You don't need to calculate EB and FD. Once you know that MO=8 and ON=10 and AB=MN you can calculate that AB=MN=MO+ON=8+10=18. Likewise AD=PO+OH=6+10=16.
6:08 The "next step" after step 2 is to recognise that HO is a radius of the semicircle and we already know that OP = MA = 6. so the height of the rectangle is 10+6 = 16. Similarly, ON =10 and OM = AP = 8, so the width of the rectangle is 10+8 = 18. Calculating DF and EB is a waste of time.
How dang frustrating.... He had the numbers written there on the middle lines, but insisted on finding the lengths of the inconsequential line segments. He didn't make it clearer, as some have suggested; he made it seem way harder than it really is!
Awesome question, excellent way solving the problem, many thanks! fast lane: tan(φ) = 3/4 = AF/AE = OP/EP = FM/OM = AM/AP → EF = 20 = 2r → r = FO = EO = 10 → (1/2)πr^2 = 50π AM = FM = 6 → EP = AP = OM = 8 → area ∆AEF = 2∙6∙8 = 96 → AD = 6 + r = 16 → AB = 8 + r = 18 → (6 + r)(8 + r) = 288 → 16(18 - 6) - 50π = 16(12) - 50π = 2(96 - 25π) ≈ 34,92 square units A slightly different approach (a bit complicated): extend CD to the left, and EF to the upper left, intersection point of both lines = D‘. Extend EF to the lower right, and BC downwards, intersection point = B‘. Now you’ve got a rectangled triangle B’CD‘ with tan(φ) = 3/4 and sin(φ) = 3/5 . Obviously: CD‘ = 18 + 16/3; and CB' = 10 + 15/2 → area ∆D'B'C = (1/2)(35/2)(70/3) = (25/6)49
Now subtract the areas of the two smaller similar triangles (upper left, lower right) → (1/6)(25∙49 - 73) = 192 → 192 - 50π ≈ 34,92 square units 🙂
By Pythagorean theorem we have the radius of the circle as √(12^2+16^2)=20. It's radius is 10. By triangles similarities the whole rectangle has length 16/2+10=18 and width 12/2+10=16. So it's area is 18*16=288 sq units. The area of the triangle inside is 16*12/2=96 sq units and the area of the inner half circle is π*10^2/2=157.08 sq units. Hence the area of the shaded region in the rectangle will be 288-157.08-96=34.92 sq units given that π=3.1416. Thank you very much!
In triangle AEF, line (OP) is parralel to side [AF] and passes through side [EF] middle : So P is the middle of [AE] . Hence AP=16/2=8 and OP=12/2=6. ==> AB=8+r=18 and AD=PH=6+r=16.
By observation, the ratio of the short to long leg on triangle ∆FAE is 12:16 = 3:4. As ∆FAE is a right triangle, it is a 4:1 ratio 3:4:5 Pythagorean Triple triangle and EF = 5(4) = 20. As EF is also the diameter of semicircle O, EO = OF = 10, which is the radius. Let G be a point on AB and H a point on CD where GH is perpendicular to both and passes through O. As it is a radius of semicircle O, OH = 10. As ∠GEO is shared between them and FA and OG are parallel, ∆OGE is similar to ∆FAE. OG/EO = FA/EF OG/10 = 12/20 = 3/5 OG = 30/5 = 6 GH = GO + OH = 6 + 10 = 16 As BC and DA are parallel to GH, and all intersect and are perpendicular to AB and CD, BC = DA = GH = 16. Let J be a point on DA and K a point on BC where KJ is perpendicular to both and passes through O. As it is a radius of semicircle O, OK = 10. As ∠OFJ is shared between them and AE and OJ are parallel, ∆FJO is similar to ∆FAE. OJ/OF = AE/EF OJ/10 = 16/20 = 4/5 OJ = 40/5 = 8 KJ = OK + OJ = 10 + 8 = 18 As AB and CD are parallel to KJ, and all intersect and are perpendicular to BC and DA, AB = CD = KJ = 18. The green enclosed area is the total area of the rectangle minus the areas of semicircle O and triangle ∆FAE: A = lw - πr²/2 - bh/2 A = 18(16) - π(10²)/2 - 16(12)/2 A = 288 - 50π - 96 A = 192 - 50π ≈ 34.92
I believe that's a property of any square triangle whose segments are cut at their midpoints, since any isosceles triangle is basically two square ones standing back-to-back and we already know that all the triangles formed by the points (A, P, E, M, F, O) are similar and that most of them are actually perfectly identical or reflections.
This is a beautiful piece of work! Please keep making more like this, these videos are incredibly instructive and inspirational. I'm a huge fan. Thank you.
Obrigado pelo vídeo!! Com certeza existe mais de uma maneira de calcular algumas medidas mas, o fato de você postar e mostrar sua forma de resolver merece só agradecimentos.
