Very easy. EF = (1/2).DA = (1/2).40 = 20 is the basis of the pink triangle. Distance from E to (AD) = (1/2).(DC + AB) = (1/2).(20 +50) = 35 is the height of the triangle. So, its area is (1/2).(20).(35) = 350.
Extend line FE downwards to H where points D, C, H are points on a straight line. EFB and ECH are congruent triangles (ASA) and area of trapezium ABCD is equal to area of rectangle AFHD which equals (1/2)*(40)*(50+20) = AF*AD = AF*40 so AF = 35. For congruent triangles FEB and ECH, FE = EH =20. So triangle FGE has base of 20 and height of 35 so the area of triangle FGE = (1/2)*Base*height = 0.5*20*35 = 350.
Let's find the area: . .. ... .... ..... Let's add point H to AB such that ADCH is a rectangle. In this case the triangle BCH is a right triangle and we can apply the Pythagorean theorem: BC² = BH² + CH² = (AB − AH)² + CH² = (AB − CD)² + AD² = (50 − 20)² + 40² = 30² + 40² = 50² ⇒ BC = 50 The triangles BCH and BEF are obviously similar, so we can conclude: BF/BH = EF/CH = BE/BC = (BC/2)/BC = 1/2 ⇒ BF = BH/2 = (AB − AH)/2 = (AB − CD)/2 = (50 − 20)/2 = 30/2 = 15 ∧ EF = CH/2 = AD/2 = 40/2 = 20 Now we are able to calculate the area of the pink triangle: A(EFG) = (1/2)*EF*h(EF) = (1/2)*EF*AF = (1/2)*EF*(AB − BF) = (1/2)*20*(50 − 15) = 10*35 = 350 Best regards from Germany
Drop a perpendicular from C up to AB at H. As CH is perpendicular to DC and AB, all four internal angles of ADCH are 90°, so ADCH is a rectangle, CH = AD = 40, and HA = DC = 20. As ∠BFE = ∠BHC = 90° and ∠B is common, ∆BHC and ∆BFE are similar triangles. Let CE = EB = x. FE/EB = HC/CB FE/x = 40/2x FE = 40/2 = 20 BF/EB = BH/CB BF/x = (50-20)/2x BF = 30/2 = 15 Triangle ∆FGE: Area = bh/2 = FE(AF)/2 Area = 20(50-15)/2 Area = 10(35) = 350 sq units
Let dc extended meet fe extended at point r. The triangles fbe and ecr are congruent.So fe=er and fr is parallel to ad so fe=er=20.Drop a perp. from c and let it meet ab at point t. The triangles tcb and fbe are similar. So 40/20= tb/fb and since tb=30 fb=15 and af=35. Therefore area= half20mult. by35=350
Point E Draw a horizontal line to AD, it divides AB into half. Draw a perpendicular from E to AB, the length of 30 will be cut in half. So 20 is base line of the Triangle and 35 ist height, so Area= 350.
1/ Drop CH perpendicular to AB we have EF//= 1/2 CH= 1/2 AD= 20 2/ Consider the right triangle CHB, we have HE=EB-> EF is the perpendicular bisector of HB-> HF=1/2 HB = 15-> AF= 20+15=35 Area of the pink triangle= 1/2 AFxEF =350 sq units😊
Midline of trapezoid ABCD=(50+20)/2=35.Expand the triangle FBE, poit B coicide with point C, ADFF`=1400. F`DGE - trapezoid, area F`DGE=(15+20)*35/2=612,5, areq triangle GAF=25*35/2=437,5, areav triangle GFE=1400-612,5-437,5=350!
I'm unclear how to proceed, so will wing it: Extend DC and make a rectangle of 50 by 40. CB = 50 due to 30,40,50 right triangle newly formed on the right, so EC and EB are each 25. Call the lower right vertex of the new triangle, H CHB is similar to FBE 30/50 = (FB)/25 so FB = 15. Height of FEG is 35 (height is sideways). FE (the base) is sqrt(25^2 - 15^2) = 20 Pink area is (20*35)/2 = 350 sq un. Yes, that was easier than it first appeared. Thank you again.
