Amazing solution and explanation )) Many thanks 🤩

@PreMath

Жыл бұрын

Glad you think so! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

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

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

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

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.

@User-jr7vf

Жыл бұрын

Does this make use of the theorem presented in the video?

@petrusneacsu

Жыл бұрын

@@User-jr7vf yes, but it's easier.

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

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.

Thanks for video.Good luck sir!!!!!!!!!!

Great explanation👍👍 Thanks for sharing😊😊

thankyou so much dear professor 🙏

@PreMath

Жыл бұрын

You are very welcome! Thanks for your continued love and support! You are awesome, Rishu. Keep it up 👍 Love and prayers from the USA! 😀

Very good solution

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 Russia.

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😅

Very nice

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.

U r the best❤

nice job

Good night sir I like it I proud of your talent

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

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

@PreMath

Жыл бұрын

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! 😀

@dyalaaleid7838

Жыл бұрын

@@PreMath thanks❤❤

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

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

@theoyanto

Жыл бұрын

😅

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

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

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?🤔

Thanks from🇦🇿

@PreMath

Жыл бұрын

You are very welcome! Thank you! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@programmer229

Жыл бұрын

@@PreMath How can I find the math questions in the university entrance exams in America?

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

@flash24g

Жыл бұрын

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.

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I would have guessed 55.

Supersonic

@PreMath

Жыл бұрын

Thanks for your feedback! Cheers! You are awesome, Erick. Keep it up 👍 Love and prayers from the USA! 😀

First

@PreMath

Жыл бұрын

Excellent! Thank you! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

好复杂啊！好复杂啊！！come from china,,.哈哈！！