It seems to me that your solution is rather more belabored than necessary, although your logic is flawless. I found the solution without calculating DF or EB. Once you find the length of EF (20) via the Pythagorean Theorem and note that O bisects FE, you continue to construct the two smaller similar triangles. Because O bisects EF, by similarity, MO and PO must be half AE and AF, respectively. The necessary dimensions necessary for your solution strategy are AF, AE, OF (or OE), AB, and AD. At that point you have them all. Glad to say I got the same solution as you did -- otherwise I'd be really embarrassed. ;-)
If ∠FAE=90º ⇒ Inscribed in semicircle ⇒ FE=2Radio=2R ⇒ 4R²=12²+16² ⇒ R=10 The legs FA and AE are chords and the radii perpendicular to them are perpendicular bisectors that divide them into two halves ⇒ Area of the rectangle = BxH = [R+(AE/2)]x [R+(AF/2)]=(10+8 )(10+6)=18x16=288 → Green area=Rectangle - ∆FAE - Semicircle = 288 - (12x16/2) - (π10²/2) = 34.92036 Another solution by symmetry.- If we call B the green area on the left, C the upper right and D the lower → The symmetrical areas of B with respect to the vertical radius and D with respect to the horizontal radius, overlap area C and leave uncovered a rectangular area with base=[R-(16/2)]=10-8=2, and height=[R-(12/2)]=10-6=4 ⇒ B+D=C- (2x4)=C-8 ⇒ Green area = 2C-8 = 2[R²-(πR²/4)]-8 = 2(100-25π)-8 = 34.92036
What's the rule/theorem again that confirms the point N, for the perpendicular line MN is the exactly point the circle connects with the square? Yes, it implies that 'visually' but how do we know the image is to scale etc and that we can trust the visual? Same for line HP.
That was a given of the problem. There is no other definiton of what the line CB is other than that it is tangent to the circle at a right angle to the line DC.
Man, you got me to do math in my summer holidays after a schoolyear in which my math teacher had annoyed me with difficult math problems like crazy. Congrats, great video!
Midpoint of segment A(0,12) B(16,0) Is (8,6). The radio Is 10. The measures of the rectangle are 18 large and16 wide. Greetings from BOGOTA D.C. COLOMBIA
6:16 Triangle FAE is under Thales Theorem or Circle's Double Angle Theorem. Thus point A is also on the circle's circumference! FM = MA (Chord Bisector Theorem) = 12/2 = 6 AP = PB (Chord Bisector Theorem) = 16/2 = 8 PB = HO = radius = 10 (Pythagorean Theorem) Area = Area of Square - Area of Triangle - Area of Half-Circle Area = (AP+PB)(HO+OP) - .5(AE)(FA) - .5*π(F0)² Area = (8+10)(10+6) - .5(16)(12) - .5*π(10)² Area = 192 - 50π
That was a cool review of transversals. I used triangle mid segment theorem to find the 6 and 8. It's a nice heuristic approach. I didn't have any paper or a pen. With it half of 12 is 6, and half of 16 is 8. Both added to the length of the radius, gives 16x18, subtract half of 12x16, subtract half of 314.15= 34.925. Fun project. Thank you!
In triangle AEF, line (OP) is parralel to side [AF] and passes through side [EF] middle : So P is the middle of [AE] . Hence AP=16/2=8 and OP=12/2=6. ==> AB=8+r=18 and AD=PH=6+r=16.
I wondered why you were calculating EB and DF when you had already discussed 3:4:5 triangles, so OP must be 3x2 and OM must be 4x2 so no new learning required.
Triangle Area = 12×16/2= 96 Hypothenuse of the rectangle is Sqrt(12^2+16^2)=20. So the half circle area has a radius of 10, therefore the area 10^2× pi/2 = 50 pi. To get the rectangle area: The center of the circle is 6 above the bottom and 8 to the right of the left of the triangle, because it is also the middle of the Hypothenuse Adding the circle radius of 10 to each, we get 16×18 = 288 as rectangle area. Now green area = rectangle area - triangle area - half circle area = 288 - 96 - 50 pi = 192 - 50 pi
I knew how to calculate the area of the initial triangle and the area of the semi circle, but was dead stuck on how to find the full sides of the rectangle! Thanks for putting the theorems and laws into the video!
I drew pretty much the same diagram. That triangle EFA is 3-4-5 almost goes without saying, so the big hypoteneuse is 20, and the radius of the circle is 10. I had a feeling that lines HP and MN would bisect lines AF and AB, respectively. If they do, then EB = PB - PE = 10 - (16/2) = 10 - 8 = 2; and FD = DM - FM = 10 - (12/2) = 10 - 6 = 4. Then it occurred to me that triangles OFM and EOP are not just similar, but congruent, by the angle-side-angle rule. That means FM = OP = MA, and PE = OM = AP. So HP and MN do indeed bisect AF and AB; and the sides of the rectangle AD and AB are indeed 16 and 18. So the area of the shaded portion is the area of the rectangle, minus the area of the triangle EFA, minus the area of the semicircle; or (16 * 18) - ((1/2) * 12 * 16) - ((1/2) * pi * 10^2); collecting terms, simplifying, and using a little rearrangement trick to avoid using a calculator by converting to numbers in the 12 x 12 multiplication table so you can do it in your head, we have: area = (18 * 16) - (6 * 16) - (50 * pi) = ((18 - 6) * 16) - 50 * pi = (12 * 16) - 50 * pi = (12 * (12 + 4)) - 50 * pi = (12 * 12) + (12 * 4) - 50 * pi = 144 + 48 - 50 * pi = 192 - 50 pi. voila! Thank you, ladies and gentlemen; I'm here all week.
