :√2
given :
Hypotenuse = 8
Calculate the other side --> 8 / √2 since it's a 45° 45° 90° with the sides ratio of 1 : 1 : √2
the length of the sides are 8 / √2 : 8 / √2 : 8
The area of a square is side x side = (8 / √2 ) x (8 / √2) = (8 / √2 )² = (64 / 2) = 32 some square units
Let’s draw 2 lines, each joining the opposite corners of the square, passing through the center of the circle.
Then, the square has 4 right angled triangles inside it.
Each of the right angled triangle has two equal sides of 4 units each.
In a right angled triangle,
a^2+b^2= c^2,
where a and b are two sides and c is the hypotenuse.
a = b = 4
So, a^2 = 4^2= 16
b^2 = 4^2 = 16
a^2+b^2= 16+16=32
Therefore, c^2=32
c = each side of the square
Area of the square = c^2 = 32 units.
Parse que
c = 32^(1/2)
I thought of this first as well. 🤓
That's the way I did it too! I think it's simpler and more elegant.
Come to think of it, if we draw two diagonals we have four right-angled triangles, each with base 4 and height 4, so:
4(2 × 4) = 32
diameter in the square = 8, means the ara must be 64/2....= 32
why show us all the s a unness math ?
like reputation of being a gig-fessor ?
take the easy problems and make them diiiiiificult.
I didn't even think of a^2 + b^2 = c^2.
I got it this way:
Since the other two angles are 45 degrees, I took the SINE of 45 degrees and multiplied it by 8 and got 5.65685..., squared it,and got 32. The COSINE works, too.
Yep, overthinking a problem as usual...
Or you could just use multivariable calculus for the volume of solid revolution for the square and circle as both there functions are easy to determine (well the circles function is x^2+y^2=r^2 so x^2+y^2=16
Divide the square into two triangles by bisecting it from corner to opposite corner. The bisect will pass through the center of the circle is a diameter and equals radius x 2 or 8 and it is the hippo of the new triangle. Square that to 64 and then half the 64 to 32 and since that 32 is the square of one of the sides of the triangle, it is the area of the square. As Pythagoras said 'The squaws of the hippopotamus is equal to the sum of the squaws of the other two hides.
An easy way to calculate it is as follows.
1. Draw two diagonals creating 4 small triangles.
2. Label the top edge of the square C and the two legs of the small triangle, A and B.
3. We know that A=4. B=4
4. Using a^2 + b^2 = c^2
5. 4^2 + 4^2 = c^2 = 32
6. We also know that c^2 is the area of the square. Area = 32.
At the thumbnail, solved in my head, haven't gone back to check on paper or with calculator, I think the answer is 32 units.
My thinking:
The radius of the circle is 4, so the diagonal of the inscribed square is twice that, 8.
Working the Pythagorean theorem backwards for a triangle forming half the square, and plugging 8 in for C, we get c^2 = 64.
A^2 and B^2 are each half of C^2, or 32.
We could then get the square root of 32 for A & B, which would be 5.?????, but htis isn't necessary, since we can figure the area of the square as A * B, or in this case A^2 or B^2 since they're equivalten, so the result takes us back to 32.
Diagonal of the square = 2r = 8
Let the side of the square be x and the area of the square be a.
2x² = 8²
2x² = 64
x² = 32
∴ a = 32 units²
Very abstract. Good one.
@@louf7178 Thank you, but it's just Pythagoras.
@@dazartingstall6680 I see it plain as day now, but seems evasive as a chemical double reaction.
@@dazartingstall6680It is just Pythagoras, but it rather neatly solves for x² directly, rather than getting to x² by solving for x first.
@@gavindeane3670 On the other hand, I realised a while back that the half × base × height method yields a nice little proof. If we inscribe a circle within a square, and a second square within the circle, the outer square will always have twice the area of the inner.
As noted elsewhere in the thread, the area of the inner square simplifies to the diameter × the radius.
The area of the outer square is d × d
We can restate that as 2 × r × d
QED.
The radius can be used as 2 known sides of a triangle with one side of the square being the 3rd.
4² + 4² = side² 32 = side²
we know that the area of a square is the side squared so we have the answer 32
the square's CORNERS touch the circle. Therefore the square's DIAGONAL = 2r.
A square's diagonal is 45° to its side. Therefore the 45/45/90 triangle has sides of ratio: 1/1/sqrt(2). Or the square's side
= (1/sqrt(2))×Diagonal
= (1/sqrt(2))×2R
= (1/sqrt(2))×8
Area of square
= S × S
= [(1/sqrt(2))×8]^2
= (1/2)(64)
= 32 units^2
2r^2 should do it, unless I'm missing something.
you missed some school lessons?
