Easier way: let r = circle's radius. Construct radius OF. For 2 right triangles, ADO & ADC, use Pythagorean formula to calculate (& set equal) side AD. Let X = DO r² - X² = 5² - (r + X)². r² & X² cancel out giving 2r² + 2X² = 25. Area ABCD = r(r + X). r² + rX = 25/2. r(r + X) = area = 25/2.
Thank you for another nice Geometry problem Letting BC = r, noticing that r also equals 1/2 × EC , and that FD^2 +DC^2 =25 If angle DCF = arctan (t ) , then t^2 = FD^2/ DC^2 With EO= OC = r , there is also an isosceles triangle FOC with angle FOC = 180 degrees - 2× angle DCF In FOC , the cosine rule gives: cos(FOC) = (2×r^2-25)/ (2×r×r) = 1- 25/ (2×r×r) cos (2A)== 2 cos^2 (A) - 1 so 2cos^2 (DCF) - 1 = 1 - 25/(2×r×r) so cos^ 2 (DCF) = 1/2(2 - 25/(2×r×r) = 1 - (25 /(4×r×r)) sin^2 (DCF) = 25/(4×r×r) * so sin (DCF) = 5/(2r) and sin (DCF) = FD/5 2 r= 25/ FD Drawing FE to form a right-angled triangle on EC gives cos(ECF) = 5/2r * Putting these together sin (DCF) = cos (DCF ) so FD=DC =r *Now, one of the starred equations is wrong, because in the diagram, F is on AD and not on AB. so now I will watch the video and read comments .
AD=r, radius of semicircle Put AB = a, DF = h In triangle CDF h^2 + a^2 = 5^2 OF = a - r In triangle OFD h^2 + (a-r)^2 = r^2 h^2 + a^2 - 2ar + r^2 = r^2 h^2 + a^2 = 2ar 2ar = 5^2 Area ABCD = ar = 25/2
Let ∠FCD=θ. AB=CD=5cosθ…(1). If BC=X, then X is equal to the radius of the circle. If E and F are connected, ⊿FEC becomes a right triangle, so ECcosθ=FC=5. Since EC=2X, FC=5, 2Xcosθ=5. ∴X=5/(2cosθ）…(2) The area S of rectangle ABCD is S=AB×BC, so according to (1) and (2), S=(5cosθ)×(5/2cosθ)=25/2…(Answer)
Extremely easy A = b.h = 5 * (½5) = 12,5 cm² Position of point F is not defined, we can choose its position, and the original conditions are still fulfilled. I chose point F be equal to point E Chord "5' becomes the diameter of semicircle and base of rectangle, and the radius of semicircle is the height of rectangle. The rectangle increased its base, reduced its height, but the area is the same !!
Other option is to equalize point F and P Chord "5" becomes at 45° Rectangle becomes a square, and all original conditions are still meeting. A = ½d² = ½5² = 12,5 cm²
@@marioalb9726 (Although the diagram shows F on AD not F on AB) . Well done that you show that only the required area is dependent on original conditions, and that theta and radius are not.
@@kateknowles8055 Point F always is on AD, that doesn't change. In the border condition I wrote (rectangle becomes a square), F is on AD and on AB, because points F, A and P are the same point.
The method is clear but very long drawn out explanation with too many now etc and gulps . Between TH-cam ads and the explanation I had to watch it twice before it was clear what was happening. The maths is not difficult but you need to realise TH-cam intersperses it so it is far too long to drag out explanations. Say it simply like I am going to use Cos theta in the two triangles and compare the value of each. I will be able to obtain the length of the rectangle . Thus … . Do not spend time telling people that rectangles have four right angles and opposite sides are equal as that is the definition. If you must do it say it separately and move on.
youtube compensation is tied to video length, and hence this prolonging. you can solve this problem without trigonometry. Consider the △FEC, which is a right-triangle (because ∠EFC is 90° by virtue of being inscribed in a semicircle). We know that an altitude (such as FD) of a right-triangle divides the triangle into similar triangles. i.e., △CFD ~ △CEF Thus, CF/CE=CD/CF i.e., CF²=CE.CD 5²=CE.CD -- (i) But CE=diameter of semi-circle=2*BC (Because BC=AD=radius of circle, as from diagram) Thus, from (i), we get: 5²=(2*BC).CD => BC.CD=5²/2=25/2 But BC.CD is the area of rectangle. i.e., Area=25/2 square units.
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The triangle FCE is a right triangle, so CD * CE = CF², i.e. 2AD * CD = 25, so AD * CD = 25/2, which is the area of the rectangle.
Easier way: let r = circle's radius. Construct radius OF. For 2 right triangles, ADO & ADC, use Pythagorean formula to calculate (& set equal) side AD. Let X = DO
r² - X² = 5² - (r + X)². r² & X² cancel out giving 2r² + 2X² = 25. Area ABCD = r(r + X).
r² + rX = 25/2. r(r + X) = area = 25/2.
