And after that, the smaller triangle 6 - a - r has the same angles as the bigger triangle and so is similar, thus R can be found just by the ratio of sizes.
My method was the same until 1.30 in to the video. The area of the triangle is 6x8/2 =24, but can also be found by doing 10 x r /2 which can be written as 5r. In other words 5r=24 so the radius is 4.8. Having done that, the area of the circle is 4.8^2 multiplied by pi= 23.04 pi
That's how i did it. Felt easier I'd'ev also left it as 4.8^2*pi m2 since it's easier to read and avoid going down the rabbit hole of decimal points, thus avoiding approximating pi and so on and so forth.
Nice one, I did something similar. I recognized that when you draw the height of a right triangle from its right angle, you end up with two smaller triangles that are similar to the original. So we can do (long leg)/(hypotenuse) = 8/10 = r/6 and we end up with effectively the same equation.
There is a simpler way! Notice how in the 6 8 10 triangle, the radius is perpendicular to the hypotenuse. Therefore the area of the triangle is equal to 6x8/2 or 10r/2. So, 10r/2 = 6x8/2, r = 48/10 = 4.8m. So area = pi x r^2 = (4.8)^2 x pi = 23.04 pi as in the video
1:33 There's an easier way. There's a theorem called "right triangle altitude theorem". With it, we can write three equations that are simpler than the Pythagorean theorem ones. I. _a * 10 = 6²_ II. _(10 - a) * 10 = 8²_ III. _a * (10 - a) = r²_ With the first, we have _'a'_ immediately, that we can substitute in the third to get _r²_ .
It's simpler than that, the height of a right angle triangle perpendicular to the hypotenuse is just the product of the two shorter sides divided by the hypotenuse. In a 6-8-10 triangle that's just (6*8)/10, or 4.8. Then you don't need to split the hypotenuse and deal with 'a' or '10-a' at all.
@@heavybrambles You mean _T = 6 * 8 = 10 * r_ from which _r_ is _6 * 8 / 10_ ? I mean, it's a simple insight, but it wasn't obvious to me at first. Nice!
Following up on what others have noted: the radius of the inscribed circle is the height of a 6-8-10 right triangle. For any right triangle, the height is ab/c, so in this case 6*8/10 = 4.8 is the radius. In the general case of an circle inscribed in a rhombus of diagonals d1 and d2, the circle’s area will be given by A = pi * d1^2 * d2^2 / [4 (d1^2 + d2^2)]. In this case, A = pi * 12*16 / [4 (144 + 256 )] = 23.04 pi. It’s easier to remember in words: the area of a circle inscribed in a rhombus in terms of the diagonals is the product of their squares divided by the sum of their squares, times pi/4.
At 1:22 we have a right triangle with sides 6 and 8 and hypotenuse 10. We can compute its area by treating one side as the base and the other as the height, so A = (1/2)(8)(6) = 24. We can also compute the same area by treating the hypotenuse as the base and then the circle's radius becomes the height, A = (1/2)(10)(r) = 5r. Equate the areas and 5r = 24, so r = 4.8. The area of the circle is πr² = π(4.8)² = 23.04π, as AndyMath also found.
You could also use the area of the rhombus. It is equal to the product of the diagonals divided by 2 so it is 16x12/2 = 96 Then, use the Pythagorean theorem to figure out the side length 10 After that, since a rhombus is a special case of a parallelogram its area is also equal to base x height. The height of the rhombus is also a diameter of the circle so 10 x diameter = 96. Therefore, the diameter is equal to 96/10 = 9.6 so the radius is 4.8. Plugging that into the formula for the area of a circle πr² gives the same result 23.04π
Easier way to find the diagonal of the 6-8 triangle. It's a right triangle with one side a multiple of 3, and another of 4, both by the same value. So the last side is 5×2.
Here's my approach: calculate the altitude of any of the right triangles. This altitude is equal to the radius of the circle. This is because both the altitude and the radius form a right angle with the tangent line. Radius = Altitude = (ab/c) = (6*8)/10 = 4.8. With this, we can calculate the area of the circle.