@@pippodeclown I am sure I did. Let me explain my reasoning.
r^2=4*4=16 This gives you a square area that covers half the area. If you want to visualize, you can draw it starting with one corner in the centre of the circle. Then half the r^2 square covers 1/4 of the target square. If you then cut the r^2 square diagonally and put the half that was outside inside the target square. You now cover half the target square. Now you have to multiply by two to get the answer.
2x16=32
So 2r^2 should always give you the largest square area you can put into a circle.
@@KaB-d3n Great one 😃
The consensus seems to be that this solution was much simpler for the class than Mr. M expected it to be. Agreed. Before the presentation was finished I began exploring the difference between total area of the circle and the square, easy enough. 50.272 - 32 = 18.272. Then the ratio, which will be constant, and the circumference. Child's play so far, right? Now wondering what is the area between the intersections of one side of the square and the arc between them. Does this require calculus?
And before anyone answers 18.272/4 = 4.568, that is not what I'm asking.
ed: bad student here. I started from the rabbit hole we were told not to go down. 8/1.414 to solve for a or b.
Or is it quadratic?
I want a smile face sticker
I've been thinking of drawing it on a T shirt, including A+++ and the ticks.
If you draw a line from each corner of the square to the opposite corner
you then have four triangle shapes that can simply be arranged to make two squares
and these two squares are both 4 x 4 4 x 4 = 16 and 16 x 2 = 32
The square has an area of 32 sq cm
Sides are x, diagonal of square =8; For the triangle x^2 +x^2 = 8^2; 2x^2 = 8^2; x^2 = 32, area of square. Done.
This sound much more complicated than it is. Using the KISS principle (Keep It Simple Stupid) Here's the simplest explanation of the simplest solution -
If a radius is drawn from the centre of the circle to each corner, the square becomes divided into four right-angle triangles each with two sides of 4. These can be simply rearranged into a rectangle with an area of 4 x 8. Job done.
Diameter = 8 = hypotenuse.
Square so bisected angles are 45.
8 Cos 45 = 4 root 2.
4 root 2 squared = 32.
Using COS is quicker.
Interesting, without all the complicated maths I thought a square that fits exactly inside a circle is half the area of the circle.
Radius 4 or Diameter 8 x 3.14 ÷ 2
Correct me if I'm wrong please.
Diameter=8= diagonol of square ,2a sqr=64,a sqr=32=Area ofsquare
OMG great lesson 👍
Wow 😂 thanks Mr J 👍👏🙏💪😎🌎
Rats! I used r to get 32 via the smaller triangle. D would have been so much easier.
got 32 Diam (Hypot) = 8 Pyth is 64 = 2X^2 so 32 = X^2 so side = sqrt 32 so B x H = 32 thanks for the fun
If I'm correct that means a square that fits exactly around the circle will be double the area of the square inside the circle.
Sin (45 degrees) x hypotenuse
Dang! I got 31.99....I used too many decimal places. Mr Mathman is that still a pass??
((32)power 1/2)power2) = 32 power 1 , 32 units squared
Cos 45 x 8 = 5.6568
Ans squared = 32
Draw a diagonal and you've got two trangles with a base of 8 and a height of 4 ...
I did half base x height for area of one triangular half of square so half 8=4 height is radius too so 4 so half the square is 4x4=16 add the other half also 16 so tot area 32
A nice reminder that we don't need to automatically reach for Pythagoras when we see a right-angle triangle. I like this.
I just made 4 right triangles bxh/2 then multiply times 4
this looked complicated. the area of the triangle is 8*4/2 or 16 and the square has two triangles.
A=((2r)^2)/2
S=32
Area of square 25.12
32
32sq.units
16+16=32
Sorry 12.56😊
I come up with thirty two.
I got 32.0096
Why, oh why, do these videos take so long to explain something so simple? The average high school student would solve problems like this in a few seconds. Why so much repetition and unnecessary verbiage? Does UA-cam reward such stuff?
I like that you are challenging people to use their brains to solve problems but I have never been able to watch a single one of your videos to the end. The condescending style is like nails on a chalkboard to me. Is it really necessary to talk to your audience as if they are 6 years old?
I change video speed to 2X and it is still very understandable. So half the regular time. It's like removing a band-aid by pulling it off quickly. Doesn't hurt as much.
Maybe you're not the target audience
This channel is famous for too much talk for even mental sums....ask him to show three sums in one lot !
Please share to the point. Too much yak!
I lost the will to live after just 1 minute. I do hope the narrator doesn't teach maths anywhere because his students will be old men by the time he gets to a testable level.
32
32
32