A= 34.92 sq unit ( pi= 3.1416) As for calculating the height and base of the rectangle ABCD at the step 4, we notice that the center O is the midpoint of FE, OM is perpendicular to AD so OM//AB so M is the midpoint of AD, therefore AM=6. Similarly, P is the midpoint of AE so AP=8 Thus the height PH=6+10=16, the base AB=8+10=18
I found it much simpler to complete the circle and the square around it. It is trivial to find the difference in area between the square and the circle. You can then cut that difference in half based on extending your line EF to bisect the figure. The bisected square includes a triangle and a trapezium not in the original figure, so they must be subtracted out. With closer inspection, the trapezium and triangle complete a simple 2x4 rectangle. I still use the arithmetic to calculate r, but the main calculating steps are contained as (20^2-10^2pi)/2 -8.
I would have stuck to some very basic tools and worked out the position of the center of the circle and added the radius to get the lengths of the rectangle. It would have worked but would have been annoying. Sometimes I need someone like you to remind me that there's elegant and easier techniques
People ask me: Why do you watch this kind of videos? you could be watching something fun, then I come back with an insulting response: The mere fact that you are asking that question, tells me you wouldn't understand the answer. PreMath, thank you for your videos.
Why did you look for the EB and DF segments? You literally went "18-16=2, great, now I can figure out that 16+2=18 so I can find the area" You already had the 18 for the segment AB, it was in the form of the segment MN. You already had the 16 for the segment AD, it was in the form of the segment FH. You did not need to substract anything only to add it back in the next step.
The the third side of the triangle = 20 ( sqrt (12sq + 16sq )). Drop a perpendicular from the centre to both 12 and 16. So E to edge = 10 - 8 = 2, So also on verticle F to edge = 10 - 6 = 4. So the length of rectangle = 16 + 2 = 18 & its height = 12 + 4 = 16. Area of the rectangle = 18*16 = 288. Area of half circle = 0.5*pi*10sq = 157.08 Area of triangle = 0.5*12*16 = 96. So Green shaded area = 288 - 157.08 - 96 = 34.92
Can be solved more easily with basic trig. Its evident that ∆EOP is similair to EAF as they share 2 angles and one side length. Seeing as point O is the midpoint/center of the semicirlce, OE=FO=10 from dividing the hypotenuse length 20 by 2. From that point sin and cos^-1 gives u lengths of OP and PO. From which you can see that one side of the rectangle is OP (6) + radius of semi circle (10) and the other is PE (8) + radius (10). From there its the trivial question of subtracting the areas of a right triangle and a semicirlce from a rectangle, no variables present. Assuming that a calculator is allowed, 192-50π jumps out as the answer. An interesting question with a few possible solutions!!
Great to hear! Thanks Philip for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
Too complex. It could be simplified. △AEF is a right triangle, AF=12 and AE=16, so EF=20. Curve EF is semicircle O is center of circle, so EF is diameter and OE=OF=10=radius. (This is important.) ABCD is a rectangle, line ON has to parallel AB and it = radius = 10, so AB = radius +1/2 AE = 18, as ON and BC verticals. line OH has to parallel AD and it = radius =10, so AD = radius+ 1/2 AF = 16, as OH and CD verticles. Area of green = rectangle - △AEF -π * 10^2 = 16*18 - 1/2 * 12 * 16 -50π = 288 - 96 - 50π =192 - 50π.
If you phycally check dimension first one is of 2 units and result is 35 units and assume left upper green 5 units compairing ist green portion and third upto 10 units then total would be 17
Third side of triangle= dia of semicircle = 20 units Radius of semicircle=10 Height of centre of semi circle from base of triang=6. Hence side of Rectangle= 6 + 10 =16 Similarly other side of rectangle = 18 Total area = 16X18=288 Area of triangle=96 A of semicircle = 10x10x(22/7)/4 =78+4/7 Green A=288 -96-78-4/7 =113+3/7 sq units
Thanks dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
I like to watch this channel. When I see his question then I go straight trying to solve it myself without his explanation. After that I come back to check my result.
Step 1. Find the length of FE by Pytagorean triples : 20 Radius of Semi cirles is 10. Area of the semi cirle is 50*phi Step 2 : Find the Area of triangle FAE 16*12/2=24 Step 3 Draw the total circle and total square. You can easily find lengthe of DF is 4 because 2*DE + 12 = 20(Diameter) Same way you can find length of EB = 2. step 4. area of rectanglar ABCD is (12+4)*(16+2) = 288 Steo 5 area of (rectanglar ABCD - triangle FAE - semicircle) = 288-96-50*phi = 192-50*phi
@@ritikadubey3531 We don't need to know EB to find AB. AB = AP + PB. By rectangle OPBN we know PB = ON = radius = 10. From Rectangle APOM we know AP = MO = 8 therefore AP + PB = 18. Similarly, we don't need to know FD to find AD. AD = AM + MD. From rectangle MOHD we know MD = OH = radius = 10 and from rectangle APOM we know AM = PO = 6. Therefore AM + MD = 16.
Hello wait man it's simple question Just use Pythagoras theorem in triangle AFE U will get FE =20 which is diameter of semicircle and r=10 Area of semicircle=πr×r/2 =3.14×10×10/2=157 Now area of rectangle=l×b=16×12=192 Area of shaded portion = Area of rectangle- Area of semicircle =192-157=35 square unit
Thank you so much my dear friend for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
In this demonstration you have not demonstrated that ON is a radius. You just said that it a parallel to AB. Did I miss something or is it just an observation
Thank you so much Govinda for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
DM=HO=radius=10 And, DM+FA-DA=FM → 10+12-DA=6 → DA=16 Similarly, PB+AE-AB=PE → 10+16-AB=8 → AB=18 Then we get the both sides of the rectangle. That is what we want!!
