CB = 4. Enclosing the figure in a rectangle with D at the top. Call top left corner F, and top right corner G. In triangle EFD, EF = root 18 = 4.24264. Area of triangle EFD = (root 18)^2 / 2 = 9. Moving to triangle DGC. DG = 10 - 4.2464 = 5.75736. Area of triangle DGC = 0.5 x 5.75736^2 = 16.5736. Moving to height of rectangle. GC = DG = 5.75736. Therefore height of rectangle = 4 + 5.75736 = 9.75736. Area of rectangle = 9.75736 x 10 = 97.5736. Area of green region = area of rect. - areas of the 3 triangles. 97.5736 - 9 - 16.5736 - 20. 52.
I added points G (top left) and F (top right) to make a rectangle ABFG. Called DF = x Angle GED is 45 as DEA is 135 so GED is isosceles. GD = GE. Called GD 10-x. Used Pythagoras to get x^2 -20x + 164 = 0 Solved using formula to get x = 5.75736. CDF is also 45 degrees as EDC is 90. So DCF is also isosceles with base and height both 5.75736. So the area ABFG = 10 x 9.75736. Now we know the base and heights of the three triangles. Subtract the areas of the three right triangles which you can now work out as you know x. This leaves 52cm^2
Connect a perpendicular line from point C to point F in AE Connect a line from point E to point G in CD to get an isosceles triangle DEG, and EG=6*sqrt(2) since the angles are 45 degrees We now have 3 regions, ACG with Area = 20, DEG with area = 0.5*6*6=18, and the middle trapazoid whose area = 0.5(10+6sqrt(2))*(10-6sqrt(2))=0.5*(100-72)=0.5*28=14 Adding the areas we get 20+18+14=52 square units
You can simplify your method by extending AE and CD to meet at P, as Premath did. Add area of ΔACG = 20 to area of ΔCFP = (1/2)(10)(10) = 50 and deduct area of ΔDEP = (1/2)(6)(6) = 18. So green area = 20 + 50 - 18 = 52 square units.
This very nice problem got me thinking about a general approach to solving any quadrilateral when two opposite sides and all angles are given or can be calculated 🤔
Height of the lower triangle ABC is (20/10)·2 = 4. Draw a horizontal line through point D to build an outlining rectangle. The upper left triangle has both legs 6/√2, therefore area (6/√2)²/2 = 36/4 = 9. The upper right triangle has both legs (10 - 6/√2), therefore area (10 - 6/√2)²/2 = (100 - 120/√2 + 36/2)/2 = (118 - 120/√2)/2 = 59 - 60/√2 The heigth of the outlining rectangle is 4 + (10 - 6/√2) = 14 - 6/√2 Area of the whole rectangle is therefore 10·(14 - 6/√2) = 140 - 60/√2 Green area then is A = 140 - 60/√2 - (59 - 60/√2) - 9 - 20 = 52 The root terms nicely cancel.
area of the triangle ABC=1/2.AB.BC 20=1/2.(10)(BC( BC=40/10=4 Angle BCD=540-(90+90+90+135)=135° area of the green region=1/2(14^2)-1/2(4^2)-1/2(6^2)-20=98-8-18-20= 52square units. ❤❤❤ Thanks
I made a right pig's ear of this one and used a far more complicated method as I didn't figure out extending the lines. I had a rectangle of 40 at the bottom, the green part of that being 20. I had an isosceles triangle at the top whose area was 10. 20+18=38/ I then proceeded to calculate a thin strip across the diagram from around point E to point C. I went wrong somewhere because I made its thin height 10 - 6*sqrt(2) and had a very final answer of approx 50.85cm^2. Ho hum.
