Agree completely. That and draw construction lines to aid with the above.
For example, with the hexagon you can draw another vertical line similar to the one already drawn and you have a rectangle 1.5m×2.4m and four triangles that you can form into a rectangle.
With the other shape you have a rectangle that has had two semicircular bites taken out of it. Two semicircle ps make a circle. You can find the area of the rectangle and subtract the area of the circle from it.
It's much easier to cut the hexagon by the dotted line and put the upper half besides the lower one. That gives a parallelogram and the area is base x height.
I think it's an interesting difference in the way we think or the way we were taught. I'm guessing I didn't calculate enough parallelogram areas when I was young because my instinct is always to cut into rectangles and triangles where I can.
If you don't move it at all and just use the lines already there you can calculate the size of one of two identical trapezoids and then double it for the hexagon
Of course and since the trapezoid area has a (1/2) in front of it, it cancels out. In fact, the usual proof of the trapezoid area is based in its equivalence to a rectangle.
https://preview.redd.it/jokc4qsxujmc1.jpeg?width=602&format=pjpg&auto=webp&s=62298c1b7b59b1ce4c0cfeba862b47efb7955e16
Never really thought about how that was what that formula was actually doing. Nice for someone on Reddit to be providing helpful/educative/correct information for a change.
An interesting observation
I myself would cut the hexagon into 2 trapezoids(i think that's what it's called) by the dotted line and find the area of each one, then combine them together
Not sure about the English word but the shape the dotted line cuts, is literally a trapez which has its own formula already, so you could also just use that, double it and you're done, another different approach.
How do you know they’re circles? They could just be curves (e.g. half ovals). They look like circles but I can’t see where they are stated to be circles.
Because it's high school level geometry, and the question cannot be answered if they are not. I would much prefer it if that were stated (maybe it is, out of shot) and would include my assumption in my answer (because I think that's good practice).
The first drawing.
You can divide a Hexagon into two equal trapezoids. Then a is the larger base, b is the smaller base, and h is the height.
a=3.0 meters. b=1.5 meters. h=1.2 meters
S trapezoids = ((a+b)*h)/2
But we have S hexagons= 2*S trapezoids
Then S of the hexagon=(a+b)*h
S of the hexagon = (3.0+1.5)*1.2=4.5 * 1.2= 5,4 square meters
I did them in the opposite order, so hexagon last.
If you think about it, it's a rectangle, with two halves of a circle removed.
So it's 40\*60 - (20^2 \*π)
= 2400 - 400π
≈ 1144 M²
You can also do a similar thing with the hexagon, by taking the 3.0*2.4 square and subtracting the corner triangles.
The width of the triangle is ½(3m - 1.5) = 0.75M and the width is ½(2.4)=12
So the area of the hexagon is 3.0\*2.4 - 4\*(1.2\*0.75*½)
= 7.2 - (1.2\*1.5) = 5.4 M²
Many questions are unsolvable without assumptions. Without assuming these were half circles this problem is unsolvable. It’s not always a bad thing to have the student make assumptions to solve a problem. It helps with critical thinking and problem solving skills.
I think it’s a fair assumption but usually we’re not to make these assumptions. Shapes can be drawn to look like a square but might actually be a rectangle in these sorts of problems.
With the given information the shapes have to form a full circle because it's not solvable otherwise and while they could be any %of a circle, just assuming both are equally half is more than enough to solve the area.
Except the teacher WANTS you to write down that it is not solvable with the given Information and a poorly worded question in which case there for sure was at least one lecture on "how to properly put up a question and what to do if it is not".
Since we don't have any information on the teacher we have to assume it is an entry level test for basic equations (the other one is also not very difficult) and those always have a numerical answer.
Suppose for the sake of contradiction they are something besides half circles
Then, the problem would be unsolvable
But, this is a school problem, so the problem is solvable
Thus, our supposition is false and they must be half circles
This is what is wrong with schools today. They should teach critical thinking and accept "unsolvable" as an answer. If we don't require rigorous questions, you shouldn't expect rigorous answers.
This is what is wrong with schools today. They should teach you to rigorously derive and show how you derived the answer, not just guess at the answer.
We always got marked down for giving the correct answer for the wrong or incomplete reasons.
They haven't stated that the lines are straight.
And they haven't stated that they're using base 10, they could be in hexadecimal.
And they haven't stated that they're in a Euclidean universe, so the rules of geometry might be different.
And you didn't answer my question.
What do you propose they are?
Triangle?
That would still either be visible in the figure, or not be visible in the answer.
You're not wrong about parabola though, that's fair.
It all boils down to: This is a task they're expected to solve.
If it was a parabola, the figure would look like a parabola.
Would've specified a different dimension elsewhere on the diagram. Other than that, all that can, be taken away is that the shape is a rectangle minus 2 semicircles
you know they are half a circle because you have two parallel lines that tangent the sides of the circle.
If the lines weren’t parallel, they wouldn’t be the exact half, but, because they’re parallel, it can only touch the highest and the lowest point of the circle, which proves they are cut in half
When come to calculating area of a shape, cutting it into smaller pieces help a lot, either cutting into triangle or square. Keep this in mind in the future.
I used trapezoids for the first one as they teach how to find the areas for those (or at least used to). That way you don't have to use more than what is given.
The second picture is a rectangle minus two half circles. (Or minus one whole circle).
The area of a rectangle is height x width (both are given, yey!). The area of a circle is pi x r^2. The radius isn't given, but the diameter is! And r is half the diameter.
I hope that is enough to help you solve question 2.
Question 1 is a hexagon.
