Find the missing area | A Very Nice Geometry Problem

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال BC=l DC=L , FD=t X=lL - 8 EF^2/L^2 =1/4 FE=L/2 X+1=1/2 l (t+L/2) X+4 =1/2 l (t+L) X+4 -(X+1) = 1/4 lL 3 = 1/4 lL 12 = lL X= lL-8 X=4

[@Ramkabharosa]

Let MN be the vertical line through O, parallel to AD, that meets AB in M & CD in N. Now ΔCOD and ΔEOF are similar (since AB is parallel to CD). So |ON| = 2.|ON|, |CD| = 2.|EF| & (½).|OM|.|EF| = 1. ∴ Area (ABCD) = |AB|.|CD| = (|ON|+|OM|).(2.|EF|) = (3.|OM|).(2.|EF|) = 12.(½).(|OM|.|EF|) = 12. Thus Area(BCOF) = 12 - 4 - 3 - 1 = 4.

[@santiagoarosam430]

s²=1/4---> s=1/2---> Si bh/2=1---> 2b*2h/2=4bh/2=4---> (2b*3h/2)*2 =(6*bh/2)*2 = 12 = Área del rectángulo ---> BCOF =12-1-4-3 =4. Gracias y saludos

[@denisrenaldo3506]

To demonstrate that triangles OEF and OCD, you don’t need to use angles, you should simply use the fact that AB and CD are parallel (Thales Theorem).

[@WahranRai]

May be he does not know the theorem of Thales as important as that of Pythagoras. He persists in not using it : elementary geometry without use fundamental theorems is incomplete I already made the remark to him about Thales' theorem.

[@santiagogonzalez9797]

A simple solution: DOC height its double of EOF height (because DOC area its 4 times EOF). So the height of ABCD square its 3/2 DOC height. And the base its the base of DOC. So, the area of ABCD its 12. Then the area of FOCB its just 12-(4+3+1)=4.

[@hongningsuen1348]

[EOF] = 2 is a mistake. Method using area-side ratio property of equal height triangles and of trapezium divided into 4 triangles by its diagonals: 1. Join ED and FC to form trapezium EFCD as AB//CD which are sides of rectangle ABCD. Area-side ratio property for the 4 regions of trapezium is top area: left area: right area:bottom area = top side^2:top side x bottom side:top side x bottom side:bottom side^2. As top area: bottom area = 1:4, sides EF:DC ratio is 1:2 and area of the left and right regions of trapezium = 1x2 = 2. 2. Area of triangle DFC = bottom area + right area = 4 + 2 = 6 3. Area of triangle AFD = 3 + 1 = 4 (given) 4. Triangles DFC and AFD are equal height triangles with area ratio = base side ratio. Hence side ratio AF:DC = 4:6 Hence AF = DC x 4/6 = 2 x 4/6 = 4/3 FB = AB - AF = DC - AF = 2 - 4/3 = 2/3 5. Triangles FBC and AFD are equal height triangles with base side ratio 2/3:4/3. Hence area ratio of FBC:AFB = 1:2. Hence area of triangle FBC = 4/2 = 2. 6. Area of BCOF = area of FBC + area of FOC (which is lateral area of trapezium = 2 from step 1) = 2 + 2 = 4.

[@Antony_V]

Since DC is twice EF, area of rectangle ABCD is 12 and area of FBCO = 12-1-3-4 = 4 squ

[@hongningsuen1348]

Please explain how to get area of rectangle ABCD = 12.

[@soli9mana-soli4953]

@@hongningsuen1348I suppose this result follows by the similarity between Triangles OEF and DOC

[@hongningsuen1348]

@@soli9mana-soli4953 Thanks for your help. I finally get the explanation after studying other comments. Triangles OEF and DOC are similar with side ratio of 1:2 from their area ratio of 1:4. It means height of rectangle:height of triangle DOC is 3:2. Let the width and height of rectangle be b and h. We have area of triangle DOC = (1/2) b(2/3)h = 4 as given. Hence bh = 12.

[@Antony_V]

@@hongningsuen1348 Sorry, I didn't see your comment. Here's my explanation: if CD is the base "b" of the rectangle, from similarity of triangles EOF and DOC we have EF = b/2. So the remain segments on AB, AE+FB = b/2 too. Let's draw triangles AOE and EOB: their areas add up to 1 (same height of EOF). Area of triangle AOB = 1+1 =2. This area added to COD area = 2+4 = 6. And that's half the area of the rectangle.

