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

Learn how to find the area of the Blue shaded Quadrilateral in a big triangle. The areas of other three triangles are 22, 33, and 44. Important Geometry skills are also explained. Step-by-step tutorial by PreMath.com
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Calculate area of the Blue shaded Quadrilateral | Important Geometry skills explained
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[@beta700a]

Amazing solution and explanation )) Many thanks 🤩

[@PreMath]

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

[@اممدنحمظ]

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

[@joserubenalcarazmorinigo9540]

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

[@pwmiles56]

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

[@petrusneacsu]

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]

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

[@petrusneacsu]

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

[@theoyanto]

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

[@rishudubey1533]

thankyou so much dear professor 🙏

[@PreMath]

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

[@HappyFamilyOnline]

Great explanation👍👍 Thanks for sharing😊😊

[@spiderjump]

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

[@KAvi_YA666]

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

[@timc5768]

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.

[@EnnioPiovesan]

Very good solution

[@rachidmeknassi3709]

Very nice

[@chanpangchin9744]

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😅

[@tusharmahesh7396]

U r the best❤

[@PhysicsMaths-z3l]

Thanks from Russia.

[@alster724]

Tricky but got the concept

[@KipIngram]

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

[@misterenter-iz7rz]

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?🤔

[@mehulpunia6174]

that's the mayajal of maths

[@shadmanhasan4205]

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

[@fokokoster]

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

[@flash24g]

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.

[@张建-w5d]

好复杂啊！好复杂啊！！come from china,,.哈哈！！

[@sanchezking6188]

I would have guessed 55.

[@kibeterick5776]

Supersonic

[@PreMath]

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

[@dyalaaleid7838]

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

[@PreMath]

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]

@@PreMath thanks❤❤

[@张建-w5d]

好复杂啊！好复杂啊！！come from china,,.哈哈！！

[@Waldlaeufer70]

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

[@flash24g]

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.

[@pralhadraochavan5179]

Good night sir I like it I proud of your talent

[@honestadministrator]

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

[@murphygreen8484]

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

[@theoyanto]

😅

[@dakeypunchar6930]

First

[@PreMath]

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

[@charlesbromberick4247]

nice job

[@programmer229]

Thanks from🇦🇿

[@PreMath]

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

[@programmer229]

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