Draw CF, where F is the point on EA where CF is perpendicular to EA. As ∠FAB = ∠ABC = ∠CFA = 90°, ∠BCF is also 90°, ABCF is a rectangle, BC = FA, and CF = AB = 10. As AC is the diagonal of ABCF, ∆ABC and ∆CFA are congruent, so the area of ∆CFA also equals 20 cm². Triangle ∆ABC: A = bh/2 20 = 10BC/2 = 5h BC = 20/5 = 4 Draw DG, where G is the point on CF where DG is perpendicular to CF. Draw EH, where H is the point on DG where EH is perpendicular to DH and parallel to CF. As EH and GF are parallel, and HG and FE are parallel, and ∠EHG = ∠HGF = ∠GFE = 90°, EHGF is a rectangle, ∠FEH is also 90°, HG = FE, and EH = GF = x. As ∠AED = 135° and ∠AEH = 90°, ∠HED = 135-90 = 45°. As ∠DHE =90°, ∠EDH = 90-45 = 45°, ∆DHE is an isosceles right triangle, and DH = EH = x. As ∠EDC = 90° and ∠EDH = 45°, ∠GDC = 45°. As ∠CGD = 90°, ∠DCG = 90-45 = 45°, ∆CGD is an isosceles right triangle, and CG = GD. As CF = AB = 10, CG = 10-GF = 10-EH. Triangle ∆DHE: DH² + HE² = ED² x² + x² = 6² 2x² = 36 x³ = 18 x = √18 = 3√2 GD = CG = 10-x CG = 10-3√2 By observation, FE = HG = GD - DH HG = (10-3√2)-3√2 HG = 10-6√2 The green area consists of right triangles ∆DHE and ∆CGD, rectangle EHGF, and right triangle ∆CFA. A = EH(DH)/2 + CG(GD)/2 + GF(FE) + CF(FA)/2 A = (3√2)²/2 + (10-3√2)²/2 + 3√2(10-6√2) + 20 A = 18/2 + (100+18-60√2)/2 + 30√2 - 36 + 20 A = 30√2 - 7 + 59 - 30√2 = 52 cm²
I see it differently: AE and BC are of unequal length, as are ED and EC! Consequently, the angle BCD is not 135 and the further conclusions drawn from it are incorrect
CB = 4.
Enclosing the figure in a rectangle with D at the top.
Call top left corner F, and top right corner G.
In triangle EFD, EF = root 18 = 4.24264.
Area of triangle EFD = (root 18)^2 / 2 = 9.
Moving to triangle DGC.
DG = 10 - 4.2464 = 5.75736.
Area of triangle DGC = 0.5 x 5.75736^2 = 16.5736.
Moving to height of rectangle.
GC = DG = 5.75736.
Therefore height of rectangle = 4 + 5.75736 = 9.75736.
Area of rectangle = 9.75736 x 10 = 97.5736.
Area of green region = area of rect. - areas of the 3 triangles.
97.5736 - 9 - 16.5736 - 20.
52.
Thanks❤️
I added points G (top left) and F (top right) to make a rectangle ABFG.
Called DF = x
Angle GED is 45 as DEA is 135 so GED is isosceles.
GD = GE.
Called GD 10-x.
Used Pythagoras to get x^2 -20x + 164 = 0
Solved using formula to get x = 5.75736.
CDF is also 45 degrees as EDC is 90. So DCF is also isosceles with base and height both 5.75736.
So the area ABFG = 10 x 9.75736.
Now we know the base and heights of the three triangles. Subtract the areas of the three right triangles which you can now work out as you know x.
This leaves 52cm^2
Since GED is an isosceles right triangle, GD = 6/√2 = 3√2. So, DF = 10 − 3√2.
Thanks❤️
شكرا لكم
نستعمل المثليين APC وEPD
نجد (جذر2)PC=10
ومساحة APC هي 70
ومساحة EPD هي18
وبالتالي مساحةACDE هي18- 70=52
Thanks dear ❤️
Connect a perpendicular line from point C to point F in AE
Connect a line from point E to point G in CD to get an isosceles triangle DEG, and EG=6*sqrt(2) since the angles are 45 degrees
We now have 3 regions, ACG with Area = 20, DEG with area = 0.5*6*6=18, and the middle trapazoid whose area = 0.5(10+6sqrt(2))*(10-6sqrt(2))=0.5*(100-72)=0.5*28=14
Adding the areas we get 20+18+14=52 square units
You can simplify your method by extending AE and CD to meet at P, as Premath did. Add area of ΔACG = 20 to area of ΔCFP = (1/2)(10)(10) = 50 and deduct area of ΔDEP = (1/2)(6)(6) = 18. So green area = 20 + 50 - 18 = 52 square units.
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sum of the interior angles of a polygon of n sides = (n - 2) × 180° = 3× 180°= 540° if n = 5
The missing (unlabeled) angle (
This very nice problem got me thinking about a general approach to solving any quadrilateral when two opposite sides and all angles are given or can be calculated 🤔
Thanks ❤️
Irregular Pentagon, Pentagon, Convex Pentagon Interior Angles always add up too 540⁰. 🙂
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very nice video sar❤❤❤❤
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Height of the lower triangle ABC is (20/10)·2 = 4.