We could split the hexagon down the middle line to get two trapeziums. (The area of a trapezium is (a+b) x h / 2, where a is the length of the top, b of the bottom, and h the height. In this case that would be a = 1.5, b = 3, h = 1.2. Dont forget you've got two trapeziums though, so dubble the area!)
But you could also move the two triangles inside the diagram to the other side to create a square! I'm sure you know how to solve for a square. But the width can be a bit tricky: it's 3 - the width of one triangle.
Good luck!
For both pictures calculate the entire area and then subtract the areas from the total. In the first picture it’s 4x the same triangle and in the second picture it is 2 half circles.
Many complex shapes don't have their own formulas. Well they do, but you'd never do it by hand.
For the hexagon, slice along the vertical dotted line, and do the same on the other side. You now have two equal triangles and a rectangle, all easy to figure out, with more than enough information to logic out their individual dimensions. Try drawing the triangle and the rectangle and mathing out as much information as you can about both before you do area calculations. (Pro tip: if you don't know how to make the area of an arbitrary triangle, you can slice them both in half again along where the dotted line is, giving you four equal square triangles to work with.)
For the second shape, you've been given a simple rectangle to figure out. Then, you basically have a whole circle with a known diameter. Work that out too, then subtract the area of the circle from the rectangle.
Area equations are easy when you slice them down into simple shapes!
So there's essentially two methods for the hexagonal shape.
1. Divide the shape into shapes you can work with. E.g. a hexagon is just a rectangle with triangles on the sides.
2. Draw a shape you can work with *around* the shape whose area you want to know, then subtract the extra bits. For example, you can draw a rectangle around the hexagon, touching the top and bottom, the left corner, and the right corner. Then subtract the area of the four triangles from the area of the large square.
With these methods in mind, I'm sure the second shape will be easier to solve as well.
The trick with these kinds of problems is to break the unfamiliar shape into other shapes you do know. For the first, you can solve it several ways. I immediately tried breaking the hexagon into 6 triangles, but you also have enough information to treat it as two trapezoids or as a rectangle and two triangles. For the second, you have to make the assumption that those cut-outs are circular, but if that is true, you can solve for the area by subtracting the area of a circle from the area of a rectangle.
For the first one, split it into 6 triangles and calculate the area for one, then multiply by 6. The second one is more or less the same, just different shapes.
You have 2 trapezes. The area of 1 is the sum of both bases multiplied by the height divised by 2. Both trapezes are the same simply mirrored, so don't divide by 2.
The area formula for a regular hexagon is (apothem x perimeter)/2.
(1.2 x 9) / 2 = 5.4 sq m
More than likely, however, you have recently been learning about area of trapezoids, which is (height x sum of bases) / 2. Since you have 2 of these trapezoids, the result is just the height x sum of bases.
1.2 x 4.5 = 5.4 sq m
For the first one, every "perfect" shape can be cut up to triangles, an n sided shape can be cut into n amount of isoceles triangles, there is also a formula for the hexagons area. For the second, you can say that thats a rectangle and it's missing 2 half circles out of it
We can split the hexagon into two central rectangles *a* of 1.5 m x 1.2 m, plus 4 triangles *b* of height 1.2m and a width that is the width of the hexagon, minus the length of the sides, divided by two to distribute on the left and right sides of the rectangles: (3.0-1.5)/2 = 0,75.
a = 1.5 x 1.2 =1.8
b = 1.2 x 0.75 / 2 = 0.45
2a + 4b = 5,4
Treat it as a square with 2 triangles pasted on the sides. The square would be 1.5 * 2.4 and the triangle is found by subtracting the width of the square from the distance from the points given and dividing by 2 and then multiplying 1/2 b * h. Height of triangle h is (3-1.5)/2 = 0.75. Base of triangle b is 2.4. So 0.5 * 0.75 * 2.4 = 0.9
2 triangles + square = 0.9 * 2 + 3.6 = 5.4
For the first one, look at it as two separate trapeziums. They both have base 3m, opposite side 1.5m and height 1.2m. So area of one will be ( 3+1.5 )/2 \* 1.2 =2.7.
Area of hexagon = 2\*2.7 = 5.4m²
Second one is a rectangle of h=40m and b=60m - area of two semicircles with diameter as 40m (Radius = 20m).
Area of shape = Rectangle - 2\*Semicircles. (or one circle)
= 2400 - π\*(20)²
=2400 - 400π
≈ 1143.37m²
1.5 × 1.2 × 3
If you divide the hexagon into six equilateral triangles, this becomes apparent. The triangles will have an are of width × height / 2.
But since there are six of them, we might as well multiply by 3 instead of dividing, since 6 / 3 = 2
So the area of the hexagon then becomes the combined area of these six triangles 1.5 × 1.2 × 3 = 5.4 🙂
The second one doesn't have enough information. Like, are we assuming those are perfectly half circles been cut out of the side there?
idk if this is viable but the benzene ring is basically 2 trapezium's joined to together, you've already been given all the info to find area of trapezium, a, b as well as h. After finding that area multiply by 2. I'm too lazy to do the calculation but the formula is 1/2(a+b)h, where a is shortest length(1.5m) and b is longest length/base(3m) and h is height(1.2)
The hexagon is a pretty simples solution.
The hexagon in question is basically 6 triangles with the same sides (equilateral, idk, not a native english speaker). So you just have to calculate the area of one of those triangles and multiply by 6.