Kinda need you for something. Edexcel paper 3 had the last question where: Theres 2 hexagons and a circle such that the circle is in between those two hexagons. The larger hexagon touches the circle by its tangent, and the smaller hexagon touches the circle by its chord. "Considering perimeters, show that 3 < pi < 2√3" Give tutorial pls
ok in 1:34 you could use the formula a^-2+b^-2+h^-2(h is the alternate height in right angle triangle where c is the base) (these are also inverse squares) or u can use r times c is equal to a times b since the tringales area can be calculated with two different bases and heights , anyways love ur vids
you place center of the circle and rhombus at the orego, drawing lines through the verticies there is a vector [x=8,y=-6] from the top corner of the rhomb to the right corner. This gives a function downslope(x)=6 - 3/4x, which is intercepted at a 90deg angle from origo by another function upslope(x) = 4/3x. The distance from origo to the point where downslope and upslope intercept is the radius of the circle, which gives the area Alternatively I think it should be possible to declare a function distance(x)=(x^2 +upslope(x)^2 )^(1/2), solve for the value which makes the derivative of distance 0 and find the interception point with this x value. Does this check out? I didnt do the math
There are two ways this probably wouldve been easier Either 1: Theres a formula for the height in a right triangle, its cathetus 1 * cathetus 2 divided by hypothenuse, so you couldve instantly just did 6 * 8 / 10 = 4.8 which is the radius Or 2: Analytic geometry Consider the center of the circle O(0,0) so that means the corner above it will be A(0,6), the corner to its left will be B(8,0) which means its line formula is 3x + 4y = 24 Use the formula for the distance from a point to a line and you get that the distance is 4.8, that distance being the radius
By symmetry, you consider a quarter of the shape which is a right triangle of sides 6-8-10 (by Pythagore), and then you juste express the area of this triangle 2 ways to get the radius of the circle : 10*r = 6*8 r = 4.8 From there you get the area of the circle with pi*rˆ2 = 23,04pi
You could also use the area of the triangle to identify the “r”. You know that r is the height to the “c” side since those are in a right angle. That means that if you identify lenght of c (=10) you can then devide the area of the triangle by that and get r=4,8. Result is the same. (6*8=48 -> 48/10) i think this is much cleaner since you don’t need a calculator for this. 😊 I like your content, keep up with the good work. ❤
I would calculate the length of a side by pythagorean theorem and the area A_r=12×16/2=96. Since the circle is inscribed we know that 2r is the length of the height of the rombus and we get an equation 20r=96 so r=24/5 from that A_c=576/25 pi
Instead of using Pythagorean Theorem to find the altitude in the right triangle, you could instead treat the altitude as the height in the triangle and the hypotenuse as the base. (1/2)*b*h is the area of a triangle where "h" is the height and "b" is the base, but (1/2)*a*b is also the area of the triangle (only right triangles) so you can set them equal and cancel the "(1/2)" terms so you'll get: b*h=a*b.
1:33 from here you can just use Euclidean theorem for right triangle They are three First (10-a)*a=(r)^2 Second (10-a)*8=(8)^2 And the third that we are gonna use is a*10= (6)^2 So a=3.6 Now skip to 3:00
The area of the rhombas is (d₁⋅d₂)/2=(12⋅16)/2=96. But the area of the rhombas is also base ⋅ height or s⋅2r, where s is any side length and 2r is the height constrained to the minimally inscribed circle's diameter, so 96=s⋅2r, r=48/s. The bisection of the diagonals gives the pythagorean relation: s²=6²+8², s=10. With r=48/10=4.8, the area of the circle is (4.8)²(Pi) A=23.04(Pi).
