Can you find the area of the Purple triangle? | (Important Geometry skills explained) |

[@robertbourke7935]

Got it! Many thanks. A clever exercise.

[@PreMath]

Thanks❤️

[@ОльгаСоломашенко-ь6ы]

You can use the chord theorem. Extend the DO to the intersection with the circle and get the diameter. And the diameter is a chord. AD*DC=1*3. CD=3/√5. According to the Pythagorean theorem, we find CB=4/√5. S=0.5*(3/√5)*(4/√5)=1.2.

[@howardaltman7212]

Beautiful, simple solution. Bravo!

[@ОльгаСоломашенко-ь6ы]

@@howardaltman7212 Спасибо

[@Jack_Callcott_AU]

Neat!

[@PreMath]

Thanks❤️

[@m.h.6470]

indeed a much better and simpler solution than the one in the video!

[@zdrastvutye]

either calculate the intersection with a thales circle or calculate the coordinate product repeatedly: 10 print "premath-can you find the area of the purple triangle" 20 l1=sqr(5):r=2*l1/sqr(5):dim x(1,2),y(1,2):sw=r/(l1+r):w=sw 30 @zoom%=@zoom%*1.4:xd=0:yd=r/2:xb=r:yb=0:n=r*r+l1^2:goto 60 40 xc=r*cos(rad(w)):yc=r*sin(rad(w)):dgu1=(xd-xc)*(xb-xc)/n:dgu2=(yd-yc)*(yb-yc)/n 50 dg=dgu1+dgu2:return 60 gosub 40 70 dg1=dg:w1=w:w=w+sw:w2=w:gosub 40:if dg1*dg>0 then 70 80 w=(w1+w2)/2:gosub 40:if dg1*dg>0 then w1=w else w2=w 90 if abs(dg)>1E-10 then 80 100 print w:la=sqr((xd-xc)^2+(yd-yc)^2):lb=sqr((xc-xb)^2+(yc-yb)^2) 110 x(0,0)=0:y(0,0)=0:x(0,1)=r:y(0,1)=0:x(0,2)=0:y(0,2)=r/2 120 x(1,0)=0:y(1,0)=r/2:x(1,1)=r:y(1,1)=0:x(1,2)=xc:y(1,2)=yc 130 ages=la*lb/2:print "die flaeche=";ages:mass=8E2/r:goto 150 140 xbu=x*mass:ybu=y*mass:return 150 for a=0 to 1:gcol8+a:x=x(a,0):y=y(a,0):gosub 140:xba=xbu:yba=ybu:for b=1 to 3 160 ib=b:if ib=3 then ib=0 170 x=x(a,ib):y=y(a,ib):gosub 140:xbn=xbu:ybn=ybu:goto 190 180 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 190 gosub 180:next b:next a:gcol8:xba=0:yba=0:gosub 140:circle xba,yba,r*mass premath-can you find the area of the purple triangle 53.1301024 die flaeche=1.2 > run in bbc basic sdl and hit ctrl tab to copy from the results window

[@spafon7799]

Alternatively: find CD and CB in order to get the area of the triangle as 1/2 * CB*CD. As in the video solution, DO= r/2. Pythagorean on OBD gives (r/2)^2+r^2=5, thus OB=2 and OD=1. Also as in the given solution, extend CD to A, the left point of the circle. Now note that ABC and ADO are similar triangles, since angle AOD and ACB are right angles and OAD and CAB are the same acute angle. Let us call x= DC and y=CB. We have by similar triangles OA/DA=CA/BA. 2/sqrt(5)=(sqrt(5)+x)/4. This gives x=2/sqrt(5). Now you apply pythagorean to triangle BCD to get x^2+y^2=5. Thus 9/5+y^2=5=25/5. Or y^2=16/5 Thus y=4/sqrt(5). Area of the triangle is (1/2)*x*y= (1/2)*(3/sqrt(5))*(4/sqrt(5))= 6/5.

