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.
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
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.
Great suggestion! Keep watching. You are very welcome! Thanks for your feedback! Cheers! You are awesome, Dyala. Keep it up 👍 Love and prayers from the USA! 😀
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.
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
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
Height with respect to a given base is, by definition, the perpendicular distance between the base and the apex. That said, we don't really care about heights. If two triangles have the same apex and collinear bases then the ratio of their areas is the same as the ratio of their base lengths.
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😅
Amazing solution and explanation )) Many thanks 🤩
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تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين .
Wow... Brilliant example... Awesome , baffled me at first.went right over my head 😃 Thanks 👍🏻
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
thankyou so much dear professor 🙏
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Thanks for video.Good luck sir!!!!!!!!!!
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.
Does this make use of the theorem presented in the video?
@@User-jr7vf yes, but it's easier.
Great explanation👍👍
Thanks for sharing😊😊
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
Very good solution
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.
Very nice
U r the best❤
Thanks from Russia.
Thanks😍
Can you give us lessons to prepare to iranian geometry olympiad
Great suggestion! Keep watching.
You are very welcome!
Thanks for your feedback! Cheers!
You are awesome, Dyala. Keep it up 👍
Love and prayers from the USA! 😀
@@PreMath thanks❤❤
nice job
Good night sir I like it I proud of your talent
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.
Do those lines bisect the angles they're drawn from? They look like they do.
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
Thanks from🇦🇿
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@@PreMath How can I find the math questions in the university entrance exams in America?
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
Tricky but got the concept
We can use the given Triangles:
Green-Yellow = 77 cm-square
Green-Pink = 66 cm-square.
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
that's the mayajal of maths
Legend has it he's still creating more equations to compare
😅
You assumed the "heights" intersect the sides at 90 degrees.
Height with respect to a given base is, by definition, the perpendicular distance between the base and the apex. That said, we don't really care about heights. If two triangles have the same apex and collinear bases then the ratio of their areas is the same as the ratio of their base lengths.
好复杂啊！好复杂啊！！come from china,,.哈哈！！
I would have guessed 55.
Supersonic
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You are awesome, Erick. Keep it up 👍
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First
Excellent!
Thank you! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
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😅
好复杂啊！好复杂啊！！come from china,,.哈哈！！