Easier way would have been to work with streching or ratios and 3-4-5 triangles. Obviously each tangent is 10m long and is the hypotenuse in the "3-4-5" or better "6-8-10" triangle. From here find a triangle where r is one of its sides. There are 2 options. Each option reuses an angle from the initial 6-8-10 triangle and also contains a 90° angle where r hits the tangent line piece. Therefore we know that the new triangle with r is equal to the initial one. From here use ratios: Option 1: hypotenuse / leg_1 = leg_2 / r = 10m / 6m = 8m / r => r = 4,8m Option 2: hypotenuse / leg_1 = leg_2 / r = 10m / 8m = 6m / r => r = 4,8m Finally r²*pi
Not sure if its easier, possibly shorter: due to the opposite sides of rhombus being parallel, we know that the points of the circle intersecting opposite sides of the rhombus must have the diameter of the circle being the shortest line between them as only points on opposites sides of a circle will have a parallel tangent line, which is also at right angles to those sides as as the sides of the rhombus are tangents to the circle. Now we just have to find the "perpendicular height" of the rhombus which is equal to diameter of the circle. Area of rhombus = base x height. Area is also equal to (diagonal1xdiagonal2)/2 = 96. Base = sqrt ( 8^2 + 6^2) = 10. Height = 96/10 = 9.6. So Radius = 4.8, area = pi*4.8^2
I can't believe you guys in the US don't know simple maths. Here in the UK we learn the following: First figure out the area of the rhombus which is easy: (12*16)/2 = 96. Then figure out a side length which is 10 in this case. Half of the perimeter of the rhombus = 20. Radius of the circle is 96/20 = 4.8. Then just calculate the area as normal using the radius. Simple, quick and easy. If you need any more details, let me know.
I didnt do the same thing, after drawing the radius i knew that the 3 triangles were similar and like that i was able to figure out that 10-a =6.4 then i just used the law of sines to determine the radius and just figured out the area as you usually would.
@1:30 you could have used the similarity of the 3/4/5 triangles, having r/8=6/10, thus r=48/10, thus A=(48/10)^2*pi (48/10)^2 is btw the 23,04 you got fro a double Pythagorean, but similarity is faster here imho
I think this is an easier way, since it requires less thinking. You can use r=2P/(a+b+c+d) becouse of the circle inscribed in the qudrilateral, and you get r=4.8
I found r in a much easier way: after dividing the rhombus in 4 right triangle, we have a^2+b^2=c^2 where a=5 and b=8, so c=10 (we can see that 6=2*3 and 8=2*4, so the triangle is just a 3-4-5 in disguise) after this, we have 2 ay to calculate the area: a*b and c*r, a*b=6*8=48, so c*r=48, but c=10, so r=4.8, and 4.8^2=23.04, and from this we obtain the same result as in your video.
I had the same until 1:30, I calculated the area of the right angled triangle by doing 6*8*0.5. I then calculated the r with the area and the base by doing: r=96/10/2=4.8. Then I calculated the area of the circle as usual by doing pi*4.8^2=72.38m².
When you have a system of equations you can add or subtract them to get a variable on its own. In this case it was easier to subtract the equations than to solve for one variable in terms of the other and substitute it into the other equation. It's called the elimination method
An equation is an expression of relationships between the values of two or more things. In a system of equations, the premise is that the equations are equal to each other in the values expressed in their respective relationships, having the same solution set: the same set of values equally applies to both equations. In the same way that we can perform equivalent operations on both sides of an equation without changing the value expressed by the relationships in the equation, including subtracting the same value from both sides of an equation, we can subtract one equation from another in a system of equations without changing the value of the shared solution set. ~ To demonstrate this equivalence, another way to solve this system of equations would be to set both equations equal to '0', then through substitution set the values for '0' equal to each other, as thus: r² - 20a + a² = -36 r² - 20a + a² + 36 = 0 ... r² + a² = 36 r² + a² - 36 = 0 ... r² - 20a + a² + 36 = r² + a² - 36 Alternately, we could set each equation equal to 'a²' then employ substitution, as thus: r² - 20a + a² = -36 a² = - (r² - 20a) -36 a² = -r² + 20a - 36 ... r² + a² = 36 a² = -r² + 36 ... -r² + 20a - 36 = -r² + 36 Regardless of the path taken, the end result is the same after simplifying. Subtracting equations is a "shortcut" method to substitution.
you get the same result as if you did (which is the regular way for somone that doesn't know that method would work it): a^2 + r^2 = -a^2 + 20a - r^2 (because 36 = - (-36)) which were the results at the time but it's a different method to get the same a = 3,6
I don't know if i was wrong but i just assumed that the height of the triangle = r, so i calculated by doing 6*8/10, which equals to 24/5, and if you square it equals to 576/25=23,04.
@AndyMath This has been bugging me for a while, but if the length of each of the sides is 10, and the circle is tangential with each side of the rhombus, shouldn’t the diameter of the circle be 10, meaning that the area should be 25*pi? Probably making a logical mistake here but it’s been bugging me for about an hour
your idea works when the quadrilateral has 90 degree corners, but a rhombus does not. Imagine if you kept the side lengths, of the rhombus, all 10m but squished the rhombus shorter.... the circle would definitely get smaller. Hope that helps!!