[@phungpham1725]

1/The radius of the circle= 2 ( it is easy) 2/Extend CD to the left to build the right triangle ABC and the diameter AB. Notice that the angle CDB = 2 the angle DAO. We have tan DAO= 1/2 so tan CDB= 1/ (1-1/4) = 4/3----> BC/DC= 4/3 ------> the triangle BCD is an 3-4-5 triple-----> BC=4/5 BD and DC=3/5 BD The area of the purple triangle = 1/2x 4/5xsqrt5 x 3/5x sqrt5= 1.2 sq units 3/ We can also use the Pythagorean theorem: Let CD and BC be a and b respectively. We have sq(sqrt5+a) + sqb = 16----> 5+sqa+ 2a sqrt5+sqb=16 ---> 5+5+2a sqrt5=16-----> a= 3/sqrt5 and b= 4/sqrt5-----> area= 1/2 x 3/sqrt5 x 4/sqrt5 = 1.2 sq units

[@PreMath]

Thanks❤️

[@jimlocke9320]

We note that ΔABC and ΔADO are similar. The ratio of sides for ΔADO is (short-long-hypotenuse) is 1:2:√5 or 1/√5:2/√5:1. Applying that ratio to ΔABC and knowing that its hypotenuse is length 4, the two sides are (4)(1/√5) and (4)(2/√5). The area of ΔABC = (1/2)bh = (1/2)(4)(1/√5)(4)(2/√5) = 16/5. The areas of ΔADO and ΔBDO are (1/2)bh = (1/2)(1)(2) = 1 each. Area ΔBCD = Area ΔABC - Area ΔADO - Area ΔBDO = 16/5 - 1 - 1 = 6/5, as PreMath also found.

[@tombufford136]

At a quick glance, started solving this in a similar way to the video forming a second Chord using Thales theorem. Then found working arduous without a drawing of the the semi circle. Thank you for the video.

[@saltydog584]

It would have been helpful if the fact that it is a quarter circle at the beginning by indicating angle EOB was a right angle on the original diagram at the beginning.

[@johnsavard7583]

The first useful thing I note is that the vertical radius of the quadrant is divided into two equal parts by point D. So the white triangle below the purple triangle has sides r/2, r, and sqrt(5). That tells me r=2, and the area of that triangle is 1. Now if we extend the quadrant on the left to become a semicircle, line CD, when extended, will (because of the right angle) intercept the point opposite B on the diameter at the bottom; let's call that point X. Angle DBO and angle DXO are equal, and so the triangle BXC is similar to triangle DBO. Triangle DXO and triangle DBO have area 1 each. Triangle BXC has a hypoteneuse of 4 instead of sqrt(5), so its area is 1 times 16/5 because it is scaled up by a factor of 4 over sqrt(5). So the purple triangle has area 6/5, 16/5 minus 2 (which is 10/5).

[@linzhaoxu]

CD=a,BC=b,so a=root(5-bsqr); (a+root5)sqr+b sqr=4sqr; {(root(5-b sqr)+root5}sqr+b sqr=16; so b=4/root5; so a=3/root5, so the sqr about purple zone is 6/5

[@zsoltszigeti758]

The (semi)circle is x^2+y^2=4, AC line is y=x/2+1. The solution of these are x=-2, y=0 (A point); x=6/5, y=8/5 (C point).

[@DB-lg5sq]

شكرا لكم DC=a AD=جذر5 BC=(5-a^2)جذر AB=4 ..... a=3/(5جذر) S=1/2 CB CA =6/5

[@jakkima1067]

AD*DC=(R+1)*1=Sqrt(5)*DC=3. DC=3/Sqrt(5). CB^2=DB^2-DC^2=5-9/5=16/5. CB =4/Sqrt(5). Area=DC*CB=3/Sqrt(5)*4/Sqrt(5)/2=12/10=1,2.

