Another neat method is to recognise the shaded figure is a cyclic quadrilateral. Then use sum of the products of opposite sides equals the product of the diagonals
AD=√24²+7²)=25 → Según eje OB trazamos la figura simétrica de la propuesta → AA´=2r→ CD+DA´=7+25=32→ El triángulo ACA´es tipo 3/4/5→ AA´=8*5=40→ r=20→ El triángulo AOD es 3/4/5→ OD=5*3=15→ Área AODC =(20*15+24*7)/2 =234 Gracias y saludos.
20:50 One easier method: AD is joined. Also OE is joined where E is the mod pt of AC.OE meets AD at F. Now, E is the mid pt of AC and O is the centre of the circle with chord AC. Therefore OE is perpendicular to AC like DC. Since E is the mid pt of AC, F is the mid pt of AD and EF is half CD. Also AODCis a cyclic quadrilateral because it's two opposite angles are right angles. Hence AD is a diameter of this circle. So F is its centre. So FA equals FO equals half AD. Obviously, AD equals 25, OF equals 12.5, FE equals half DC equals 3.5. So OE equals 16. Now the area of the quadrilateral AODC is the sum of the area of the trapezium OECD and that of the triangle AOF which can be calculated out to be 138 plus 96 i.e 234.
In method 2, there is no need to find R. Area ADM is simply 25*24/2=300, and area of OAD is just half of that which is 150. Final area = 150+24*7/2 = 234.
@@rick57hart I'm not sure how @spacer999 arrived at that formula, but from diagram, we can tell that Area(ADM) = Area(ACM) - Area(ACD) = 1/2(24*32) - 1/2(24*7) = 1/2(24*25) = 300
In method2 Let Area of triangle OAD=S= Area of triangle ODM then Area of triangle ACM= 2S+84 Area of triangle ACM=1/2×32×24=384 therefore S=150 Then ?=S+84=150+84=234.
Проводится АД . Вокруг четырехугольника , у которого сумма противоположных углов = 180* можно описать окружность . АД является диаметром окружности ., с точки О проводится перпендикуляр на хорду АС , который и делит хорду АС пополам , отрезок перпендикуляра на хорду АС является срединой линией треугольника АСД и равен 1/2 ДС=7/2=3,5 . После вычислений по теореме Пифагора - АС=24 . СД=7 , ДО = 15 , АО= 20 . Площадь четырехугольника , вокруг которого можно описать окружность , вычисляется по формуле Брахмагупты S=\|(р-а)(р-в)(р-с)(р-d) , р - полупериметр = (24+7+15+20)/2=33 , S=\|(33-24)(33-7)(33-15)(33-20) = \|9х26х18х13 = \|9х2х13х9х2х13=9х2х13=234 .
Draw AD. Triangle ∆DCA: DC² + CA² = AD² 7² + 24² = AD² AD² = 49 + 576 = 625 AD = √625 = 25 Mirror arc AB, radius OA, and line segment AD about OB to create a semicircle with diameter AE. As ∠C = 90°, CD and the mirrored AD will form CE, so CE = CD+AD = 7+25 = 32. Triangle ∆ECA: CA² + EC² = AE² 24² + 32² = AE² AE² = 576 + 1024 = 1600 AE = √1600 = 40 Triangle ∆DOE: OD² + OE² = DE² OD² + 20² = 25² OD² = 625 - 400 = 225 OD = √225 = 15 The orange area is the area of ∆ECA minus the area of ∆DOE. Orange kite(?) ODCA: A = BH/2 - bh/2 A = 32(24)/2 - 20(15)/2 A = 16(12) - 10(15) A = 384 - 150 = 234 sq units
2nd Method Can Be More Easy, Because ΔADM is A isosceles triangle. So ⊿AOD's Area Is Half Of ΔADM's Area. And ΔADM's Area Is ⊿ACM's Area Minus ΔACD's Area.So When We Use Pythagorean theorem Found AD's length And Made semicircle O And ⊿ACM, We Can calculated quadrilateral ACDO's Area Easily.
Another neat method is to recognise the shaded figure is a cyclic quadrilateral. Then use sum of the products of opposite sides equals the product of the diagonals
You are indeed a prof. Hope to see more videos.