An even easier way that I did in my head. r divides the 6 8 10 triangle into 2 - 345 triangles. Therefore r is 4/5s of 6.....(24/5). area is pi*(24/5)^2 = 576pi/25.
I tried this once in grade school (albeit without the sandwich), but all it got me was a chuckle from the teacher and a '0' for the problem... but the chuckle was worth it....
im slightly confused. isnt the diameter of the circle just 10m bcuz you can draw a diameter parallel to 2 of the sides which would make a parallelogram. this means the diameter is equal to the side length, 10 meaning the radius would be 5. this means r^2 would be 25 and the area would be 25pi
There is in fact a simplier way, but I guess you know that and used the question as a little clickbait...🤣 OK, Rhgombus with diagonals e = 16 and f = 12. Get the sidelength of the rhombus with Pythagoras s = √(8² + 6²) = 10. Then you know the area of the rhombus is given by A = (e·f)/2 = (16·12)/2 = 96. And the area is also given by A = s·h, with the height h beeing the diameter of the circle, so h = 2r, therefore we have A = 10·2r = 96 => r = 96/20 = 4.8. The area of the circle is then 4.8²·π = 23.04·π.
I would have loved if you had shown the inverse pythagoras theorem in the video or atleast hinted at it, since the problem can be directly solved using it. Still wonderful video as always.
Because sides of rhombus are 10 due to hypotenuse of triangle 345 and doesn't tangent the circle in the middle so one part is a(4.8) and rest is 10_a(5.2) smaller half is equal to radius of circle 😊
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you had a perfect opportunity to describe and show a 3-4-5 triangle
Exactly
And after that, the smaller triangle 6 - a - r has the same angles as the bigger triangle and so is similar, thus R can be found just by the ratio of sizes.
And just notice (by triangle area) that 6×8=10×r.
@@daniel_77. That's what i thought, 3-4-5 triangle and 6
*8=10*r
IT HURTS
My method was the same until 1.30 in to the video. The area of the triangle is 6x8/2 =24, but can also be found by doing 10 x r /2 which can be written as 5r. In other words 5r=24 so the radius is 4.8. Having done that, the area of the circle is 4.8^2 multiplied by pi= 23.04 pi
That's how i did it. Felt easier
I'd'ev also left it as 4.8^2*pi m2 since it's easier to read and avoid going down the rabbit hole of decimal points, thus avoiding approximating pi and so on and so forth.
Nice one, I did something similar. I recognized that when you draw the height of a right triangle from its right angle, you end up with two smaller triangles that are similar to the original. So we can do
(long leg)/(hypotenuse) = 8/10 = r/6
and we end up with effectively the same equation.
I did the same
*Yeah, this is **_waaay_** easier, I don't see how he didn't see it*
@mr.d8747 smart people look for a hard way 😂
There is a simpler way! Notice how in the 6 8 10 triangle, the radius is perpendicular to the hypotenuse. Therefore the area of the triangle is equal to 6x8/2 or 10r/2. So, 10r/2 = 6x8/2, r = 48/10 = 4.8m. So area = pi x r^2 = (4.8)^2 x pi = 23.04 pi as in the video
I've never been more excited
Sex is better :)
the small triangle you created inside the 6-8-10 triangle is equal in shape so by comparing sides, 10/6 = 8/r
1:33 There's an easier way. There's a theorem called "right triangle altitude theorem". With it, we can write three equations that are simpler than the Pythagorean theorem ones.
I. _a * 10 = 6²_
II. _(10 - a) * 10 = 8²_
III. _a * (10 - a) = r²_
With the first, we have _'a'_ immediately, that we can substitute in the third to get _r²_ .
It's simpler than that, the height of a right angle triangle perpendicular to the hypotenuse is just the product of the two shorter sides divided by the hypotenuse. In a 6-8-10 triangle that's just (6*8)/10, or 4.8.
Then you don't need to split the hypotenuse and deal with 'a' or '10-a' at all.
@@heavybrambles You mean _T = 6 * 8 = 10 * r_ from which _r_ is _6 * 8 / 10_ ? I mean, it's a simple insight, but it wasn't obvious to me at first. Nice!
aint this just euclid
How exciting!