[@quigonkenny]

Another, potentially easier way to measure the area of the larger triangle: We can tell by complementary angles that ∆ABC and ∆ADO are similar. By Pythagorean Theorem: AD² = OD² + OA² AD² = 1² + 2² = 5 AD =√5 CB/OD = AB/AD CB/1 = 4/√5 CB = 4/√5 AC = 2CB = 8/√5 A = ½bh = ½(8/√5)(4/√5) = 16/5 Aₚ= ∆ABC - ∆ABD = 16/5 - 2 = 6/5

[@misterenter-iz7rz]

I see, I have think out of box, that is extending the quarter circle to semicircle, so there is a right angled triangle inscribed in a semicircle, let x be DC, 2/sqrt(5)=(sqrt(5)+x)/4, 8=5+sqrt(5)x, x=3/sqrt(5)=3/5 sqrt(5), and BC=4× 1/sqrt(5)=4/5 sqrt(5), thus the triangle is 3/5 sqrt(5)× 4/5 sqrt(5), and the area is 5× 3/5×4/5×1/2=6/5.😊

[@ProfessorDBehrman]

Nice problem. Thanks.

[@awandrew11]

BC=2, DC=1, therefore DB=Square root 5, Area opf triangleBCD=CBxCD/2=2x1/2=1!?

[@LuisdeBritoCamacho]

1) Finding the Radius of the Quarter of a Circle: x^2 + 2x^2 = 5 5x^2 = 5 5x^2 - 5 = 0 5*(x^2 - 1) = 0 x^2 - 1 = 0 x = 1 Radius = 2 lu (linear units) 2) Finding the Slope of the Straight Line passing the points C and D and points B and C: First Slope: m = (1 - 0) / (0 + 2) = 1/2 m * m' = -1 So, the 2nd Slope is m' = - 2 3) Equations of the Straight Lines a) y = x/2 + 1 b) y = - 2x + 4 4) Point of Intersection (Point C); coordinates: x = 1,2 y = 1,6 4) Finding the Distances between Point C and D and Point B and C: CD ~ 1,342 BC ~ 1,789 5) Finding the Purple Area: PA = CD * BC / 2 PA = 1,342 * 1,789 / 2 PA = 2,400 / 2 PA ~ 1,2 su Final Answer: Purple Area equal 1,2 su

[@alokranjan4149]

Very Beautifully solved by using co-ordinate geometry. So nice 👍

[@bigm383]

Thanks Professor for a very nice problem.

[@PreMath]

Thanks❤️🌹

[@uwelinzbauer3973]

This one I also was able to find out. Again I used a way different from the video. This question was a bit challenging to me. But that's what we need to improve our skills. Thanks for the interesting video! Greetings 🙏

[@giuseppemalaguti435]

CB=b...intanto calcolo r...(r/2)^2+r^2=5...r=2...poi imposto l'equazione risolutiva,..b=√5cos(arccos(b/2r)-arctg(1/2))..quindi ,calcolo l'altro cateto CD...e quindi l'area..... l'equazione diventa..b/2+√(1-b^2/16=b...b=4/√5...CD=√(5-16/5)=3/√5...A=(4/√5*3/√5)/2=6/5.... alleluia

[@hermannschachner977]

bravo seppe, so denke auch ich (Österreich): 2,5 x sin 36,87.... x cos 36,87.... = 1.2

[@PreMath]

Thanks❤️

[@marcgriselhubert3915]