Once you have identified the values of MO and OP, then finished. Area of rectangleABCD- area of triangleFAE - area of semi circleFE= area of green region
Thank you so much Temyra for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
Why did you do the subtraction.. imran weren't you supposed to be adding the area of the semi circle and the area of the triangle and them subtracting this area from the area of the rectangle
My calculation: The length of EF is the root of AF²+AE² = 20. O is halfway EF, so the radius=OE=10. AD=OE+AF/2=16, AB=OE+AE/2=18, the area ABCD=16*18, AEF=16*12/2,=16*6, the semicircle is 10²π/2=50π, so the area is (16*18)-(16*6)-50π=16*12-50π=192-50π. Done.
MN and ON are both part of the same line. ON is a radius. This radius also touches that side of the rectangle at N. That means that point N is on a tangent line BC. Tangent lines are always at right angles to the corresponding radii. So that means that both MN and AB both perpendicular to AD and BC, which makes them parallel to each other.
Way too many unnecessary steps. You had all the info required half way through the video to solve it. Once you had OP and OM, rectangle dimensions were easily found to calculate its area, then simply subtract area of semicircle and triangle, done.
Once you had the lengths of OM & OP this was solvable. As you had the lengths of AB & CD to get the area of the whole rectangle. You had the area of the triangle from the given, and when you found the radius of the circle you had the area of the semicircle. There is no point in finding DF and EB
Think you took a long approach, once you had diameter of the circle, the area of triangle AEF, and length of MN and HP, you solve by calculating area of ABCD - 1/2 of circle- area of AEF.
Excuse me but why is the right angled triangle a 3-4-5 triangle? The teacher made that statement ("triangle AEF is a 3-4-5 triangle") without showing why.
Ahhh... I know why: Use Pythagoras' theorem to get the length of the hypotenuse (20), then use the common factor of 4 for each side of triangle: ratios is seen to be 3-4-5.
Calculating the length of lines DF and BE was an unnecessary step. From the calculation of lines MN and HP you already have the length and height of the rectangle.... My calculation ended up with the green area being 34.92036 square units. My calculation was at 3 seconds in passing the vid, the comment was after seeing the whole vid. Good challenge, good explanation, good vid, thanks.
You don't need to calculate EB and FD. Once you know that MO=8 and ON=10 and AB=MN you can calculate that AB=MN=MO+ON=8+10=18. Likewise AD=PO+OH=6+10=16.
Thanks for the feedback
Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards
Brilliant thinking ... Bravo
Exactly.
Nice
Nice
I always needed someone who can teach step by step calmly . Thank you sir :D
Love how you use angle theorem! Super great, how you simplify anything that looks complicated! Keep up the great work!
6:08 The "next step" after step 2 is to recognise that HO is a radius of the semicircle and we already know that OP = MA = 6. so the height of the rectangle is 10+6 = 16. Similarly, ON =10 and OM = AP = 8, so the width of the rectangle is 10+8 = 18. Calculating DF and EB is a waste of time.
He was just explaining why it was so...
You are absolutely right
For someone like me, it is very helpful to do it step by step, because since I am not an expert, that little extra step helps me in other situations.
How dang frustrating.... He had the numbers written there on the middle lines, but insisted on finding the lengths of the inconsequential line segments. He didn't make it clearer, as some have suggested; he made it seem way harder than it really is!
Awesome question, excellent way solving the problem, many thanks!
fast lane:
tan(φ) = 3/4 = AF/AE = OP/EP = FM/OM = AM/AP →
EF = 20 = 2r → r = FO = EO = 10 → (1/2)πr^2 = 50π
AM = FM = 6 → EP = AP = OM = 8 →
area ∆AEF = 2∙6∙8 = 96 → AD = 6 + r = 16 → AB = 8 + r = 18 →
(6 + r)(8 + r) = 288 →
16(18 - 6) - 50π = 16(12) - 50π = 2(96 - 25π) ≈ 34,92 square units
A slightly different approach (a bit complicated):
extend CD to the left, and EF to the upper left, intersection point of both lines = D‘.
Extend EF to the lower right, and BC downwards, intersection point = B‘.
Now you’ve got a rectangled triangle B’CD‘ with tan(φ) = 3/4 and sin(φ) = 3/5
.
Obviously: CD‘ = 18 + 16/3; and CB' = 10 + 15/2 →
area ∆D'B'C = (1/2)(35/2)(70/3) = (25/6)49
Now subtract the areas of the two smaller similar triangles (upper left, lower right) →
(1/6)(25∙49 - 73) = 192 → 192 - 50π ≈ 34,92 square units 🙂
Good example of how to make something simple as hard as possible.
By Pythagorean theorem we have the radius of the circle as √(12^2+16^2)=20. It's radius is 10. By triangles similarities the whole rectangle has length 16/2+10=18 and width 12/2+10=16. So it's area is 18*16=288 sq units. The area of the triangle inside is 16*12/2=96 sq units and the area of the inner half circle is π*10^2/2=157.08 sq units. Hence the area of the shaded region in the rectangle will be 288-157.08-96=34.92 sq units given that π=3.1416. Thank you very much!