Following up on what others have noted: the radius of the inscribed circle is the height of a 6-8-10 right triangle. For any right triangle, the height is ab/c, so in this case 6*8/10 = 4.8 is the radius. In the general case of an circle inscribed in a rhombus of diagonals d1 and d2, the circle’s area will be given by A = pi * d1^2 * d2^2 / [4 (d1^2 + d2^2)]. In this case, A = pi * 12*16 / [4 (144 + 256 )] = 23.04 pi.
It’s easier to remember in words: the area of a circle inscribed in a rhombus in terms of the diagonals is the product of their squares divided by the sum of their squares, times pi/4.
At 1:22 we have a right triangle with sides 6 and 8 and hypotenuse 10. We can compute its area by treating one side as the base and the other as the height, so A = (1/2)(8)(6) = 24. We can also compute the same area by treating the hypotenuse as the base and then the circle's radius becomes the height, A = (1/2)(10)(r) = 5r. Equate the areas and 5r = 24, so r = 4.8. The area of the circle is πr² = π(4.8)² = 23.04π, as AndyMath also found.
How exciting…. Love that statement
You could also use the area of the rhombus. It is equal to the product of the diagonals divided by 2 so it is 16x12/2 = 96
Then, use the Pythagorean theorem to figure out the side length 10
After that, since a rhombus is a special case of a parallelogram its area is also equal to base x height. The height of the rhombus is also a diameter of the circle so 10 x diameter = 96. Therefore, the diameter is equal to 96/10 = 9.6 so the radius is 4.8. Plugging that into the formula for the area of a circle πr² gives the same result 23.04π
This is fairly similar to my way. I got 96 = 20r and went from there.
Easier way to find the diagonal of the 6-8 triangle.
It's a right triangle with one side a multiple of 3, and another of 4, both by the same value.
So the last side is 5×2.
Here's my approach: calculate the altitude of any of the right triangles. This altitude is equal to the radius of the circle. This is because both the altitude and the radius form a right angle with the tangent line. Radius = Altitude = (ab/c) = (6*8)/10 = 4.8. With this, we can calculate the area of the circle.
Kinda need you for something.
Edexcel paper 3 had the last question where:
Theres 2 hexagons and a circle such that the circle is in between those two hexagons.
The larger hexagon touches the circle by its tangent, and the smaller hexagon touches the circle by its chord.
"Considering perimeters, show that 3 < pi < 2√3"
Give tutorial pls
ok in 1:34 you could use the formula a^-2+b^-2+h^-2(h is the alternate height in right angle triangle where c is the base) (these are also inverse squares) or u can use r times c is equal to a times b since the tringales area can be calculated with two different bases and heights , anyways love ur vids
you place center of the circle and rhombus at the orego, drawing lines through the verticies
there is a vector [x=8,y=-6] from the top corner of the rhomb to the right corner.
This gives a function downslope(x)=6 - 3/4x, which is intercepted at a 90deg angle from origo by another function upslope(x) = 4/3x.
The distance from origo to the point where downslope and upslope intercept is the radius of the circle, which gives the area
Alternatively I think it should be possible to declare a function distance(x)=(x^2 +upslope(x)^2 )^(1/2), solve for the value which makes the derivative of distance 0 and find the interception point with this x value.
Does this check out? I didnt do the math
There are two ways this probably wouldve been easier
Either 1: Theres a formula for the height in a right triangle, its cathetus 1 * cathetus 2 divided by hypothenuse, so you couldve instantly just did 6 * 8 / 10 = 4.8 which is the radius
Or 2: Analytic geometry
Consider the center of the circle O(0,0) so that means the corner above it will be A(0,6), the corner to its left will be B(8,0) which means its line formula is 3x + 4y = 24
Use the formula for the distance from a point to a line and you get that the distance is 4.8, that distance being the radius
By symmetry, you consider a quarter of the shape which is a right triangle of sides 6-8-10 (by Pythagore), and then you juste express the area of this triangle 2 ways to get the radius of the circle : 10*r = 6*8 r = 4.8
From there you get the area of the circle with pi*rˆ2 = 23,04pi
You could also use the area of the triangle to identify the “r”. You know that r is the height to the “c” side since those are in a right angle. That means that if you identify lenght of c (=10) you can then devide the area of the triangle by that and get r=4,8. Result is the same. (6*8=48 -> 48/10) i think this is much cleaner since you don’t need a calculator for this. 😊
I like your content, keep up with the good work. ❤
I would calculate the length of a side by pythagorean theorem and the area A_r=12×16/2=96. Since the circle is inscribed we know that 2r is the length of the height of the rombus and we get an equation 20r=96 so r=24/5 from that A_c=576/25 pi
Instead of using Pythagorean Theorem to find the altitude in the right triangle, you could instead treat the altitude as the height in the triangle and the hypotenuse as the base. (1/2)*b*h is the area of a triangle where "h" is the height and "b" is the base, but (1/2)*a*b is also the area of the triangle (only right triangles) so you can set them equal and cancel the "(1/2)" terms so you'll get: b*h=a*b.