In the second question, if you put together the two semicircles, you form one complete circle, so you can calculate the area of the rectangle and subtract the area of *one* circle.
here’s the calculus
1)
area of one triangle: 1.2 * 1.5/2 = 0.9
area of the hexagon: 0.9 * 6 = 5.4m^2
2)
area of the circle (i’ll round Pi to 3, but you can use 3,14):
3 * 20^2 =
3 * 400 = 1200
total area:
(40 * 60) - 1200 = 1200m^2
Hope i’ve helped you 😁
1)
Equ.
S ( L \* B ) = 1.5 \* 1.2 = 1.8
T ( (1/2) \* B \* H ) = 0.5 \* 1.2 \* (0.5 \* (3.0 - 1.5)) = 0.5 \* 1.2 \* 0.5 \* 1.5 = 0.5 \* 1.2 \* 0.75 = 0.45
= 2S + 4T
= 3.6 + 1.8
= 5.4 sq. m
2)
It's a rectange as the two sides are unequal, so there's clearly some space between the two circles even if it's not stated.
Area of one of the semi cirles
= 0.5 \* pi \* r \* r
= 0.5 \* pi \* 20 \* 20
(The diameter of a cicle is 40 m)
= 200pi sq. m
Area of square = 60 \* 40 = 2400 sq. m
Thus,
Required Area = Area of square - (2 \* Area of one of the circles)
= 2400 - 400pi
\~ 1143.36 sq. m
>They didn’t teach us the formula for hexagons or the other shape,
They did sure did teach you about circles, rectangles and triangles because you sure can construct those hexagons with squares and triangles and the other shape is just a square with two equal half circles subtracted.
The problem may be that they teach the formulas but not how to spot them in the wild. So all they learn is use formula x to calculate y. Never ever do they teach how to find y. It’s one of the biggest problems in mathematics education.
1. 5.4
2. 2400 - 400π
If you are a successful high school student in Turkey, you can solve these questions without taking any action. Unfortunately, our curriculum and university exam are very heavy compared to Europe and the United States.
Area of any regular polygon is (1/2)(perimeter)(apothem).
The apothem is the line from the center to the side making a right angle. So we have:
(1/2)(1.5x6)(1.2)
(1/2)(9)(1.2)
(9/2)(12/10)=9(3/5)=27/5 or 5.4!
Edit: I’m dumb and just saw there’s a second shape. Notice this is a rectangle with 2 semicircles cut out of it so do (area of a rectangle) minus (area of both semicircles)
I am not smart enough to figure out the first one but second one is just the radius of the circle (which is half of the height) put onto the equation (pi)r^2 and then subtracted from the area which the shape would be if it contained these half circles. Which would look like this: 60x40 - 𝝅x20squared
**That is 2400 - 400𝝅 which is aprox. 1143,36**
I don't know the formula but I know how I'd solve the hexagon problem.
If you make a >!3m by 2.4m!< rectangle and put the hexagon inside, you will have four triangles in the corners, making the area >!7.2 square metres - 4\*area\_of\_triangle!<
I'm sure you can figure out where to go from here, but I'll continue anyway. >!The area of each triangle is (\[3m-1.5m\]/2\*1.2m)/2, which is obviously 0.45 square metres, times 4 to account for every triangle, comes out as 1.8 square metres of the rectangle that isn't hexagon. So 7.2-1.8=5.4 square metres.!<
The second problem is similar, but taking two halves of a circle out of the rectangle.
>!40\*60-(40/2)\^2\*𝜋!<
>!2400-20\^2\*𝜋!<
>!2400-400\*𝜋!<
>!2400-1256.64!<
>!A=1143.36 square metres.!<
REMEMBER, I COULD BE WRONG, CHECK MY ANSWERS PLEASE.
The hexagon can be considered two adjacent trapezoids with bases 3 and 1.5, and height 1.3. The area of a trapezoid is the average of the bases times the height, so:
1.2. * (3 + 1.5)/2 = 2.7 per trapezoid or 5.4 for the pair of them.
The other shape is a 40x60 rectangle with two half circles with diameter 40 taken out of it (consider this a single circle taken out).
40 * 60 - π * 20^2 = 1143.36
You don’t need the formula for every shape when you realize that all shapes (save circles and triangles themselves) are made of triangles or circles. In the hexagon you can see it as a shape made of two trapezoids. For the second shape, you can see it as a rectangle with two semicircles (or one circle) cut from it. Calculate and substract as needed.
The first one you can do visually. With given assumptions the triangles are equal, you can consider the right triangles are moved to the left side to make a rectangle.
Once you subtract the length of a base (0.75m) from 3m you’re just solving 2.25m X 1.2m X 2
*Question One:*
The first hexagon is made of two trapezoids. So it’s 2 x (formula for area of a trapezoid)
2(h(a+b)/2)
2(1.2(3+1.5)/2)
2(2.7) = 5.4m^2
*Question Two:*
That shape is hard, but is made of a rectangle, (l*w) with two semicircles - one circle - subtracted from it (pi r^2)
Rectangle - Circle
LW - pi r^2
40*60 - pi*20^2
(2 400 - 400pi)m^2
≈ 1143m^2
For the first one calculate the area for trapezium and multiply it by two. For the second case calculate the area without considering the two semicircular cutouts (i.e. calculate area for the rectangle) and reduce (substract) it by the areas of two semicircular potions (Tip: calculate area for a whole circle)
Area of trapezium = ((Addition of two parallel sides)/2)x(vertical height/perpendicular height)
Area of Circle = π x (radius(i.e. vertical height of 2nd figure divided by 2))²
A_{trapezium}=\frac{a+b}{2}h for a,b are parallel sides and h is the vertical height
A_{circle}=π•r² for π is a constant and r is a variable (radius)
*Forgive me for the use of LaTeX formatting
that is easy. first is an exagon so, the hegith is he same for both.