In triangle AEF, line (OP) is parralel to side [AF] and passes through side [EF] middle :
So P is the middle of [AE] .
Hence AP=16/2=8 and OP=12/2=6.
==> AB=8+r=18 and AD=PH=6+r=16.
Ur excellent out performance 👏👏👏👏👏👏👏👏👏
By observation, the ratio of the short to long leg on triangle ∆FAE is 12:16 = 3:4. As ∆FAE is a right triangle, it is a 4:1 ratio 3:4:5 Pythagorean Triple triangle and EF = 5(4) = 20. As EF is also the diameter of semicircle O, EO = OF = 10, which is the radius.
Let G be a point on AB and H a point on CD where GH is perpendicular to both and passes through O. As it is a radius of semicircle O, OH = 10. As ∠GEO is shared between them and FA and OG are parallel, ∆OGE is similar to ∆FAE.
OG/EO = FA/EF
OG/10 = 12/20 = 3/5
OG = 30/5 = 6
GH = GO + OH = 6 + 10 = 16
As BC and DA are parallel to GH, and all intersect and are perpendicular to AB and CD, BC = DA = GH = 16.
Let J be a point on DA and K a point on BC where KJ is perpendicular to both and passes through O. As it is a radius of semicircle O, OK = 10. As ∠OFJ is shared between them and AE and OJ are parallel, ∆FJO is similar to ∆FAE.
OJ/OF = AE/EF
OJ/10 = 16/20 = 4/5
OJ = 40/5 = 8
KJ = OK + OJ = 10 + 8 = 18
As AB and CD are parallel to KJ, and all intersect and are perpendicular to BC and DA, AB = CD = KJ = 18.
The green enclosed area is the total area of the rectangle minus the areas of semicircle O and triangle ∆FAE:
A = lw - πr²/2 - bh/2
A = 18(16) - π(10²)/2 - 16(12)/2
A = 288 - 50π - 96
A = 192 - 50π ≈ 34.92
The point A is sited on the semicircle opposed to the red semicircle FMNE; the segment OA is a radius, so the triangles OAF and OAE are isosceles.
Excellent Massimo
You are awesome 👍
Take care dear and stay blessed😃 Love and prayers from the USA!
I believe that's a property of any square triangle whose segments are cut at their midpoints, since any isosceles triangle is basically two square ones standing back-to-back and we already know that all the triangles formed by the points (A, P, E, M, F, O) are similar and that most of them are actually perfectly identical or reflections.
OA = √(OP² + PA²) = √(64+36) = 10 = OE
*∴ ΔOAE is an isosceles triangle.*
OA = OF
*∴ ΔOAF is an isosceles triangle.*
This is a beautiful piece of work! Please keep making more like this, these videos are incredibly instructive and inspirational. I'm a huge fan. Thank you.
Obrigado pelo vídeo!! Com certeza existe mais de uma maneira de calcular algumas medidas mas, o fato de você postar e mostrar sua forma de resolver merece só agradecimentos.
It seems to me that your solution is rather more belabored than necessary, although your logic is flawless. I found the solution without calculating DF or EB. Once you find the length of EF (20) via the Pythagorean Theorem and note that O bisects FE, you continue to construct the two smaller similar triangles. Because O bisects EF, by similarity, MO and PO must be half AE and AF, respectively. The necessary dimensions necessary for your solution strategy are AF, AE, OF (or OE), AB, and AD. At that point you have them all. Glad to say I got the same solution as you did -- otherwise I'd be really embarrassed. ;-)
This solution was like a travel of a road by watching the views of the road.
Good luck.
Thanks once again dear. Cheers
If ∠FAE=90º ⇒ Inscribed in semicircle ⇒ FE=2Radio=2R ⇒ 4R²=12²+16² ⇒ R=10
The legs FA and AE are chords and the radii perpendicular to them are perpendicular bisectors that divide them into two halves ⇒ Area of the rectangle = BxH = [R+(AE/2)]x [R+(AF/2)]=(10+8 )(10+6)=18x16=288 → Green area=Rectangle - ∆FAE - Semicircle = 288 - (12x16/2) - (π10²/2) = 34.92036
Another solution by symmetry.- If we call B the green area on the left, C the upper right and D the lower → The symmetrical areas of B with respect to the vertical radius and D with respect to the horizontal radius, overlap area C and leave uncovered a rectangular area with base=[R-(16/2)]=10-8=2, and height=[R-(12/2)]=10-6=4 ⇒ B+D=C- (2x4)=C-8 ⇒ Green area = 2C-8 = 2[R²-(πR²/4)]-8 = 2(100-25π)-8 = 34.92036
What's the rule/theorem again that confirms the point N, for the perpendicular line MN is the exactly point the circle connects with the square? Yes, it implies that 'visually' but how do we know the image is to scale etc and that we can trust the visual? Same for line HP.
That was a given of the problem. There is no other definiton of what the line CB is other than that it is tangent to the circle at a right angle to the line DC.
Man, you got me to do math in my summer holidays after a schoolyear in which my math teacher had annoyed me with difficult math problems like crazy. Congrats, great video!
Midpoint of segment A(0,12) B(16,0) Is (8,6). The radio Is 10. The measures of the rectangle are 18 large and16 wide.