interesting but good
1:33 from here you can just use Euclidean theorem for right triangle
They are three
First (10-a)*a=(r)^2
Second (10-a)*8=(8)^2
And the third that we are gonna use is
a*10= (6)^2
So a=3.6
Now skip to 3:00
Calculate the area of the triangle and then divide by the hypotenuse
Multiply diagonals and divide by 2 for rhombus area of 96 m^2. A rough estimate has circle area as
Man, I'd love to see you ramp it up a bit. The geometry has been fun, but I want to see a bigger math challenge
The area of the rhombas is (d₁⋅d₂)/2=(12⋅16)/2=96.
But the area of the rhombas is also base ⋅ height or s⋅2r, where s is any side length and 2r is the height constrained to the minimally inscribed circle's diameter, so
96=s⋅2r, r=48/s.
The bisection of the diagonals gives the pythagorean relation:
s²=6²+8², s=10.
With r=48/10=4.8, the area of the circle is (4.8)²(Pi)
A=23.04(Pi).
I used substitution instead of elimination and a 3-4-5 triangle to get the length was 10 quickly
b = 10 - a
r^2 + a^2 = 36
r^2 = 36 - a^2
r^2 + b^2 = 64
r^2 = 64 - b^2
64 - b^2 = 36 - a^2
a^2 - b^2 = -28 then use difference of squares
(a+b)(a-b) = -28 then substitute b = 10 - a
(a+10-a)(a-(10-a)) = -28
(10)(-10+2a)= -28
-10 + 2a = -2.8
2a = 7.2
a = 3.6
r^2 = 36 - (3.6)^2
r^2 = 23.04
Easier way would have been to work with streching or ratios and 3-4-5 triangles. Obviously each tangent is 10m long and is the hypotenuse in the "3-4-5" or better "6-8-10" triangle. From here find a triangle where r is one of its sides. There are 2 options. Each option reuses an angle from the initial 6-8-10 triangle and also contains a 90° angle where r hits the tangent line piece. Therefore we know that the new triangle with r is equal to the initial one. From here use ratios:
Option 1:
hypotenuse / leg_1 = leg_2 / r
= 10m / 6m = 8m / r
=> r = 4,8m
Option 2:
hypotenuse / leg_1 = leg_2 / r
= 10m / 8m = 6m / r
=> r = 4,8m
Finally r²*pi
Not sure if its easier, possibly shorter: due to the opposite sides of rhombus being parallel, we know that the points of the circle intersecting opposite sides of the rhombus must have the diameter of the circle being the shortest line between them as only points on opposites sides of a circle will have a parallel tangent line, which is also at right angles to those sides as as the sides of the rhombus are tangents to the circle.
Now we just have to find the "perpendicular height" of the rhombus which is equal to diameter of the circle. Area of rhombus = base x height. Area is also equal to (diagonal1xdiagonal2)/2 = 96. Base = sqrt ( 8^2 + 6^2) = 10. Height = 96/10 = 9.6. So Radius = 4.8, area = pi*4.8^2
Please upload more frequently i am improving my geometry with your videos
I can't believe you guys in the US don't know simple maths. Here in the UK we learn the following:
First figure out the area of the rhombus which is easy: (12*16)/2 = 96. Then figure out a side length which is 10 in this case. Half of the perimeter of the rhombus = 20. Radius of the circle is 96/20 = 4.8. Then just calculate the area as normal using the radius. Simple, quick and easy. If you need any more details, let me know.
