It is much simpler than that. Draw a line from the upper end of the line with length 10 to the left corner of the semicircle. Now we have two similar right-angle triangles, one with hypotenuse length 10, the other with hypotenuse length 2r. 10/2r = x/10, xr = 50.
@@chrisjfox8715 They share the acute angle on the right, and they both have a right angle (a triangle on the diameter touches the circle at a right angle). QED.
A very similar problem to this has already been presented elsewhere on you tube this is just a repeat. Let’s label so points A and B are the right and left ends, respectively, of the diameter of the semicircle with origin 0. AC is a chord of length 10 u with C a point between A and B on the semicircle. Let’s label two more points D, E and F. D is point (0,r), E is point (r,r) and F is point (r- horizontal traverse of AC,r). Length EF then equals (r,r) - (r- horizontal traverse of AC) = horizontal traverse of AC. How do we define a chord in terms of radius. r * chord θ (on unit circle) = length In this case r must be greater than 5 so that r * chord θ = 10 How do we define r there are two ways. Chord θ calculation of Chord θ = 2 sin (θ/2) Therefore r = 10/ chord θ or 10/(2 sin (θ/2)) Another useful value are halfchords and bisectors r = 10/( 2 * halfchord) The horizontal traverse of a chord with one point on the horizon is r * 2 * halfchord/ r)^2 = 2*(chord/(2r))^2 = 2*chord^2/4r^2 2*chord^2 / 4r^2 = 0.5 chord^2/ r = 0.5 * 10^2 / r = 0.5 * 100/r = 50/r So the height of the rectangle of base at y = 0 is D - O = (0,r) - (0,0) = r length Height times width = r u * 50u / r = 50 Domain of this answer is for all r >=5 and for all 0° < θ
I solved it this way: According to the restrictions of the problem, the top left corner of the rectangle may be on any point of the semicircle, and the answer shouldn’t depend on it. Therefore I can draw the rectangle such that the top left corner falls not on a random point, but on a more convenient point such as the leftmost or the topmost point of the semicircle. In both cases the solution is trivial.
I see that the angle of the slope is not given, so the angle does not matter, i set the angle to 0 degrees, now you have a rectangle with side lengths 10 and 5, finding the area is trivial
Here's a geometric proof: Label the end-points of the diameter as A and B. Thus, B is also the end-point of the line-segment of length=10. Let its other end-point be labeled C. △ABC is oobviously a right-triangle because it's inscribed in a semicircle. Let the altitude from C to AB meet AB at D. We know that the altitude CD divides △ABC into triangles which in turn are similar to △ABC. △CBD ~ △ABC i.e., CB/AB=BD/BC But CB=BC=10 Thus, 10/AB=BD/10 => (AB)(BD)=(10)(10)=100 -- (i) But AB=diameter of the circle=2*the height of the rectangle and BD=the widdth of the rectangle Thus (i) gives 2*(height of rectangle)(width of rectangle)=100 or (height)(width)=100/2 i.e., area=50 square units
Extend semi-circle to circle, let x = base of rectangle & r = radius of circle. Then by Pythagoras theorem and by the intersecting-chord theorem, x.(2r-x) = √(100-x²).√(100-x²) = 100-x². So 2xr - x² = 100 - x². ∴ 2xr = 100. Hence area of rectangle = xr = 50 square units.
It is much simpler than that. Draw a line from the upper end of the line with length 10 to the left corner of the semicircle. Now we have two similar right-angle triangles, one with hypotenuse length 10, the other with hypotenuse length 2r. 10/2r = x/10, xr = 50.
Brilliant
How do you know for certain those two triangles are similar?
@@chrisjfox8715 They share the acute angle on the right, and they both have a right angle (a triangle on the diameter touches the circle at a right angle). QED.
😮 excellent
Exactly how I did.
A very similar problem to this has already been presented elsewhere on you tube this is just a repeat. Let’s label so points A and B are the right and left ends, respectively, of the diameter of the semicircle with origin 0. AC is a chord of length 10 u with C a point between A and B on the semicircle. Let’s label two more points D, E and F. D is point (0,r), E is point (r,r) and F is point (r- horizontal traverse of AC,r). Length EF then equals (r,r) - (r- horizontal traverse of AC) = horizontal traverse of AC.
How do we define a chord in terms of radius. r * chord θ (on unit circle) = length
In this case r must be greater than 5 so that r * chord θ = 10
How do we define r there are two ways. Chord θ calculation of Chord θ = 2 sin (θ/2)
Therefore r = 10/ chord θ or 10/(2 sin (θ/2))
Another useful value are halfchords and bisectors r = 10/( 2 * halfchord)
The horizontal traverse of a chord with one point on the horizon is r * 2 * halfchord/ r)^2 = 2*(chord/(2r))^2 = 2*chord^2/4r^2
2*chord^2 / 4r^2 = 0.5 chord^2/ r = 0.5 * 10^2 / r = 0.5 * 100/r = 50/r
So the height of the rectangle of base at y = 0 is D - O = (0,r) - (0,0) = r length
Height times width = r u * 50u / r = 50
Domain of this answer is for all r >=5 and for all 0° < θ
I solved it this way:
According to the restrictions of the problem, the top left corner of the rectangle may be on any point of the semicircle, and the answer shouldn’t depend on it. Therefore I can draw the rectangle such that the top left corner falls not on a random point, but on a more convenient point such as the leftmost or the topmost point of the semicircle. In both cases the solution is trivial.
I see that the angle of the slope is not given, so the angle does not matter, i set the angle to 0 degrees, now you have a rectangle with side lengths 10 and 5, finding the area is trivial
Here's a geometric proof:
Label the end-points of the diameter as A and B. Thus, B is also the end-point of the line-segment of length=10. Let its other end-point be labeled C. △ABC is oobviously a right-triangle because it's inscribed in a semicircle. Let the altitude from C to AB meet AB at D.
We know that the altitude CD divides △ABC into triangles which in turn are similar to △ABC.
△CBD ~ △ABC
i.e., CB/AB=BD/BC
But CB=BC=10
Thus, 10/AB=BD/10 => (AB)(BD)=(10)(10)=100 -- (i)
But AB=diameter of the circle=2*the height of the rectangle
and BD=the widdth of the rectangle
Thus (i) gives
2*(height of rectangle)(width of rectangle)=100
or (height)(width)=100/2
i.e., area=50 square units
Extend semi-circle to circle, let x = base of rectangle & r = radius of circle.
Then by Pythagoras theorem and by the intersecting-chord theorem,
x.(2r-x) = √(100-x²).√(100-x²) = 100-x². So 2xr - x² = 100 - x². ∴ 2xr = 100.
Hence area of rectangle = xr = 50 square units.
nice 👌
A second method : The that see the diameter is 90 degrees. Form there it has eucledean. (I dört havle enough english but you will probably get that)
Sorry man, the other guy shows and explains it better.
If I see the intersecting chords theorum can be used, I look no further..
Wouldnt it be easier to do a 6 8 10 triangle?
by proportion
10/2r=s/10
2sr=100
sr=50 sq units
GOOD, very simple
看起來很難，結果莫名其妙很簡單就解出來了
= R = x = y 8 direction of next week so that we use this video is that 49 days of time of the kids aj se the same thing
Area of rectangle equals 50 square units…now, what is area of semicircle ?
Can't be found?
32pi
@@user-st3di6yn9gplease show calculation… it doesn’t seem it can be about twice area of rectangle
50
nice 👌