Greetings from BOGOTA D.C. COLOMBIA
6:16 Triangle FAE is under Thales Theorem or Circle's Double Angle Theorem. Thus point A is also on the circle's circumference!
FM = MA (Chord Bisector Theorem)
= 12/2
= 6
AP = PB (Chord Bisector Theorem)
= 16/2
= 8
PB = HO = radius = 10 (Pythagorean Theorem)
Area = Area of Square - Area of Triangle - Area of Half-Circle
Area = (AP+PB)(HO+OP) - .5(AE)(FA) - .5*π(F0)²
Area = (8+10)(10+6) - .5(16)(12) - .5*π(10)²
Area = 192 - 50π
That was a cool review of transversals.
I used triangle mid segment theorem to find the 6 and 8. It's a nice heuristic approach. I didn't have any paper or a pen. With it half of 12 is 6, and half of 16 is 8. Both added to the length of the radius, gives 16x18, subtract half of 12x16, subtract half of 314.15= 34.925.
Fun project. Thank you!
I think you don't need to solve EB and DF because OH = ON = R = 10 so that you can calculate HP = AD and MN = AB
It is very simple solution,
and not confusing new leaning person's. THANKS
In triangle AEF, line (OP) is parralel to side [AF] and passes through side [EF] middle :
So P is the middle of [AE] .
Hence AP=16/2=8 and OP=12/2=6.
==> AB=8+r=18 and AD=PH=6+r=16.
I wondered why you were calculating EB and DF when you had already discussed 3:4:5 triangles, so OP must be 3x2 and OM must be 4x2 so no new learning required.
Love the problems and love the solutions. Trying to solve without first checking. Really clear explanation. 👍👍
Why did you calculate lenght of AD 2 times? When you know lenght of AB you don't need to calculate this AGAIN.
Triangle Area = 12×16/2= 96
Hypothenuse of the rectangle is
Sqrt(12^2+16^2)=20.
So the half circle area has a radius of 10, therefore the area
10^2× pi/2 = 50 pi.
To get the rectangle area:
The center of the circle is 6 above the bottom and 8 to the right of the left of the triangle, because it is also the middle of the Hypothenuse
Adding the circle radius of 10 to each, we get 16×18 = 288 as rectangle area.
Now green area = rectangle area - triangle area - half circle area
= 288 - 96 - 50 pi
= 192 - 50 pi
I knew how to calculate the area of the initial triangle and the area of the semi circle, but was dead stuck on how to find the full sides of the rectangle! Thanks for putting the theorems and laws into the video!
I drew pretty much the same diagram. That triangle EFA is 3-4-5 almost goes without saying, so the big hypoteneuse is 20, and the radius of the circle is 10.
I had a feeling that lines HP and MN would bisect lines AF and AB, respectively. If they do, then
EB = PB - PE = 10 - (16/2) = 10 - 8 = 2; and FD = DM - FM = 10 - (12/2) = 10 - 6 = 4.
Then it occurred to me that triangles OFM and EOP are not just similar, but congruent, by the angle-side-angle rule. That means FM = OP = MA, and PE = OM = AP.
So HP and MN do indeed bisect AF and AB; and the sides of the rectangle AD and AB are indeed 16 and 18.
So the area of the shaded portion is the area of the rectangle, minus the area of the triangle EFA, minus the area of the semicircle; or
(16 * 18) - ((1/2) * 12 * 16) - ((1/2) * pi * 10^2);
collecting terms, simplifying, and using a little rearrangement trick to avoid using a calculator by converting to numbers in the 12 x 12 multiplication table so you can do it in your head, we have:
area = (18 * 16) - (6 * 16) - (50 * pi)
= ((18 - 6) * 16) - 50 * pi
= (12 * 16) - 50 * pi
= (12 * (12 + 4)) - 50 * pi
= (12 * 12) + (12 * 4) - 50 * pi
= 144 + 48 - 50 * pi
= 192 - 50 pi. voila!
Thank you, ladies and gentlemen; I'm here all week.
A= 34.92 sq unit ( pi= 3.1416)
As for calculating the height and base of the rectangle ABCD at the step 4, we notice that the center O is the midpoint of FE, OM is perpendicular to AD so OM//AB so M is the midpoint of AD, therefore AM=6.
Similarly, P is the midpoint of AE so AP=8
Thus the height PH=6+10=16, the base AB=8+10=18
I found it much simpler to complete the circle and the square around it. It is trivial to find the difference in area between the square and the circle. You can then cut that difference in half based on extending your line EF to bisect the figure. The bisected square includes a triangle and a trapezium not in the original figure, so they must be subtracted out. With closer inspection, the trapezium and triangle complete a simple 2x4 rectangle. I still use the arithmetic to calculate r, but the main calculating steps are contained as (20^2-10^2pi)/2 -8.
Area Rectangle = 288
Area Triangle = 96
Area Semi-Circle = 157.1
Area Green = 34.9
I would have stuck to some very basic tools and worked out the position of the center of the circle and added the radius to get the lengths of the rectangle. It would have worked but would have been annoying. Sometimes I need someone like you to remind me that there's elegant and easier techniques
Themi el jadheja explain miraculously.
Thank u.
People ask me: Why do you watch this kind of videos? you could be watching something fun, then I come back with an insulting response: The mere fact that you are asking that question, tells me you wouldn't understand the answer. PreMath, thank you for your videos.
