Find the missing area | A Very Nice Geometry Problem
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/ @mathbooster
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).
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.
[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.
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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
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.
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
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.
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
Since DC is twice EF, area of rectangle ABCD is 12 and area of FBCO = 12-1-3-4 = 4 squ
Please explain how to get area of rectangle ABCD = 12.
@@hongningsuen1348I suppose this result follows by the similarity between Triangles OEF and DOC
@@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.
@@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.
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
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
△EDO=2
△FCO＝2
△ADE=1
△EDF=3
AE:EF=1:3
EF:DC=1:2
AE:EF:FB=1:3:2
△FCB=2
EOF is 1 or 2? Please make it clear.
It is 1. Sorry for mistyping.
i would guess the missing area=3 because it looks symmetric
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 √ )
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.
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Is [EOF] 2 or 1 ?
It is 1, sorry for mistyping.
Mental calculation in 30 sec, 4
4
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.
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