The raduis of the circle is 2 (easy, as evrerybody says). In an adapted orthonormal we have O(0;0), B(2;0) E(0;2) D(0;1) and C(2 cos(x); 2 sin(x)) where x is unknown between 0 and 90° Then VectorDC (2 cos(x); 2 sin(x) -1) and Vector BC (2 cos(x) -2; 2 sin(x)) . These vectors are orthogonal, so we heve: (2 cos(x)). (2 cos(x) -2) + (2 sin(x) -1). (2 sin(x)) = 0. We develop, use the fact that cos(x)^2 + sin(x)^2 = 1, and simplify. We obtain: 2 cos(x) + sin(x) = 1. This is a well known trigonometric equation. Let's divide by sqrt(5) and consider x0 between 0° and 90° as cos(x0) = 2/sqrt(5) and sin(x0) = 1/sqrt(5). We get cos(x0).cos(x) + sin(x0).sin(x) = cos(x0), or cos (x-x0) = cos(x0) Then x - x0 = x0 (mod 360°) or x -x0 = -x0 (mod 360°), giving that x = 2x0 is the only solution between 0° and 90° So, cos(x) = cos (2. x0) = 2 (cos(x0))^2 - 1 = 2. (4/5) - 1 = 3/5, and sin(x) = sin(2.x0) = 2. sin(x0). cos x0) = 2.(1/sqrt(5)). (2/sqrt(5)) = 4/5. Now we have point C (6/5; 8/5) and then Vector DC (6/5; 3/5), giving DC = sqrt ((36/25) + (9/25)) = sqrt(45)/5 = 3.sqrt(5)/5 and also Vector BC (-4/5; 8/5), giving BC _ sqrt ((16/25) + (64/25)) = sqrt (80)/5 = 4.sqrt(5)/5 The area of the triangle is (1/2). DC. BC = 6/5 when simplified.

[@PreMath]

Thanks❤️

[@ANNUANNU-b7f]

In which grade these questions. Come

[@deepaagarwal8743]

Sir can you plz upload some trigno and geometry high level questions

[@nunoalexandre6408]

Love it!!!!!!!!!!!!

[@PreMath]

Thanks❤️

[@afshinfarzaadi1371]

👍

[@misterenter-iz7rz]

radius r is easy to determine, r^2+r^2/4=5/4 r^2=5, r^2=4, r=2, but it is difficult to determine the area of the right angled triangle inscribed in the upper part of sector.😅

[@PreMath]

Thanks❤️

[@HeywoodUmanoff]

How do we know that C, D and A are colinear?

[@hemalathar8842]

Is it possible for them to ask this in 10th board exam?

[@frooooo7896]

Why is the area not just 0.5 x 1 x 2. Since the sides of the pink triangle must be 2 and 1.?

[@HeywoodUmanoff]

As it turns out, the sides of the pink triangle don't have to be 2 and 1. In fact, the legs turn out to be 1.74 and 1.40, with the angle between the shorter leg and the hypotenuse = 51.14 degrees. This is different from the 63.43 degree angle between the shorter leg and hypotenuse of the 1, sqrt5 and 2 side lengths of right triangle DOB. I was thrown off by this issue initially as well.

[@xsilata]

Triangles AOD and ABC are similar. The calculation is easy.

[@PreMath]

Thanks

[@vimsriani]

❤

[@PreMath]

Thanks❤️

[@RazvanMihaeanu]

Thank God for the right angle 'cause otherwise... we would have been screwed!

[@pralhadraochavan5179]

Good evening sir

[@PreMath]

Hello dear ❤️

[@devondevon4366]

1

[@Weizsaecker]

0:23 Stop! How does he know that DE = DA? Is there a theorem describing a pythagorean triangle in a quarter circle?

[@wackojacko3962]

If Schrodinger's Cat could have thought outside the box while inside the box, could the cat find the square root of a tree? Just curious...🙂

[@billycox475]

Well done, dad jokes always welcome!

[@PreMath]

Thanks❤️

[@edsznyter1437]

You're really overcomplicating things. Triangles ADO and ACB are similar. The ratio of the hypotenuses is 4:Sqrt[5]. Thus, the ratio of the areas is the square of that, 16:5. So [ACB]=16/5. [DCB]=[ACB]-2[ADO]=16/5-10/5=6/5.

[@Dinhnguyen-km6zd]

S(ABD)=2=1/2 BD x BC. => BC=4/√ 5. Therom pytago=> CD=√ (9/5). End S(BCD) =√ (144/25) = 1,2

[@edsznyter1437]

You're really overcomplicating things. Triangles ADO and ACB are similar. The ratio of the hypotenuses is 4:Sqrt[5]. Thus, the ratio of the areas is the square of that, 16:5. So [ACB]=16/5. [DCB]=[ACB]-2[ADO]=16/5-10/5=6/5.