Thank you for another nice Geometry problem
Letting BC = r, noticing that r also equals 1/2 × EC , and that FD^2 +DC^2 =25
If angle DCF = arctan (t ) , then t^2 = FD^2/ DC^2
With EO= OC = r , there is also an isosceles triangle FOC with angle FOC = 180 degrees - 2× angle DCF
In FOC , the cosine rule gives: cos(FOC) = (2×r^2-25)/ (2×r×r) = 1- 25/ (2×r×r)
cos (2A)== 2 cos^2 (A) - 1 so 2cos^2 (DCF) - 1 = 1 - 25/(2×r×r) so cos^ 2 (DCF) = 1/2(2 - 25/(2×r×r) = 1 - (25 /(4×r×r))
sin^2 (DCF) = 25/(4×r×r) * so sin (DCF) = 5/(2r) and sin (DCF) = FD/5
2 r= 25/ FD
Drawing FE to form a right-angled triangle on EC gives cos(ECF) = 5/2r * Putting these together sin (DCF) = cos (DCF ) so FD=DC =r
*Now, one of the starred equations is wrong, because in the diagram, F is on AD and not on AB.
so now I will watch the video and read comments .
Interestingly: the area would be squarey and would still be 25/2 if PC were equal to 5, and, in that problem, r would be 5/ (2^(1/2)).
AD=r, radius of semicircle
Put AB = a, DF = h
In triangle CDF
h^2 + a^2 = 5^2
OF = a - r
In triangle OFD
h^2 + (a-r)^2 = r^2
h^2 + a^2 - 2ar + r^2 = r^2
h^2 + a^2 = 2ar
2ar = 5^2
Area ABCD = ar = 25/2
I like doing it without trigonometry. Use the fact that triangles DEF and DFC are similar.
Let ∠FCD=θ. AB=CD=5cosθ…(1).
If BC=X, then X is equal to the radius of the circle. If E and F are connected, ⊿FEC becomes a right triangle, so ECcosθ=FC=5. Since EC=2X, FC=5, 2Xcosθ=5. ∴X=5/(2cosθ）…(2)
The area S of rectangle ABCD is S=AB×BC, so according to (1) and (2), S=(5cosθ)×(5/2cosθ)=25/2…(Answer)
Extremely easy
A = b.h = 5 * (½5) = 12,5 cm²
Position of point F is not defined, we can choose its position, and the original conditions are still fulfilled.
I chose point F be equal to point E
Chord "5' becomes the diameter of semicircle and base of rectangle, and the radius of semicircle is the height of rectangle.
The rectangle increased its base, reduced its height, but the area is the same !!
Other option is to equalize point F and P
Chord "5" becomes at 45°
Rectangle becomes a square,
and all original conditions are still meeting.
A = ½d² = ½5² = 12,5 cm²
@@marioalb9726 (Although the diagram shows F on AD not F on AB) . Well done that you show that only the required area is dependent on original conditions, and that theta and radius are not.
@@kateknowles8055
Point F always is on AD, that doesn't change.
In the border condition I wrote (rectangle becomes a square), F is on AD and on AB, because points F, A and P are the same point.
(5)^2=25 {90°A+90°B+90°C+90°D}=360°ABCD/25=13.35 1^1.3^2^3 1^2^3 2^3 (ABCD ➖ 3ABCD+2).
Hacemos coincidir E con D→ CF=CE→ Radio =BC=5/2→ Área ABCD =5*5/2=25/2 =12,5.
Gracias y saludos.
This can solve sometimes easily selecting the CD=r or CD=2r case.
We join E & F
cos t = 5/ 2 r , cost = CD/5 , t is angle between diameter and CF
2 r CD = 25
area = 25/2 sq units
12.5
Brought back the old outro music! 👍👍
A = ½5² = 12,5 cm² ( Solved √ )
Can we find radius ?
Cannot. Problem does not constrain radius, but area is constrained!
The method is clear but very long drawn out explanation with too many now etc and gulps . Between TH-cam ads and the explanation I had to watch it twice before it was clear what was happening. The maths is not difficult but you need to realise TH-cam intersperses it so it is far too long to drag out explanations. Say it simply like I am going to use Cos theta in the two triangles and compare the value of each. I will be able to obtain the length of the rectangle . Thus … . Do not spend time telling people that rectangles have four right angles and opposite sides are equal as that is the definition. If you must do it say it separately and move on.
youtube compensation is tied to video length, and hence this prolonging.
you can solve this problem without trigonometry.
Consider the △FEC, which is a right-triangle (because ∠EFC is 90° by virtue of being inscribed in a semicircle).
We know that an altitude (such as FD) of a right-triangle divides the triangle into similar triangles.
i.e., △CFD ~ △CEF
Thus, CF/CE=CD/CF
i.e., CF²=CE.CD
5²=CE.CD -- (i)
But CE=diameter of semi-circle=2*BC (Because BC=AD=radius of circle, as from diagram)
Thus, from (i), we get: 5²=(2*BC).CD => BC.CD=5²/2=25/2
But BC.CD is the area of rectangle.
i.e., Area=25/2 square units.