Area is approx 72.36
Why you divided 96 by 20
I didnt do the same thing, after drawing the radius i knew that the 3 triangles were similar and like that i was able to figure out that 10-a =6.4 then i just used the law of sines to determine the radius and just figured out the area as you usually would.
@1:30 you could have used the similarity of the 3/4/5 triangles, having r/8=6/10, thus r=48/10, thus A=(48/10)^2*pi
(48/10)^2 is btw the 23,04 you got fro a double Pythagorean, but similarity is faster here imho
very interesting andy!
I think this is an easier way, since it requires less thinking. You can use r=2P/(a+b+c+d) becouse of the circle inscribed in the qudrilateral, and you get r=4.8
I found r in a much easier way: after dividing the rhombus in 4 right triangle, we have a^2+b^2=c^2 where a=5 and b=8, so c=10 (we can see that 6=2*3 and 8=2*4, so the triangle is just a 3-4-5 in disguise) after this, we have 2 ay to calculate the area: a*b and c*r, a*b=6*8=48, so c*r=48, but c=10, so r=4.8, and 4.8^2=23.04, and from this we obtain the same result as in your video.
Using similar triangles and spotting the 3-4-5 we don’t even need to worry about “a”. r=6x8/10, or 8x6/10, or 4.8
I had the same until 1:30, I calculated the area of the right angled triangle by doing 6*8*0.5. I then calculated the r with the area and the base by doing: r=96/10/2=4.8. Then I calculated the area of the circle as usual by doing pi*4.8^2=72.38m².
at 2:25 why did we subtract the two equations?
When you have a system of equations you can add or subtract them to get a variable on its own. In this case it was easier to subtract the equations than to solve for one variable in terms of the other and substitute it into the other equation.
It's called the elimination method
so you can get rid of the r^2 and have a correct equation that only has one variable
An equation is an expression of relationships between the values of two or more things. In a system of equations, the premise is that the equations are equal to each other in the values expressed in their respective relationships, having the same solution set: the same set of values equally applies to both equations. In the same way that we can perform equivalent operations on both sides of an equation without changing the value expressed by the relationships in the equation, including subtracting the same value from both sides of an equation, we can subtract one equation from another in a system of equations without changing the value of the shared solution set.
~
To demonstrate this equivalence, another way to solve this system of equations would be to set both equations equal to '0', then through substitution set the values for '0' equal to each other, as thus:
r² - 20a + a² = -36
r² - 20a + a² + 36 = 0
...
r² + a² = 36
r² + a² - 36 = 0
...
r² - 20a + a² + 36 = r² + a² - 36
Alternately, we could set each equation equal to 'a²' then employ substitution, as thus:
r² - 20a + a² = -36
a² = - (r² - 20a) -36
a² = -r² + 20a - 36
...
r² + a² = 36
a² = -r² + 36
...
-r² + 20a - 36 = -r² + 36
Regardless of the path taken, the end result is the same after simplifying. Subtracting equations is a "shortcut" method to substitution.
you get the same result as if you did (which is the regular way for somone that doesn't know that method would work it):
a^2 + r^2 = -a^2 + 20a - r^2 (because 36 = - (-36)) which were the results at the time
but it's a different method to get the same a = 3,6
I don't know if i was wrong but i just assumed that the height of the triangle = r, so i calculated by doing 6*8/10, which equals to 24/5, and if you square it equals to 576/25=23,04.
Elegant and awesome
Wait, are the horizontal and vertical lines really the diagonals? Wouldn't the diagonal lines be the diagonals?
Once I had the 10 I thought about the area of the triangle, which is 6*8/2, but also 10*r/2, if you solve for r you get 4.8
Okay every other comment also said that hahaha
What grade/year would you find this level of maths in?
@AndyMath This has been bugging me for a while, but if the length of each of the sides is 10, and the circle is tangential with each side of the rhombus, shouldn’t the diameter of the circle be 10, meaning that the area should be 25*pi? Probably making a logical mistake here but it’s been bugging me for about an hour
your idea works when the quadrilateral has 90 degree corners, but a rhombus does not. Imagine if you kept the side lengths, of the rhombus, all 10m but squished the rhombus shorter.... the circle would definitely get smaller. Hope that helps!!