Thank you for this lucid explanation.very very interesting class. Pranam guruji.
Клёвая задача, и решается в уме. Cool problem, and it is solved in mind. 18*16, the last step is to multiplicate these two numbers:)
Loved this one. I was able to do it in my head. Very rare.
How do you know that MN || AB? I mean they definitely look parallel. But can you prove it?
Infact, how can we prove ABCD is a rectangle?
Why did you look for the EB and DF segments? You literally went "18-16=2, great, now I can figure out that 16+2=18 so I can find the area"
You already had the 18 for the segment AB, it was in the form of the segment MN. You already had the 16 for the segment AD, it was in the form of the segment FH. You did not need to substract anything only to add it back in the next step.
why divide by 2 if it's a semi-circle? it doesn't look like half a circle, if it's not drawn to scale, how would I know either way?
S=192-50π=2(96-25π)≈34,86
The the third side of the triangle = 20 ( sqrt (12sq + 16sq )). Drop a perpendicular from the centre to both 12 and 16. So E to edge = 10 - 8 = 2, So also on verticle F to edge = 10 - 6 = 4. So the length of rectangle = 16 + 2 = 18 & its height = 12 + 4 = 16.
Area of the rectangle = 18*16 = 288.
Area of half circle = 0.5*pi*10sq = 157.08
Area of triangle = 0.5*12*16 = 96.
So Green shaded area = 288 - 157.08 - 96 = 34.92
Simple and clear explanation for math. You made it simple for me. Thanks
Very well done - a Genius game of puzzles - very good thinking - keep it up !!!! ..... CONGRATULATIONS (kudos) ...
Can be solved more easily with basic trig. Its evident that ∆EOP is similair to EAF as they share 2 angles and one side length. Seeing as point O is the midpoint/center of the semicirlce, OE=FO=10 from dividing the hypotenuse length 20 by 2. From that point sin and cos^-1 gives u lengths of OP and PO. From which you can see that one side of the rectangle is OP (6) + radius of semi circle (10) and the other is PE (8) + radius (10). From there its the trivial question of subtracting the areas of a right triangle and a semicirlce from a rectangle, no variables present. Assuming that a calculator is allowed, 192-50π jumps out as the answer. An interesting question with a few possible solutions!!
Fantastic working. Very well explained.
I did it exactly the same way! Great! The videotorials seem to finally pay off!
Great to hear!
Thanks Philip for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
Too complex. It could be simplified.
△AEF is a right triangle, AF=12 and AE=16, so EF=20.
Curve EF is semicircle O is center of circle, so EF is diameter and OE=OF=10=radius. (This is important.)
ABCD is a rectangle, line ON has to parallel AB and it = radius = 10, so AB = radius +1/2 AE = 18, as ON and BC verticals.
line OH has to parallel AD and it = radius =10, so AD = radius+ 1/2 AF = 16, as OH and CD verticles.
Area of green = rectangle - △AEF -π * 10^2 = 16*18 - 1/2 * 12 * 16 -50π = 288 - 96 - 50π
=192 - 50π.
Thanks Robert for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard
@@PreMath thank you.
That's literally what I was thinking as well.
I calculated it the same way. Rather easy.
If you phycally check dimension first one is of 2 units and result is 35 units and assume left upper green 5 units compairing ist green portion and third upto 10 units then total would be 17
I'm very pleased I did this one all on my own, and such fun !!... Thanks again
Go me ... Go me 😃
Once u know the triangle and semi circle its 18*16-1/2*12*16-pi*100/2. Why was eb or fd needed?
Third side of triangle= dia of semicircle = 20 units
Radius of semicircle=10
Height of centre of semi circle from base of triang=6. Hence side of Rectangle= 6 + 10 =16
Similarly other side of rectangle = 18
Total area = 16X18=288
Area of triangle=96
A of semicircle = 10x10x(22/7)/4 =78+4/7
Green A=288 -96-78-4/7
=113+3/7 sq units
Me: I don't have time to watch a 15-minute video
Also me 15 minutes later:
"That's awesome"
Keep up the good work I love your videos 👍👍
Thanks dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
I like to watch this channel. When I see his question then I go straight trying to solve it myself without his explanation. After that I come back to check my result.
@@paulc3960 Great way to learn Paul.
You are awesome. Love and prayers from the USA! 😃
Really great example. 👍
Step 1. Find the length of FE by Pytagorean triples : 20
Radius of Semi cirles is 10. Area of the semi cirle is 50*phi
Step 2 : Find the Area of triangle FAE 16*12/2=24
Step 3 Draw the total circle and total square. You can easily find lengthe of DF is 4 because 2*DE + 12 = 20(Diameter)
Same way you can find length of EB = 2.
step 4. area of rectanglar ABCD is (12+4)*(16+2) = 288
Steo 5 area of (rectanglar ABCD - triangle FAE - semicircle) = 288-96-50*phi = 192-50*phi
No need to calculate EB and DF
Dear Naeem, you are correct! There are many different ways to solve this problem. You are awesome 👍 Take care dear and stay blessed😃
It is not a different solution. Calculating them is useless....
Then how can we solve it without finding the two sides, please tell
@@ritikadubey3531 think out of the box
@@ritikadubey3531 We don't need to know EB to find AB. AB = AP + PB. By rectangle OPBN we know PB = ON = radius = 10. From Rectangle APOM we know AP = MO = 8 therefore AP + PB = 18.