Draw a horizontal line through point D to build an outlining rectangle.
The upper left triangle has both legs 6/√2, therefore area (6/√2)²/2 = 36/4 = 9.
The upper right triangle has both legs (10 - 6/√2), therefore area
(10 - 6/√2)²/2 = (100 - 120/√2 + 36/2)/2 = (118 - 120/√2)/2 = 59 - 60/√2
The heigth of the outlining rectangle is 4 + (10 - 6/√2) = 14 - 6/√2
Area of the whole rectangle is therefore 10·(14 - 6/√2) = 140 - 60/√2
Green area then is A = 140 - 60/√2 - (59 - 60/√2) - 9 - 20 = 52
The root terms nicely cancel.
Thanks ❤️
area of the triangle ABC=1/2.AB.BC
20=1/2.(10)(BC(
BC=40/10=4
Angle BCD=540-(90+90+90+135)=135°
area of the green region=1/2(14^2)-1/2(4^2)-1/2(6^2)-20=98-8-18-20= 52square units. ❤❤❤ Thanks
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i completed to a rectangle instead of a triangle and got the same answer
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I made a right pig's ear of this one and used a far more complicated method as I didn't figure out extending the lines.
I had a rectangle of 40 at the bottom, the green part of that being 20. I had an isosceles triangle at the top whose area was 10. 20+18=38/ I then proceeded to calculate a thin strip across the diagram from around point E to point C. I went wrong somewhere because I made its thin height 10 - 6*sqrt(2) and had a very final answer of approx 50.85cm^2.
Ho hum.
Thanks ❤️
Draw CF, where F is the point on EA where CF is perpendicular to EA. As ∠FAB = ∠ABC = ∠CFA = 90°, ∠BCF is also 90°, ABCF is a rectangle, BC = FA, and CF = AB = 10. As AC is the diagonal of ABCF, ∆ABC and ∆CFA are congruent, so the area of ∆CFA also equals 20 cm².
Triangle ∆ABC:
A = bh/2
20 = 10BC/2 = 5h
BC = 20/5 = 4
Draw DG, where G is the point on CF where DG is perpendicular to CF. Draw EH, where H is the point on DG where EH is perpendicular to DH and parallel to CF. As EH and GF are parallel, and HG and FE are parallel, and ∠EHG = ∠HGF = ∠GFE = 90°, EHGF is a rectangle, ∠FEH is also 90°, HG = FE, and EH = GF = x.
As ∠AED = 135° and ∠AEH = 90°, ∠HED = 135-90 = 45°. As ∠DHE =90°, ∠EDH = 90-45 = 45°, ∆DHE is an isosceles right triangle, and DH = EH = x.
As ∠EDC = 90° and ∠EDH = 45°, ∠GDC = 45°. As ∠CGD = 90°, ∠DCG = 90-45 = 45°, ∆CGD is an isosceles right triangle, and CG = GD. As CF = AB = 10, CG = 10-GF = 10-EH.
Triangle ∆DHE:
DH² + HE² = ED²
x² + x² = 6²
2x² = 36
x³ = 18
x = √18 = 3√2
GD = CG = 10-x
CG = 10-3√2
By observation, FE = HG = GD - DH
HG = (10-3√2)-3√2
HG = 10-6√2
The green area consists of right triangles ∆DHE and ∆CGD, rectangle EHGF, and right triangle ∆CFA.
A = EH(DH)/2 + CG(GD)/2 + GF(FE) + CF(FA)/2
A = (3√2)²/2 + (10-3√2)²/2 + 3√2(10-6√2) + 20
A = 18/2 + (100+18-60√2)/2 + 30√2 - 36 + 20
A = 30√2 - 7 + 59 - 30√2 = 52 cm²
La somma delle aree dei due triangoli è 24,4264+27,5736=52...ho usato il teorema dei seni per calcolare i lati ...
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It took me a while to solve it, but in the end i did, but my solution was much uglier.
50.5
ECD = 18 and ACE = 20
So the area ACDE = 38 cm2
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If AB = 10 then ED = √50
Why do you think that ECD=18 and ACE=20?
@@WernHerr
When AED and BCD = 135° then CD = ED and BC = AE
6*6/2 = 18 and AEC = ABC
I see it differently: AE and BC are of unequal length, as are ED and EC! Consequently, the angle BCD is not 135 and the further conclusions drawn from it are incorrect
Just now 😅😂
Thanks❤️