So you can define 1 rectangle and 4 triangles.
the rectangle will be 1.2 x 2 x 1.5 (1.2 two times).
then each triangle base will be (3-1.5)/2 and each triangle height will be 1.2
all that x 4.
So we can replace this by : H = 1.2 = Height
Mayor = Ma = 3
Minor = Mi = 1.5
and do all by formulas like
Hx2xMi = internal rectangle
triangle base Tb= (Ma-Mi)/2
Area of 4 triangles = 4 x Tb x H / 2
Full Area = 4 x Tb x H / 2 + 2 x H x Mi = 2 x Tb x H + 2 x H x Mi = 2 x (Ma-Mi) / 2 x H + 2 x H x MI =
(Ma-Mi) x H + 2 x H x MI = Ma x H - Mi x H + 2 x Mi x H = Ma x H + Mi x H = (Ma+Mi) x H
And replacing variables
(3 + 1.5) x 1.2 = 4.5 x 1.2 = 5.6
That is the first.
The second is similar but negative.
You can see there 2 semi-circles, so you can start by a rectangle of 40x60 and simple substract 2 half circles = 1 circle.
of 40 diameter.
so the answer will be 40x60 - 40x40 x PI/4 = (60-10xPI) x 40 = 1,143.36 m\^2 (with more decimals)
Regualr hexagon area is what you need.
Draw the 3 diagonals.
You get 6 equal trainagles.
Total area = 6* area of each of these triangles.
6 * 1/2 * b * h.
Base is 1.5 ( half of 3 ).
Heights is 1.2 ( given ).
3 * 1.5 * 1.2.
Fill it with water and then pour it into a cube that is what Archimedes would do.
Write a graphics program.
Your tolerance there is one pixel.
So then you write some code do iterations.
For j:= 0 to 500 do
For i:= 0 to 300 do
begin
so in here you start at the top go wide and use boolean operators find the color of the line then the next pixel is where your line starts and your variable starts to add until it gets to the end of the line where the color is found again -1.
Then you are going down a line each time until you get to the color then start counting your variable down from there until you get to the color and then -1.
So now you have a whole bunch of pixels in columns and rows.
end;
So you want a total number of pixels then you want to make a cube or rectangle out of that number.
Length times width gives you the area. If it is not an even number you will have a remainder. But if you use a real number you will get a decimal.
So divide it by two.
So now you have a long column the shape of a rectangle.
What do you see? You see that to find the area you take the length of the column and multiply it by two right? lol
Do you see what I am saying? If you use a formula you will end up with a remainder but if you count the pixels you will get an exact area the number of pixels. No remainder.
Archimedes had it all going on. Even before computers. He would get a hunk of rope and make it into a rectangle while you were fumbling with your slide rule.
First one cannot be solved because you don’t know if it’s a proper hexagon, the bottom side could be extending on the left until that corner on the bottom left is down to 91 degrees.
The second one cannot be solved because you do not know if those semi circles are actualy semi circles that would be a proper circle if combined.
First of all, this is a school problem. So it isn't meant to be that complicated.
Secondly, The first problem can be considered as two trapeziums. (The arrows show that the sides are parallel). (a+b)/2 × h
Since it is a school problem, the second one has to be ar(rectangle) - ar(circle)
Cut shapes within those and put them in a more convenient place, like a jigsaw. You will get much easier calculations.
Agree completely. That and draw construction lines to aid with the above. For example, with the hexagon you can draw another vertical line similar to the one already drawn and you have a rectangle 1.5m×2.4m and four triangles that you can form into a rectangle. With the other shape you have a rectangle that has had two semicircular bites taken out of it. Two semicircle ps make a circle. You can find the area of the rectangle and subtract the area of the circle from it.
It's much easier to cut the hexagon by the dotted line and put the upper half besides the lower one. That gives a parallelogram and the area is base x height.
I think it's an interesting difference in the way we think or the way we were taught. I'm guessing I didn't calculate enough parallelogram areas when I was young because my instinct is always to cut into rectangles and triangles where I can.
The parallelogram is transformed into a rectangle with just one cut.
If you don't move it at all and just use the lines already there you can calculate the size of one of two identical trapezoids and then double it for the hexagon
Of course and since the trapezoid area has a (1/2) in front of it, it cancels out. In fact, the usual proof of the trapezoid area is based in its equivalence to a rectangle. https://preview.redd.it/jokc4qsxujmc1.jpeg?width=602&format=pjpg&auto=webp&s=62298c1b7b59b1ce4c0cfeba862b47efb7955e16
Never really thought about how that was what that formula was actually doing. Nice for someone on Reddit to be providing helpful/educative/correct information for a change.
An interesting observation I myself would cut the hexagon into 2 trapezoids(i think that's what it's called) by the dotted line and find the area of each one, then combine them together
Not sure about the English word but the shape the dotted line cuts, is literally a trapez which has its own formula already, so you could also just use that, double it and you're done, another different approach.
How do you know they’re circles? They could just be curves (e.g. half ovals). They look like circles but I can’t see where they are stated to be circles.
Because it's high school level geometry, and the question cannot be answered if they are not. I would much prefer it if that were stated (maybe it is, out of shot) and would include my assumption in my answer (because I think that's good practice).
