What is the size of the rectangle? #maths #mathematics #math #mathpuzzles #geometry
Пікірлер: 37
@gibbogle7 ай бұрын
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.
@syedali99146 ай бұрын
Brilliant
@chrisjfox87154 ай бұрын
How do you know for certain those two triangles are similar?
@gibbogle4 ай бұрын
@@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.
@AlamKhan-yt9wd4 ай бұрын
😮 excellent
@Marcelo-kd9wk4 ай бұрын
Exactly how I did.
@hongningsuen134816 күн бұрын
Method using Thales theorem and trigonometric ratio: 1. Construct right-angled triangle with side of length 10 and diameter of semicircle by Thales theorem. Let the right bottom angle of this triangle be theta. 2. Let radius of semicircle be R and width of rectangle be W. Height of rectangle = R as top of rectangle is tangent to circle. 3. In original right-angled triangle W = 10 cos(theta) In constructed right- angled triangle 2R = 10/cos(theta) 4. Area of rectangle = WR = 10 cos(theta) x 5/cos(theta) = 50.
@cheesebusiness4 ай бұрын
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.
@Ramkabharosa7 ай бұрын
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.
@Masterclass_Geometry5 ай бұрын
nice 👌
@uggupuggu4 ай бұрын
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
@PS-mh8ts7 ай бұрын
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
@Darisiabgal75732 ай бұрын
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
@Eilli-lieАй бұрын
I ain't readin allat
@beatjosefbuhrer7829Ай бұрын
the result, only if the base of the rectangular is 3/4 of the diameter, or?
@rey-dq3nx6 ай бұрын
by proportion 10/2r=s/10 2sr=100 sr=50 sq units
@user-ws9zd8pn8r5 ай бұрын
GOOD, very simple
@Sg190th2 ай бұрын
Wouldnt it be easier to do a 6 8 10 triangle?
@Peter_Riis_DK2 ай бұрын
Sorry man, the other guy shows and explains it better.
@hohoy956310 ай бұрын
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)
@cowofthemonth6 ай бұрын
If I see the intersecting chords theorum can be used, I look no further..
@fish8807314 ай бұрын
看起來很難,結果莫名其妙很簡單就解出來了
@yakupbuyankara59039 ай бұрын
50
@Masterclass_Geometry5 ай бұрын
nice 👌
@AjaySivaram-by8vl4 ай бұрын
= 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
@kituluipitu16447 ай бұрын
Area of rectangle equals 50 square units…now, what is area of semicircle ?
@bobbybannerjee51566 ай бұрын
Can't be found?
@user-st3di6yn9g6 ай бұрын
32pi
@kituluipitu16446 ай бұрын
@@user-st3di6yn9gplease show calculation… it doesn’t seem it can be about twice area of rectangle