@@z1y1h401BROO IM SO STUPID i always thought rhombus is just a slanted square. Now i know it doesnt always have 90 degree corners
Look like cursed Brazil flag lmao
true
An even easier way that I did in my head. r divides the 6 8 10 triangle into 2 - 345 triangles. Therefore r is 4/5s of 6.....(24/5). area is pi*(24/5)^2 = 576pi/25.
3-4-5 right triangle
r = (4/5)(6)
From the triangle 0.5*8*6 = 0.5*10*r So r = 4.8 So area of circle = 23.04 pi
You could have used similarity of triangles to find r much quicker than doing Pyth thrm so many times
_"Is there an easier way?"_
Well, yeah. I just declared the area to be 56 and then went and made myself a sandwich. 💯💯💯
I tried this once in grade school (albeit without the sandwich), but all it got me was a chuckle from the teacher and a '0' for the problem... but the chuckle was worth it....
@@jamesarthurreed The sandwich just makes the experience, though.
_"I HAVE CONQUERED!! Now I go eat..."_
im slightly confused.
isnt the diameter of the circle just 10m bcuz you can draw a diameter parallel to 2 of the sides which would make a parallelogram. this means the diameter is equal to the side length, 10 meaning the radius would be 5. this means r^2 would be 25 and the area would be 25pi
tell me why this wouldnt work
nvm i realised the circle doesnt meet the sides in the middle
You could if the circle was inside a square not in a rhombus 😂
Can please prove that the surface area of sphere is 4πr^2 in your next video
🙏🥺
I have a question, why did you subtract one Pythagorean theorem from the other?
i am desperate please answer
Why you use longer solution when there is shorter one?
Of course he knows this, but by similarity of triangle it comes out instantly: r/6 = 8/10 => r = 4.8
a=6
b=8
c=10
r=ab/c which is 6×8/10=4.8
This would've been simpler way
Nah. (6 x 8)/2 is the area of the quarter rhombus. The base of the area is 10, the height is the radius and you know the area. Radius is 4.8 easy.
There is in fact a simplier way, but I guess you know that and used the question as a little clickbait...🤣
OK, Rhgombus with diagonals e = 16 and f = 12. Get the sidelength of the rhombus with Pythagoras s = √(8² + 6²) = 10. Then you know the area of the rhombus is given by A = (e·f)/2 = (16·12)/2 = 96. And the area is also given by A = s·h, with the height h beeing the diameter of the circle, so h = 2r, therefore we have
A = 10·2r = 96 => r = 96/20 = 4.8. The area of the circle is then 4.8²·π = 23.04·π.
How easier!
Hi, Andy. Once you draw the altitude r you have 3 similar triangles. Proportion r/6 = 8/10. Cross-multiply r=4.8, thus r² = 23.04. No algebra.
I would have loved if you had shown the inverse pythagoras theorem in the video or atleast hinted at it, since the problem can be directly solved using it. Still wonderful video as always.
quicker way for r - ((1/2)*6*8) = ((1/2)*r*10)
Yeah it’s like 7
Proof? Equation? You ask?
I eyeballed it.
Heron c'est bon!?!
Why is it 10-a on the other side?
Because sides of rhombus are 10 due to hypotenuse of triangle 345 and doesn't tangent the circle in the middle so one part is a(4.8) and rest is 10_a(5.2) smaller half is equal to radius of circle 😊
could you not have just multiplied the triangle with base 8 and height 6, by 4 because all the triangles are identical.
He's looking for the circle's area, not the rhombus'
these comments are like :
yeah my method was the same , so pi r^2 = 23 something like that lol.
senseiii r = (6 x 8) / 10
I got 576π over 25
Why is it 10 - a
Bro just Use Euclid's theorem r=(6x8)÷10
Why you had to do this hard method
There is a much easier way to do this. Whip out your tape measure and find the diameter. Square that bad boy and multiply it by 3/4. Boom.
That's too easy. Tell me the blue are of the Brazilian Flag.
Try harder ones not fun one. These are for Primary the school. make our brains burn some phosphor
Sorry but the board of the math peanut gallery cannot accept this answer, it is not 42, 69, or 420.
Hi, i'm a possum, and I find garbage
You in the wrong location bro
@sumportuguesedude2605 no, he's sounds just like him!