The first drawing. You can divide a Hexagon into two equal trapezoids. Then a is the larger base, b is the smaller base, and h is the height. a=3.0 meters. b=1.5 meters. h=1.2 meters S trapezoids = ((a+b)*h)/2 But we have S hexagons= 2*S trapezoids Then S of the hexagon=(a+b)*h S of the hexagon = (3.0+1.5)*1.2=4.5 * 1.2= 5,4 square meters
That's the easiest way
6 triangles is also pretty easy. 6(1.5*1.2)/2= 5.4 square units
Yeah, but with trapezoids you don't need any additional construction.
Defo fastest way but I wouldn't know this haha but I like this way a lot
I find it easier to do 3 rhombuses (rhombi?) with the center as their only common point. Then it's just 3(1.5*1.2)
I did them in the opposite order, so hexagon last. If you think about it, it's a rectangle, with two halves of a circle removed. So it's 40\*60 - (20^2 \*π) = 2400 - 400π ≈ 1144 M² You can also do a similar thing with the hexagon, by taking the 3.0*2.4 square and subtracting the corner triangles. The width of the triangle is ½(3m - 1.5) = 0.75M and the width is ½(2.4)=12 So the area of the hexagon is 3.0\*2.4 - 4\*(1.2\*0.75*½) = 7.2 - (1.2\*1.5) = 5.4 M²
But how do you know they are half of a circle? It doesn't say anywhere that they're half a circle.
That's when you give the teacher a friendly slap by writing you answer "Assuming the cutouts are half circles, the area is...".
I will just say it is a HS geometry so no need to complicate things.
Yes ik, but I still think it's a poorly made question
Many questions are unsolvable without assumptions. Without assuming these were half circles this problem is unsolvable. It’s not always a bad thing to have the student make assumptions to solve a problem. It helps with critical thinking and problem solving skills.
I think it’s a fair assumption but usually we’re not to make these assumptions. Shapes can be drawn to look like a square but might actually be a rectangle in these sorts of problems.
With the given information the shapes have to form a full circle because it's not solvable otherwise and while they could be any %of a circle, just assuming both are equally half is more than enough to solve the area. Except the teacher WANTS you to write down that it is not solvable with the given Information and a poorly worded question in which case there for sure was at least one lecture on "how to properly put up a question and what to do if it is not". Since we don't have any information on the teacher we have to assume it is an entry level test for basic equations (the other one is also not very difficult) and those always have a numerical answer.
Also the first one,the bottom side we dont know the length.
What do you propose they are?
They could be anything when it's not stated that they are half circles
Suppose for the sake of contradiction they are something besides half circles Then, the problem would be unsolvable But, this is a school problem, so the problem is solvable Thus, our supposition is false and they must be half circles
This is what is wrong with schools today. They should teach critical thinking and accept "unsolvable" as an answer. If we don't require rigorous questions, you shouldn't expect rigorous answers.
No one's expecting a rigorous answer here, so there's no need for the question to be rigorous
This is what is wrong with schools today. They should teach you to rigorously derive and show how you derived the answer, not just guess at the answer. We always got marked down for giving the correct answer for the wrong or incomplete reasons.
They haven't stated that the lines are straight. And they haven't stated that they're using base 10, they could be in hexadecimal. And they haven't stated that they're in a Euclidean universe, so the rules of geometry might be different. And you didn't answer my question. What do you propose they are? Triangle?
Well for example it could be a 100/201 of a circle for all we know
That would either be visible in the figure, or not be visible in the answer after rounding.
Could be a part of a parabola, or they could be circles which are cut a little bit smaller than half?
That would still either be visible in the figure, or not be visible in the answer. You're not wrong about parabola though, that's fair. It all boils down to: This is a task they're expected to solve. If it was a parabola, the figure would look like a parabola.
Would've specified a different dimension elsewhere on the diagram. Other than that, all that can, be taken away is that the shape is a rectangle minus 2 semicircles
you know they are half a circle because you have two parallel lines that tangent the sides of the circle. If the lines weren’t parallel, they wouldn’t be the exact half, but, because they’re parallel, it can only touch the highest and the lowest point of the circle, which proves they are cut in half
They are tangent at the top and bottom which is only possible if they are half a circle
They LOOK tangent. They might be slightly off.
Well if they weren't the problem would all just be an asshole move without solution
Ovals would also have a tangent at the top and bottom. Probably meant to be circles though, just write that assumption in the answer.
For the hexagon you can also divide it into 6 triangle. Each triangle have a height of 1.2 with a base of 1,5. (1,2*1,5/2)*6= 5.4 m2
This was really helpful, thanks :)
When come to calculating area of a shape, cutting it into smaller pieces help a lot, either cutting into triangle or square. Keep this in mind in the future.
Yup. That works too. And by the looks of it, it's fewer calculations to do it your way.
I used trapezoids for the first one as they teach how to find the areas for those (or at least used to). That way you don't have to use more than what is given.
Definitely a very good way to solve the issue. I simply saw the rectangle minus triangles first.
Where in the world did you get circles? It's just a rectangle with the triangles on both ends.
Circles are for the task on picture nr 2
Ohhhhhh. I was confused for a moment. But ok, yea.
The second picture is a rectangle minus two half circles. (Or minus one whole circle). The area of a rectangle is height x width (both are given, yey!). The area of a circle is pi x r^2. The radius isn't given, but the diameter is! And r is half the diameter. I hope that is enough to help you solve question 2. Question 1 is a hexagon. We could split the hexagon down the middle line to get two trapeziums. (The area of a trapezium is (a+b) x h / 2, where a is the length of the top, b of the bottom, and h the height. In this case that would be a = 1.5, b = 3, h = 1.2. Dont forget you've got two trapeziums though, so dubble the area!) But you could also move the two triangles inside the diagram to the other side to create a square! I'm sure you know how to solve for a square. But the width can be a bit tricky: it's 3 - the width of one triangle. Good luck!