Mostly the same method as I used, except I drew a line straight up from C instead of diagonal DF. Also figured you don't need to calculate the length of BC and its segments, just need to notice that it creates 2 similar triangles with a ratio of 1/2.
STEP-BY-STEP RESOLUTION PROPOSAL : 1) Divide the Red Line in two segments; one with length 20, and another with length 30, by droping a vertical line passing through point C. Call this point H. Thus : AH = 20 and BH = 30. 2) BC^2 = CH^2 + BH^2 ; BC^2 = 1600 + 900 ; BC^2 = 2.500 ; BC = 50 3) BC / 2 = 50 / 2 = CE = BE = 25 4) Draw a line parallel to AB passing through E and reaching blue line AD. Call it point I 5) By Thales Theorem we know that AI = ID = 20 6) Now, we can conclude that : EF = AI = 20 7) FB^2 = 625 - 400 ; FB^2 = 225 ; FB = 15 8) AF = 50 - 15 ; AF = 35 9) Area = 20 * 35 / 2 ; Area = 700 / 2 ; Area = 350 10) ANSWER : The Pink Triangle Area equal to 350 Square Units. Greetings from the Universal Islamic Institute for the Study of Ancient Mathematical Thinking, Knowledge and Wisdom in Cordoba Caliphate!!
Thank you!
You're welcome!🌹
Thanks ❤️
Very easy.
EF = (1/2).DA = (1/2).40 = 20 is the basis of the pink triangle.
Distance from E to (AD) = (1/2).(DC + AB) =
(1/2).(20 +50) = 35 is the height of the triangle.
So, its area is (1/2).(20).(35) = 350.
Yes, I solved it this way as well.
I solved it the same way you did . He did a lot of work that was not needed . We do not need length BC . All we need are the two lengths EF and AF .
Extend line FE downwards to H where points D, C, H are points on a straight line. EFB and ECH are congruent triangles (ASA) and area of trapezium ABCD is equal to area of rectangle AFHD which equals (1/2)*(40)*(50+20) = AF*AD = AF*40 so AF = 35. For congruent triangles FEB and ECH, FE = EH =20. So triangle FGE has base of 20 and height of 35 so the area of triangle FGE = (1/2)*Base*height = 0.5*20*35 = 350.
The way this rigid problem is solved is impressive
G can be anywhere on AD, making the solution quite obvious.
Correct!!
Let's find the area:
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Let's add point H to AB such that ADCH is a rectangle. In this case the triangle BCH is a right triangle and we can apply the Pythagorean theorem:
BC² = BH² + CH² = (AB − AH)² + CH² = (AB − CD)² + AD² = (50 − 20)² + 40² = 30² + 40² = 50² ⇒ BC = 50
The triangles BCH and BEF are obviously similar, so we can conclude:
BF/BH = EF/CH = BE/BC = (BC/2)/BC = 1/2
⇒ BF = BH/2 = (AB − AH)/2 = (AB − CD)/2 = (50 − 20)/2 = 30/2 = 15
∧ EF = CH/2 = AD/2 = 40/2 = 20
Now we are able to calculate the area of the pink triangle:
A(EFG) = (1/2)*EF*h(EF) = (1/2)*EF*AF = (1/2)*EF*(AB − BF) = (1/2)*20*(50 − 15) = 10*35 = 350
Best regards from Germany
Drop a perpendicular from C up to AB at H. As CH is perpendicular to DC and AB, all four internal angles of ADCH are 90°, so ADCH is a rectangle, CH = AD = 40, and HA = DC = 20.
As ∠BFE = ∠BHC = 90° and ∠B is common, ∆BHC and ∆BFE are similar triangles. Let CE = EB = x.