[@zdrastvutye]

there was a mistake, which was visible because one line was crooked which should have been straight instead. so "xs2" must fulfill 2 equations as written in line 40: 10 print "mathbooster-find the missing area, a very nice geometry problem" 20 dim x(5,2),y(5,2):a1=1:a2=3:a3=4:lb=3:sw=sqr(a1+a2+a3)/183:goto 120 30 xs3=2*(a1+a2)/lb:la=2*a3/ys2:xs2=ys2*xs3/lb 40 d1=la/lb:d2=la/ys2:d3=xs2/ys2:xs1=lb*(d1-d2+d3) 50 a4=lb*(la-xs1)/2:a4=a4-a1:xs2=ys2*xs3/lb 60 dg=(a1+a2+a3+a4-la*lb)/la/lb:return 70 ys2=sw:gosub 30 80 dg1=dg:ys21=ys2:ys2=ys2+sw:if ys2>20*la then return 90 ys22=ys2:gosub 30:if dg1*dg>0 then 80 100 ys2=(ys22+ys21)/2:gosub 30:if dg1*dg>0 then ys21=ys2 else ys22=ys2 110 if abs(dg)>1E-10 then 100 else return 120 gosub 70:print "a4=";a4:gosub 130:goto 260 130 x(0,0)=0:y(0,0)=0:x(0,1)=la:y(0,1)=0:x(0,2)=xs2:y(0,2)=ys2:x(1,0)=la:y(1,0)=0 140 x(1,1)=la:y(1,1)=lb:x(1,2)=xs2:y(1,2)=ys2:x(2,0)=xs2:y(2,0)=ys2:x(2,1)=la:y(2,1)=lb 150 x(2,2)=xs3:y(2,2)=lb:x(3,0)=xs2:y(3,0)=ys2:x(3,1)=xs3:y(3,1)=lb:x(3,2)=xs2:y(3,2)=lb 160 x(4,0)=0:y(4,0)=lb:x(4,1)=xs2:y(4,1)=ys2:x(4,2)=xs1:y(4,2)=lb:x(5,0)=0:y(5,0)=0:x(5,1)=xs2 170 y(5,1)=ys2:x(5,2)=0:y(5,2)=lb:masx=1200/la:masy=850/lb:if masx

[@marioalb9726]

Similarity of triangles: b₁/b₂ = h₁/h₂ =√(A₁/A₂) = √(4/1) = 2 b₁= 2.b₂ Area of rectangle: A = b₁.h = b₁(h₁+h₂) = b₁.h₁+b₁.h₂ A = b₁.h₁+2b₂.h₂ = 8+2*2 = 12 cm² Area of quadrilateral : A₄= A -A₁ -A₂ -A₃ = 12 -4 -1 -3 A₄= 4 cm² ( Solved √ )

[@ناصريناصر-س4ب]

The triangles OEF and OCD are similar and their similarity ratio is √1/4=1/2, from which OD=2OF, so the area of triangle ODE is 2 and the area of triangle OFC=2, from which the area of triangle AED is 1, which means that EF=3AE, from which FB=2AE, so the area of triangle BFC=2, so the area of quadrilateral OFBC=the area of triangle OFC+the area of triangle FBC=2+2=4

[@himo3485]

EOF∞COD 1 : 4 = 1² : 2² FO : OD = 1: 2 EOD = 2 EDC = 2 + 4 = 6 ABCD = 6 * 2 = 12 BCOF = 12 - (3 + 1 + 4) = 4

[@rooker56]

EOF is 1 or 2? Please make it clear.

[@MathBooster]

It is 1. Sorry for mistyping.

[@michaeldoerr5810]

The area is 12 units square. Also I have noticed regarding the theta and alpha criterion for the HL similarity, required the ratios of the two triangles-not drawn to scale-equaled the ratio of the triangle bases squared. I am wondering would that work without the rectangle??? I believe that this is similar to one of your videos that makes use of a similar criterion. I could be wrong.

[@oscarcastaneda5310]

Something seems to be missing in my logic... Help : ) Since EF = (1/2)L then AE = (1/4) L so AF = (3/4) L. 4 = (1/2)(3/4 L) W so LW = 32/3 = 10 2/3. [BCOF] = 10 2/3 - 8 = 2 2/3.

[@aksiiska9470]

i would guess the missing area=3 because it looks symmetric

[@oscarcastaneda5310]

Is [EOF] 2 or 1 ?

[@MathBooster]

It is 1, sorry for mistyping.

[@xz1891]

Mental calculation in 30 sec, 4

[@vasanthosangadi2896]

3🙏🇮🇳🚩

[@yakupbuyankara5903]

4