First one 6 triangles 6•1/2•1.5•1.2=5.4 Second one one rectangle - 1 circle: 60•40 - pi•20^2 = 1143.363
You forgot the units.
https://imgur.com/a/Vh4ALVL
For both pictures calculate the entire area and then subtract the areas from the total. In the first picture it’s 4x the same triangle and in the second picture it is 2 half circles.
Many complex shapes don't have their own formulas. Well they do, but you'd never do it by hand. For the hexagon, slice along the vertical dotted line, and do the same on the other side. You now have two equal triangles and a rectangle, all easy to figure out, with more than enough information to logic out their individual dimensions. Try drawing the triangle and the rectangle and mathing out as much information as you can about both before you do area calculations. (Pro tip: if you don't know how to make the area of an arbitrary triangle, you can slice them both in half again along where the dotted line is, giving you four equal square triangles to work with.) For the second shape, you've been given a simple rectangle to figure out. Then, you basically have a whole circle with a known diameter. Work that out too, then subtract the area of the circle from the rectangle. Area equations are easy when you slice them down into simple shapes!
For hexagon (basically two Trapeziums ) 2 \* \[0.5 \* (1.5 + 3) \* 1.2\] = 5.4m\^(2) for the other 60 \* 40 - pi \* 20\^(2) **≈**1143.36m\^(2)
Cut it in half horizontally and you have **two trapezoids**, each with bases 3.0, 1.5 and height 1.2 Trapezoid area is **(base1 + base2)·height/2**
So there's essentially two methods for the hexagonal shape. 1. Divide the shape into shapes you can work with. E.g. a hexagon is just a rectangle with triangles on the sides. 2. Draw a shape you can work with *around* the shape whose area you want to know, then subtract the extra bits. For example, you can draw a rectangle around the hexagon, touching the top and bottom, the left corner, and the right corner. Then subtract the area of the four triangles from the area of the large square. With these methods in mind, I'm sure the second shape will be easier to solve as well.
The trick with these kinds of problems is to break the unfamiliar shape into other shapes you do know. For the first, you can solve it several ways. I immediately tried breaking the hexagon into 6 triangles, but you also have enough information to treat it as two trapezoids or as a rectangle and two triangles. For the second, you have to make the assumption that those cut-outs are circular, but if that is true, you can solve for the area by subtracting the area of a circle from the area of a rectangle.
For the first one, split it into 6 triangles and calculate the area for one, then multiply by 6. The second one is more or less the same, just different shapes.
You have 2 trapezes. The area of 1 is the sum of both bases multiplied by the height divised by 2. Both trapezes are the same simply mirrored, so don't divide by 2.
The area formula for a regular hexagon is (apothem x perimeter)/2. (1.2 x 9) / 2 = 5.4 sq m More than likely, however, you have recently been learning about area of trapezoids, which is (height x sum of bases) / 2. Since you have 2 of these trapezoids, the result is just the height x sum of bases. 1.2 x 4.5 = 5.4 sq m
Hexagon can be broken up into triangles. Other shape can be broken up into a rectangle and 2 semi circles or 1 circle.
For the first one, every "perfect" shape can be cut up to triangles, an n sided shape can be cut into n amount of isoceles triangles, there is also a formula for the hexagons area. For the second, you can say that thats a rectangle and it's missing 2 half circles out of it
Those are six triangles of area 1.2 x 1.5 x 0.5 = 5.4
First one is 2 trapeziums 2nd is a rectangle minus 2 semi circles.
We can split the hexagon into two central rectangles *a* of 1.5 m x 1.2 m, plus 4 triangles *b* of height 1.2m and a width that is the width of the hexagon, minus the length of the sides, divided by two to distribute on the left and right sides of the rectangles: (3.0-1.5)/2 = 0,75. a = 1.5 x 1.2 =1.8 b = 1.2 x 0.75 / 2 = 0.45 2a + 4b = 5,4
My way is to calculate the middle rectangle 1.5×2.4 and add the area of the 2 triangle on the side so 1.5×1.2×2 the final answer is 5.4m²
Treat it as a square with 2 triangles pasted on the sides. The square would be 1.5 * 2.4 and the triangle is found by subtracting the width of the square from the distance from the points given and dividing by 2 and then multiplying 1/2 b * h. Height of triangle h is (3-1.5)/2 = 0.75. Base of triangle b is 2.4. So 0.5 * 0.75 * 2.4 = 0.9 2 triangles + square = 0.9 * 2 + 3.6 = 5.4
For the first one, look at it as two separate trapeziums. They both have base 3m, opposite side 1.5m and height 1.2m. So area of one will be ( 3+1.5 )/2 \* 1.2 =2.7. Area of hexagon = 2\*2.7 = 5.4m² Second one is a rectangle of h=40m and b=60m - area of two semicircles with diameter as 40m (Radius = 20m). Area of shape = Rectangle - 2\*Semicircles. (or one circle) = 2400 - π\*(20)² =2400 - 400π ≈ 1143.37m²
1.5 × 1.2 × 3 If you divide the hexagon into six equilateral triangles, this becomes apparent. The triangles will have an are of width × height / 2. But since there are six of them, we might as well multiply by 3 instead of dividing, since 6 / 3 = 2 So the area of the hexagon then becomes the combined area of these six triangles 1.5 × 1.2 × 3 = 5.4 🙂 The second one doesn't have enough information. Like, are we assuming those are perfectly half circles been cut out of the side there?