FE/EB = HC/CB
FE/x = 40/2x
FE = 40/2 = 20
BF/EB = BH/CB
BF/x = (50-20)/2x
BF = 30/2 = 15
Triangle ∆FGE:
Area = bh/2 = FE(AF)/2
Area = 20(50-15)/2
Area = 10(35) = 350 sq units
A = ½ b.h
A = ½ (40/2).(20+50)/2
A = 350 cm² ( Solved √ )
Let dc extended meet fe extended at point r. The triangles fbe and ecr are congruent.So fe=er and fr is parallel to ad so fe=er=20.Drop a perp. from c and let it meet ab at point t. The triangles tcb and fbe are similar. So 40/20= tb/fb and since tb=30 fb=15 and af=35. Therefore area= half20mult. by35=350
Interseting thing is Point G can be anywhere on AD, still the area of ∆ GFE does not change
Point E Draw a horizontal line to AD, it divides AB into half. Draw a perpendicular from E to AB, the length of 30 will be cut in half. So 20 is base line of the Triangle and 35 ist height, so Area= 350.
S=350 👏
How did you consider that FEG is right angled traingle
I did this in my head!
A=AF*FE/2=35*20/2=350
base : 40/2=20 height : (20+50)/2=35
Pink Triangle Area = 20*35/2 = 350
Me first again 😂😂
1/ Drop CH perpendicular to AB we have EF//= 1/2 CH= 1/2 AD= 20
2/ Consider the right triangle CHB, we have HE=EB-> EF is the perpendicular bisector of HB-> HF=1/2 HB = 15-> AF= 20+15=35
Area of the pink triangle= 1/2 AFxEF =350 sq units😊
Midline of trapezoid ABCD=(50+20)/2=35.Expand the triangle FBE, poit B coicide with point C, ADFF`=1400. F`DGE - trapezoid, area F`DGE=(15+20)*35/2=612,5, areq triangle GAF=25*35/2=437,5, areav triangle GFE=1400-612,5-437,5=350!
I'm unclear how to proceed, so will wing it:
Extend DC and make a rectangle of 50 by 40.
CB = 50 due to 30,40,50 right triangle newly formed on the right, so EC and EB are each 25.
Call the lower right vertex of the new triangle, H
CHB is similar to FBE
30/50 = (FB)/25 so FB = 15.
Height of FEG is 35 (height is sideways).
FE (the base) is sqrt(25^2 - 15^2) = 20
Pink area is (20*35)/2 = 350 sq un.
Yes, that was easier than it first appeared.
Thank you again.
350
Verbal calculation,
A = 20×35/2
= 350 sq. units.
Mostly the same method as I used, except I drew a line straight up from C instead of diagonal DF. Also figured you don't need to calculate the length of BC and its segments, just need to notice that it creates 2 similar triangles with a ratio of 1/2.
You did it with a longer solution
STEP-BY-STEP RESOLUTION PROPOSAL :
1) Divide the Red Line in two segments; one with length 20, and another with length 30, by droping a vertical line passing through point C. Call this point H. Thus : AH = 20 and BH = 30.
2) BC^2 = CH^2 + BH^2 ; BC^2 = 1600 + 900 ; BC^2 = 2.500 ; BC = 50
3) BC / 2 = 50 / 2 = CE = BE = 25
4) Draw a line parallel to AB passing through E and reaching blue line AD. Call it point I
5) By Thales Theorem we know that AI = ID = 20
6) Now, we can conclude that : EF = AI = 20
7) FB^2 = 625 - 400 ; FB^2 = 225 ; FB = 15
8) AF = 50 - 15 ; AF = 35
9) Area = 20 * 35 / 2 ; Area = 700 / 2 ; Area = 350
10) ANSWER : The Pink Triangle Area equal to 350 Square Units.
Greetings from the Universal Islamic Institute for the Study of Ancient Mathematical Thinking, Knowledge and Wisdom in Cordoba Caliphate!!