Super job
You are very intelligent prof
Hope to see mire Olympiad problems
AD=√24²+7²)=25 → Según eje OB trazamos la figura simétrica de la propuesta → AA´=2r→ CD+DA´=7+25=32→ El triángulo ACA´es tipo 3/4/5→ AA´=8*5=40→ r=20→ El triángulo AOD es 3/4/5→ OD=5*3=15→ Área AODC =(20*15+24*7)/2 =234
Gracias y saludos.
20:50
One easier method:
AD is joined. Also OE is joined where E is the mod pt of AC.OE meets AD at F.
Now, E is the mid pt of AC and O is the centre of the circle with chord AC. Therefore OE is perpendicular to AC like DC. Since E is the mid pt of AC, F is the mid pt of AD and EF is half CD.
Also AODCis a cyclic quadrilateral because it's two opposite angles are right angles. Hence AD is a diameter of this circle. So F is its centre. So FA equals FO equals half AD.
Obviously, AD equals 25, OF equals 12.5, FE equals half DC equals 3.5.
So OE equals 16.
Now the area of the quadrilateral AODC is the sum of the area of the trapezium OECD and that of the triangle AOF which can be calculated out to be 138 plus 96 i.e 234.
In method 2, there is no need to find R. Area ADM is simply 25*24/2=300, and area of OAD is just half of that which is 150. Final area = 150+24*7/2 = 234.
Nice dude
I don't get it. How do you calculate the area ADM?
@@rick57hart I'm not sure how @spacer999 arrived at that formula, but from diagram, we can tell that
Area(ADM) = Area(ACM) - Area(ACD) = 1/2(24*32) - 1/2(24*7) = 1/2(24*25) = 300
Just consider AC as a height and DM as a base of the triangle ADM you will get the area of this equal to 1/2*25*24=300
This is beautiful! Thahk you!
In method2
Let Area of triangle OAD=S= Area of triangle ODM
then Area of triangle ACM= 2S+84
Area of triangle ACM=1/2×32×24=384
therefore S=150
Then ?=S+84=150+84=234.
How do you get 144 + 156 = 400?
How did you get( x- 9)(x+16)
Проводится АД . Вокруг четырехугольника , у которого сумма противоположных углов = 180* можно описать окружность . АД является диаметром окружности ., с точки О проводится перпендикуляр на хорду АС , который и делит хорду АС пополам , отрезок перпендикуляра на хорду АС является срединой линией треугольника АСД и равен 1/2 ДС=7/2=3,5 . После вычислений по теореме Пифагора - АС=24 . СД=7 , ДО = 15 , АО= 20 . Площадь четырехугольника , вокруг которого можно описать окружность , вычисляется по формуле Брахмагупты S=\|(р-а)(р-в)(р-с)(р-d) , р - полупериметр = (24+7+15+20)/2=33 ,
S=\|(33-24)(33-7)(33-15)(33-20) = \|9х26х18х13 = \|9х2х13х9х2х13=9х2х13=234 .
but why not 240.25???
Excellent
Draw AD.
Triangle ∆DCA:
DC² + CA² = AD²
7² + 24² = AD²
AD² = 49 + 576 = 625
AD = √625 = 25
Mirror arc AB, radius OA, and line segment AD about OB to create a semicircle with diameter AE. As ∠C = 90°, CD and the mirrored AD will form CE, so CE = CD+AD = 7+25 = 32.
Triangle ∆ECA:
CA² + EC² = AE²
24² + 32² = AE²
AE² = 576 + 1024 = 1600
AE = √1600 = 40
Triangle ∆DOE:
OD² + OE² = DE²
OD² + 20² = 25²
OD² = 625 - 400 = 225
OD = √225 = 15
The orange area is the area of ∆ECA minus the area of ∆DOE.
Orange kite(?) ODCA:
A = BH/2 - bh/2
A = 32(24)/2 - 20(15)/2
A = 16(12) - 10(15)
A = 384 - 150 = 234 sq units
2nd Method Can Be More Easy, Because ΔADM is A isosceles triangle. So ⊿AOD's Area Is Half Of ΔADM's Area. And ΔADM's Area Is ⊿ACM's Area Minus ΔACD's Area.So
When We Use Pythagorean theorem Found AD's length And Made semicircle O And ⊿ACM, We Can calculated quadrilateral ACDO's Area Easily.
Just to write 256 in place of 156 for 16 square.
Entendido pela matemática; não pelo idioma.
👍
Like ❤🎉
234
234
234