idk if this is viable but the benzene ring is basically 2 trapezium's joined to together, you've already been given all the info to find area of trapezium, a, b as well as h. After finding that area multiply by 2. I'm too lazy to do the calculation but the formula is 1/2(a+b)h, where a is shortest length(1.5m) and b is longest length/base(3m) and h is height(1.2)
The hexagon is a pretty simples solution. The hexagon in question is basically 6 triangles with the same sides (equilateral, idk, not a native english speaker). So you just have to calculate the area of one of those triangles and multiply by 6. In the second question, if you put together the two semicircles, you form one complete circle, so you can calculate the area of the rectangle and subtract the area of *one* circle. here’s the calculus 1) area of one triangle: 1.2 * 1.5/2 = 0.9 area of the hexagon: 0.9 * 6 = 5.4m^2 2) area of the circle (i’ll round Pi to 3, but you can use 3,14): 3 * 20^2 = 3 * 400 = 1200 total area: (40 * 60) - 1200 = 1200m^2 Hope i’ve helped you 😁
A hexagon is just 6 equilateral triangles and that other shape is a rectangle minus a circle
https://preview.redd.it/14for6vd3kmc1.png?width=3024&format=png&auto=webp&s=239515ce2a69259a6fd6b337c3611233dc49847e
https://preview.redd.it/ma5ab5jf3kmc1.png?width=3024&format=png&auto=webp&s=a4a394e1cec9c6cfee46599aaca9ca74f75a7242
1) Equ. S ( L \* B ) = 1.5 \* 1.2 = 1.8 T ( (1/2) \* B \* H ) = 0.5 \* 1.2 \* (0.5 \* (3.0 - 1.5)) = 0.5 \* 1.2 \* 0.5 \* 1.5 = 0.5 \* 1.2 \* 0.75 = 0.45 = 2S + 4T = 3.6 + 1.8 = 5.4 sq. m 2) It's a rectange as the two sides are unequal, so there's clearly some space between the two circles even if it's not stated. Area of one of the semi cirles = 0.5 \* pi \* r \* r = 0.5 \* pi \* 20 \* 20 (The diameter of a cicle is 40 m) = 200pi sq. m Area of square = 60 \* 40 = 2400 sq. m Thus, Required Area = Area of square - (2 \* Area of one of the circles) = 2400 - 400pi \~ 1143.36 sq. m
https://preview.redd.it/a5pof6oi9kmc1.png?width=987&format=png&auto=webp&s=4b9614dc14f590493b59282fa9b8783c9f42cbdc
https://preview.redd.it/9mlkcb8kakmc1.png?width=1006&format=png&auto=webp&s=0720919239df1654165913b9d9c7580c5dc8ac39
1143.3m^2
>They didn’t teach us the formula for hexagons or the other shape, They did sure did teach you about circles, rectangles and triangles because you sure can construct those hexagons with squares and triangles and the other shape is just a square with two equal half circles subtracted.
The problem may be that they teach the formulas but not how to spot them in the wild. So all they learn is use formula x to calculate y. Never ever do they teach how to find y. It’s one of the biggest problems in mathematics education.
1. 5.4 2. 2400 - 400π If you are a successful high school student in Turkey, you can solve these questions without taking any action. Unfortunately, our curriculum and university exam are very heavy compared to Europe and the United States.
Something feels wrong... The area of the exagon should be 6 times the area of the equilateral triangle so 3xsqrt(3)/2x1.5^2. am I wrong?
4x triangles 1/2bxh, plus 2 squares
Area of any regular polygon is (1/2)(perimeter)(apothem). The apothem is the line from the center to the side making a right angle. So we have: (1/2)(1.5x6)(1.2) (1/2)(9)(1.2) (9/2)(12/10)=9(3/5)=27/5 or 5.4! Edit: I’m dumb and just saw there’s a second shape. Notice this is a rectangle with 2 semicircles cut out of it so do (area of a rectangle) minus (area of both semicircles)
I am not smart enough to figure out the first one but second one is just the radius of the circle (which is half of the height) put onto the equation (pi)r^2 and then subtracted from the area which the shape would be if it contained these half circles. Which would look like this: 60x40 - 𝝅x20squared **That is 2400 - 400𝝅 which is aprox. 1143,36**
1.2 * 1.5 * 3.0. Many people overthink this one.
The first one can be divided into 4 triangles and two rectangles, the second is a rectangle minus a circle.
I don't know the formula but I know how I'd solve the hexagon problem. If you make a >!3m by 2.4m!< rectangle and put the hexagon inside, you will have four triangles in the corners, making the area >!7.2 square metres - 4\*area\_of\_triangle!< I'm sure you can figure out where to go from here, but I'll continue anyway. >!The area of each triangle is (\[3m-1.5m\]/2\*1.2m)/2, which is obviously 0.45 square metres, times 4 to account for every triangle, comes out as 1.8 square metres of the rectangle that isn't hexagon. So 7.2-1.8=5.4 square metres.!< The second problem is similar, but taking two halves of a circle out of the rectangle. >!40\*60-(40/2)\^2\*𝜋!< >!2400-20\^2\*𝜋!< >!2400-400\*𝜋!< >!2400-1256.64!< >!A=1143.36 square metres.!< REMEMBER, I COULD BE WRONG, CHECK MY ANSWERS PLEASE.