Similarly, we don't need to know FD to find AD. AD = AM + MD. From rectangle MOHD we know MD = OH = radius = 10 and from rectangle APOM we know AM = PO = 6. Therefore AM + MD = 16.
شكرا على المجهودات
مساحة المستطيل18x16ناقص المثلث16x12x1/2ناقص نصف الداءرة ذات الشعاع10
نجد المساحة الخضراء 192ناقص 50pi
Hello wait man it's simple question
Just use Pythagoras theorem in triangle AFE
U will get FE =20 which is diameter of semicircle and r=10
Area of semicircle=πr×r/2 =3.14×10×10/2=157
Now area of rectangle=l×b=16×12=192
Area of shaded portion = Area of rectangle- Area of semicircle =192-157=35 square unit
Волшебник!)
Very nice. Congratulations friend 🌹❤
Thank you so much my dear friend for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
In this demonstration you have not demonstrated that ON is a radius. You just said that it a parallel to AB. Did I miss something or is it just an observation
Yes you missed something, don't give up.
Good,👍 better,👍 best👌 solution
Thank you so much Govinda for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
DM=HO=radius=10
And, DM+FA-DA=FM → 10+12-DA=6 → DA=16
Similarly, PB+AE-AB=PE → 10+16-AB=8 → AB=18
Then we get the both sides of the rectangle. That is what we want!!
i know i watch too many of these videos when i can solve this in my head before watching haha. very good lessons as always
Wow! Thank you teacher¡ 🙂
Muy bien , Sobresaliente profe!
Buen trabajo.
Rectangle is 16x18, by Pythagoras/similar ∆s, - 50π - 96 , green = 192-50π, ~35
Once you have identified the values of MO and OP, then finished.
Area of rectangleABCD- area of triangleFAE - area of semi circleFE= area of green region
Nothing it is impossible! incredible. Thanks
Thank you so much Temyra for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
How do you know OE = OF ? It is not given. It is assumed. No?
Hi, why ABCD is a rectangle?
Really like such challenging problems... 👍👍
Radius 10.
Quad ABCD=18×16
Green =18×16-50π-½.12.16
=35
Thanks for sharing Dhrubajoti
Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard
Why did you do the subtraction.. imran weren't you supposed to be adding the area of the semi circle and the area of the triangle and them subtracting this area from the area of the rectangle
My calculation: The length of EF is the root of AF²+AE² = 20. O is halfway EF, so the radius=OE=10. AD=OE+AF/2=16, AB=OE+AE/2=18, the area ABCD=16*18, AEF=16*12/2,=16*6, the semicircle is 10²π/2=50π, so the area is (16*18)-(16*6)-50π=16*12-50π=192-50π. Done.
Why MN is parallel to AB?
MN and ON are both part of the same line. ON is a radius. This radius also touches that side of the rectangle at N. That means that point N is on a tangent line BC. Tangent lines are always at right angles to the corresponding radii. So that means that both MN and AB both perpendicular to AD and BC, which makes them parallel to each other.
This is simple and i have solved this ..
I calculated 34.92 sq units.
Also proved by mid point theorem 5:23
There wasn't need to find DF and EB
Just SUBSCRIBED to this channel
Im so happy i solved this
Way too many unnecessary steps. You had all the info required half way through the video to solve it. Once you had OP and OM, rectangle dimensions were easily found to calculate its area, then simply subtract area of semicircle and triangle, done.
Excellent
Why this much long way?
But in question it is not given that of and le are radii of circle
Once you had the lengths of OM & OP this was solvable. As you had the lengths of AB & CD to get the area of the whole rectangle. You had the area of the triangle from the given, and when you found the radius of the circle you had the area of the semicircle. There is no point in finding DF and EB
Me gustaron los pasos que se hicieron para llegar a lo pedido.
rectangle (288cm²) - 1/2 circle (157,0795cm²) - triangle (96cm²) = 34,92cm²
How can we write the unit as square cm ? No unit is used in question .
Thank for premath
Thank you so much Taha for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
It could be done without evaluating the length of DF and EB
Thank you!
Good idea
Managed to get all the way to "approximately 35" in my head in a few minutes.
No unit is used , why?
Think you took a long approach, once you had diameter of the circle, the area of triangle AEF, and length of MN and HP, you solve by calculating area of ABCD - 1/2 of circle- area of AEF.
Если провести ao? Получится что это медиана будет =10,отсюда можно найти высоты op и +OH
EB было не обязательно высчитывать: PB=PN=R(10); ΔOPE вдвое меньше ΔAFE=>AP=PE=16/2=8; Т. о. AB=AP+R=8+10=18.
Excuse me but why is the right angled triangle a 3-4-5 triangle? The teacher made that statement ("triangle AEF is a 3-4-5 triangle") without showing why.
Ahhh... I know why: Use Pythagoras' theorem to get the length of the hypotenuse (20), then use the common factor of 4 for each side of triangle: ratios is seen to be
3-4-5.
It was pretty easy. Just requiring to concentrate on a proper construction
Calculating the length of lines DF and BE was an unnecessary step. From the calculation of lines MN and HP you already have the length and height of the rectangle.... My calculation ended up with the green area being 34.92036 square units. My calculation was at 3 seconds in passing the vid, the comment was after seeing the whole vid. Good challenge, good explanation, good vid, thanks.