Calculate area of the Blue shaded Quadrilateral | Important Geometry skills explained

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Calculate area of the Blue shaded Quadrilateral | Important Geometry skills explained

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Amazing solution and explanation )) Many thanks 🤩

betaa

Ladder Theorem
1/S +1/44 = 1/(22+44) + 1/(33+44)
De donde S = 924/5
Luego X = 924/5 - (22 + 33 + 44) = 85, 8

joserubenalcarazmorinigo

Another solution with direct calculation without a system of equations. Join E with D. In the triangle CAD we have FD*2=FC, because the area of AFD=22 and the area of AFC = 44. It follows that in the triangle CED we have: Area of FED*2=33 (area of CEF). Now he had the area of FED=33/2. Now we have x1/y1=Area ACD/Area DCB=Area AED/Area DEB. I note the area of DEB, with z. So we have 66/(33+33/2+z)=(22+33/2)/z The result is z=693/10. So the area of FEBD=693/10+33/2=85.8
Q.E.D.

petrusneacsu

تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين .

اممدنحمظ

Wow... Brilliant example... Awesome, baffled me at first.went right over my head 😃 Thanks 👍🏻

theoyanto

You can do it by the theorem of Menelaus. Take triangle CEF and line ADB. Then

(AE/AF)(DF/DC)(BC/BE) = 1

(7/4)(1/3)(BC/BE) = 1
(BE+EC)/BE = 12/7
EC/BE = 5/7
BE = (7/5) EC

BDFE = (7/5)(77) - 22
= (539 - 110)/5
= 429/5

pwmiles

Thanks😍
Can you give us lessons to prepare to iranian geometry olympiad

dyalaaleid

Engr. Chan Pang Chin from San Min Secondary School, Telok Intan, Malaysia aka "the Wizard" has the shortest solution. Join D to E. Let area of triangle BDE = Y .
(44+22)/ (33+16.5+Y)=(22+16.5)/Y
Y=69.3 Required area = 16.5+69.3=85.8
Thanks to Mr Oon Kee Soon, Mr Loke Siu Kow, (San Min), Mr. Ganesan, (HMSS), Mr Lim Chee Lin (NJC, Singapore), my sifus.乾杯😅

chanpangchin

Area of triangle ACD/area of triangle AFD=66/22=3
Hence area of BCD/area of BFD=3
Let area of BFD= b and area BFC=2b and area of BFE=2b—33
Area of ACE/area ofFCE=77/33=7/3
Area ABE/Area of BFE=7/3

Hence (2b—33)4/3= 22+b
b=198/5
Area of BFE=2b—33= 231/5
Area of blue part = 198/5+231/5=85.8

spiderjump

i) a : 33 = (a + b + 22) : (33 + 44)
ii) b : 22 = (a + b + 33) : (22 + 44)

i) a * 77 = 33 * (a + b + 22)
77a = 33a + 33b + 33*22
77a = 33a + 33b + 726
44a - 33b = 726

ii) b * 66 = 22 * (a + b + 33)
66b = 22a + 22b + 22*33
66b = 22a + 22b + 726
44b - 22a = 726

i) 44a - 33b = 726
- 22a + 44b = 726 | * 2
ii) - 44a + 88b = 1452

i) + ii)
55b = 2178
b = 39.6

i) 44a - 33 * 39.6 = 726
44a - 1306.8 = 726
44a = 2032.8
a = 46.2

A = a + b = 46.2 + 39.6 = 85.8 square units

Waldlaeufer

Too few information to determine the area of the region? 77/22+A=r, 66/33+A=s, two equations with three unknowns, how to calculate the value of A?🤔

misterenter-izrz

Great explanation👍👍
Thanks for sharing😊😊

HappyFamilyOnline

My approach was to realise that the distances of F, D and E from AC are in the ratio 4 : 6 : 7, based on the triangle area formula with AC as base. From this and with the aid of a custom coordinate system, I worked out how far away B is, giving me the area of the whole triangle ABC and hence the area of the blue quadrilateral.

flashg

Join left-right diagonal of desired quadrilateral to divide the region in two triangles of area X and Y respectively
Hereby
(33 /x) = ( 44 +33)/( 22 + x + y)
4 x / 3 = 22 + y
And
(22 /y) = ( 44 +22)/( 33 + x + y)
i.e. 2 y = 33 + x
44 + 33 + x = 4 x / 3
x = 77 * 3 = 231
Hereby y = (33 + 3 * 77) / 2 = 33 * 6
Hereby x + y = 7 * 33 + 6 * 33 = 429

honestadministrator

Good night sir I like it I proud of your talent

pralhadraochavan

You assumed the "heights" intersect the sides at 90 degrees.

fokokoster

Legend has it he's still creating more equations to compare

murphygreen

Do those lines bisect the angles they're drawn from? They look like they do.

KipIngram

Thanks for that tour de force! Perhaps another method would be:
Consider AC = y = base ; altitude (CFA) = 4/7 alt.(CEA) =2/3 alt.(CDA). Let G and H lie on AC so that line FG is // to CE, and line FH is// to AD. Then by sim. triangles : AG = (4/7) y, and CH = (2/3)y, so
GH = [4/7 + 2/3 - 1]y = (5/21) y.
But triangles FGH and BCA are similar so
area(BCA) = (21/5)44 = 184.8, (the bases of BCA and FCA being the same).
So req'd area = 184.8 - (44 + 33 +22) = 85.8 sq. units.

timc

We can use the given Triangles:
Green-Yellow = 77 cm-square
Green-Pink = 66 cm-square.

shadmanhasan

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