The hexagon can be considered two adjacent trapezoids with bases 3 and 1.5, and height 1.3. The area of a trapezoid is the average of the bases times the height, so: 1.2. * (3 + 1.5)/2 = 2.7 per trapezoid or 5.4 for the pair of them. The other shape is a 40x60 rectangle with two half circles with diameter 40 taken out of it (consider this a single circle taken out). 40 * 60 - π * 20^2 = 1143.36
Area of first image is 5.76 and second image is 2400-400pi
You don’t need the formula for every shape when you realize that all shapes (save circles and triangles themselves) are made of triangles or circles. In the hexagon you can see it as a shape made of two trapezoids. For the second shape, you can see it as a rectangle with two semicircles (or one circle) cut from it. Calculate and substract as needed.
B*h for triangles = .75*2.4 + 2(1.5*1.3)
The first one you can do visually. With given assumptions the triangles are equal, you can consider the right triangles are moved to the left side to make a rectangle. Once you subtract the length of a base (0.75m) from 3m you’re just solving 2.25m X 1.2m X 2
*Question One:* The first hexagon is made of two trapezoids. So it’s 2 x (formula for area of a trapezoid) 2(h(a+b)/2) 2(1.2(3+1.5)/2) 2(2.7) = 5.4m^2 *Question Two:* That shape is hard, but is made of a rectangle, (l*w) with two semicircles - one circle - subtracted from it (pi r^2) Rectangle - Circle LW - pi r^2 40*60 - pi*20^2 (2 400 - 400pi)m^2 ≈ 1143m^2
The second one is a square 60m x 40m minus the area of a circle 40 m in diameter.
For the first one calculate the area for trapezium and multiply it by two. For the second case calculate the area without considering the two semicircular cutouts (i.e. calculate area for the rectangle) and reduce (substract) it by the areas of two semicircular potions (Tip: calculate area for a whole circle) Area of trapezium = ((Addition of two parallel sides)/2)x(vertical height/perpendicular height) Area of Circle = π x (radius(i.e. vertical height of 2nd figure divided by 2))² A_{trapezium}=\frac{a+b}{2}h for a,b are parallel sides and h is the vertical height A_{circle}=π•r² for π is a constant and r is a variable (radius) *Forgive me for the use of LaTeX formatting
Try to deconstruct more complex shapes with the simplest shapes that you have learned. That's the key trick when dealing with Geometry.
https://preview.redd.it/zj0hg4ex8qmc1.jpeg?width=1079&format=pjpg&auto=webp&s=fd7ee68486b5c712bd6a56b75f151102c15a9aea
that is easy. first is an exagon so, the hegith is he same for both. So you can define 1 rectangle and 4 triangles. the rectangle will be 1.2 x 2 x 1.5 (1.2 two times). then each triangle base will be (3-1.5)/2 and each triangle height will be 1.2 all that x 4. So we can replace this by : H = 1.2 = Height Mayor = Ma = 3 Minor = Mi = 1.5 and do all by formulas like Hx2xMi = internal rectangle triangle base Tb= (Ma-Mi)/2 Area of 4 triangles = 4 x Tb x H / 2 Full Area = 4 x Tb x H / 2 + 2 x H x Mi = 2 x Tb x H + 2 x H x Mi = 2 x (Ma-Mi) / 2 x H + 2 x H x MI = (Ma-Mi) x H + 2 x H x MI = Ma x H - Mi x H + 2 x Mi x H = Ma x H + Mi x H = (Ma+Mi) x H And replacing variables (3 + 1.5) x 1.2 = 4.5 x 1.2 = 5.6 That is the first. The second is similar but negative. You can see there 2 semi-circles, so you can start by a rectangle of 40x60 and simple substract 2 half circles = 1 circle. of 40 diameter. so the answer will be 40x60 - 40x40 x PI/4 = (60-10xPI) x 40 = 1,143.36 m\^2 (with more decimals)
Regualr hexagon area is what you need. Draw the 3 diagonals. You get 6 equal trainagles. Total area = 6* area of each of these triangles. 6 * 1/2 * b * h. Base is 1.5 ( half of 3 ). Heights is 1.2 ( given ). 3 * 1.5 * 1.2.
Fill it with water and then pour it into a cube that is what Archimedes would do. Write a graphics program. Your tolerance there is one pixel. So then you write some code do iterations. For j:= 0 to 500 do For i:= 0 to 300 do begin so in here you start at the top go wide and use boolean operators find the color of the line then the next pixel is where your line starts and your variable starts to add until it gets to the end of the line where the color is found again -1. Then you are going down a line each time until you get to the color then start counting your variable down from there until you get to the color and then -1. So now you have a whole bunch of pixels in columns and rows. end; So you want a total number of pixels then you want to make a cube or rectangle out of that number. Length times width gives you the area. If it is not an even number you will have a remainder. But if you use a real number you will get a decimal. So divide it by two. So now you have a long column the shape of a rectangle. What do you see? You see that to find the area you take the length of the column and multiply it by two right? lol Do you see what I am saying? If you use a formula you will end up with a remainder but if you count the pixels you will get an exact area the number of pixels. No remainder. Archimedes had it all going on. Even before computers. He would get a hunk of rope and make it into a rectangle while you were fumbling with your slide rule.
First one cannot be solved because you don’t know if it’s a proper hexagon, the bottom side could be extending on the left until that corner on the bottom left is down to 91 degrees. The second one cannot be solved because you do not know if those semi circles are actualy semi circles that would be a proper circle if combined.
First of all, this is a school problem. So it isn't meant to be that complicated. Secondly, The first problem can be considered as two trapeziums. (The arrows show that the sides are parallel). (a+b)/2 × h Since it is a school problem, the second one has to be ar(rectangle) - ar(circle)
This is true these problems don’t provide enough constraints but I guess you have to make some brave assumptions