Ar(ADC)=Ar(DEC)+30. Let heights of both triangle be x and y . 1/2 (15)(x) = 1/2(15)(y)+30. x-y=4 . Here x-y is height of triangle ABE . Ar(AEB)= 1/2(6)(4)= 12.
Triangles ABE,CDE similar. BE/DE = AB/DE = 6/15 =2/5. Triangles ABE and AED have same height over base BD, so areas in ratio 2:5 = 12:30. Area ABE =12.
Nice solution. Here is the solution I came up with, after a few false starts: First, we establish that △AEB ~ △CED (as you did in the beginning). In similar triangles, ratio of corresponding sides are equal: BE/DE = AB/CD = 6/15 = 2/5 Let BE = 2k, DE = 5k △AED has base DE and height = altitude from A to BD = h Area(△AED) = 1/2 * DE * h = 1/2 * 5k * h = 5kh/2 △AEB has base BE and height = altitude from A to BD = h Area(△AEB) = 1/2 * BE * h = 1/2 * 2k * h = kh = 2/5 * (5kh/2) = 2/5 * Area(△AED) = 2/5 (30) = 12
I did it slightly differently, same results: We have a bunch of matched angles: AEB CED BEC AED ABD BDC DCE BAE We have similar triangles: ABE ~ CDE, their areas are proportional to square of their length ratios: ABE=CDE*(6/15)^2=4CDE/25,CDE=25ABE/4 AB=6,CD=15,ADE=30 Areas, triagles, bh/2 ABD=ABC=ABh/2=3h=ADE+ABE=30+ABE,ACD=BCD=DCh/2=15h/2, ADE=30=ABD-ABE,BCE=ABC-ABE=ABD-ABE=30 3h=30+ABE,h=10+ABE/3 ACD=15h/2=ADE+CDE ACD=15h/2=30+25ABE/4 ADE+CDE=15h/2=30+25ABE/4 30+25ABE/4=15h/2=30+25ABE/4 h=(30+25ABE/4)*(2/15) (30+25ABE/4)*(2/15)=10+ABE/3 4+25AEB/30=10+AEB/3 25AEB/30-AEB/3=6 15AEB/30=6 AEB=12
Where we have rays emanating from a point (in our case E), and where those rays intersect parallel lines, the "Intercept Theorem" tells us something about the ratios of certain lengths. In our case it tells us that the heights of the North and South triangles (i.e. the distances of E from lines AB and CD) are in the ratio 6:15 or 2:5. If h is the height of the trapezoid, then the heights of these two triangles are (2/7)h and (5/7)h. Thus their areas are (1/2)(6)(2/7)h and (1/2)(15)(5/7)h, respectively, or (12/14)h and (75/14)h. It is a property of trapezoids that their diagonals divide them into four triangles in such a way that (in our case) the East and West triangles have the same area, so we know that y=30. The area T of the whole trapezoid is equal to the product of its height h and its average width (1/2)(6+15). Therefore T = (21/2)h. It is also equal to the sum of the areas of the four triangles: T = 30 + 30 + (12/14)h + (75/14)h = 60 + (87/14)h. We can set the two expressions for T equal, expanding 21/2 to 147/14: 147(h/14) = 60 + 87(h/14) (147-87)(h/14) = 60 (60/14)h = 60 h = 14 So the North area is (12/14)14 = 12.
Similar to MarieAnne below: If 'y = altitude of AEB', then 'y/h = 2/7' , as AEB~CED, ratios of sides and altitudes =2:5. '2[Area(ADE)] = 2[Area(ADB) - Area(AEB)]', so '2(30) = 6h - (2/7)6h', 'h = 10(7/5) = 14', so 'y = 4') 'Area(AED) = (1/2) (6)(4) = 12'
Ar(ADC)=Ar(DEC)+30.
Let heights of both triangle be x and y .
1/2 (15)(x) = 1/2(15)(y)+30.
x-y=4 .
Here x-y is height of triangle ABE .
Ar(AEB)= 1/2(6)(4)= 12.
Nice work!
Triangles ABE,CDE similar. BE/DE = AB/DE = 6/15 =2/5. Triangles ABE and AED have same height over base BD, so areas in ratio 2:5 = 12:30. Area ABE =12.
Nice solution. Here is the solution I came up with, after a few false starts:
First, we establish that △AEB ~ △CED (as you did in the beginning).
In similar triangles, ratio of corresponding sides are equal:
BE/DE = AB/CD = 6/15 = 2/5
Let BE = 2k, DE = 5k
△AED has base DE and height = altitude from A to BD = h
Area(△AED) = 1/2 * DE * h = 1/2 * 5k * h = 5kh/2
△AEB has base BE and height = altitude from A to BD = h
Area(△AEB) = 1/2 * BE * h
= 1/2 * 2k * h
= kh = 2/5 * (5kh/2)
= 2/5 * Area(△AED)
= 2/5 (30)
= 12
Excellent!
Sir , you explained very well .
I did it slightly differently, same results:
We have a bunch of matched angles:
AEB CED
BEC AED
ABD BDC
DCE BAE
We have similar triangles:
ABE ~ CDE, their areas are proportional to square of their length ratios:
ABE=CDE*(6/15)^2=4CDE/25,CDE=25ABE/4
AB=6,CD=15,ADE=30
Areas, triagles, bh/2
ABD=ABC=ABh/2=3h=ADE+ABE=30+ABE,ACD=BCD=DCh/2=15h/2,
ADE=30=ABD-ABE,BCE=ABC-ABE=ABD-ABE=30
3h=30+ABE,h=10+ABE/3
ACD=15h/2=ADE+CDE
ACD=15h/2=30+25ABE/4
ADE+CDE=15h/2=30+25ABE/4
30+25ABE/4=15h/2=30+25ABE/4
h=(30+25ABE/4)*(2/15)
(30+25ABE/4)*(2/15)=10+ABE/3
4+25AEB/30=10+AEB/3
25AEB/30-AEB/3=6
15AEB/30=6
AEB=12
Great work! Thank you for sharing.
Where we have rays emanating from a point (in our case E), and where those rays intersect parallel lines, the "Intercept Theorem" tells us something about the ratios of certain lengths.
In our case it tells us that the heights of the North and South triangles (i.e. the distances of E from lines AB and CD) are in the ratio 6:15 or 2:5.
If h is the height of the trapezoid, then the heights of these two triangles are (2/7)h and (5/7)h.
Thus their areas are (1/2)(6)(2/7)h and (1/2)(15)(5/7)h, respectively, or (12/14)h and (75/14)h.
It is a property of trapezoids that their diagonals divide them into four triangles in such a way that (in our case) the East and West triangles have the same area, so we know that y=30.
The area T of the whole trapezoid is equal to the product of its height h and its average width (1/2)(6+15). Therefore T = (21/2)h.
It is also equal to the sum of the areas of the four triangles: T = 30 + 30 + (12/14)h + (75/14)h = 60 + (87/14)h.
We can set the two expressions for T equal, expanding 21/2 to 147/14:
147(h/14) = 60 + 87(h/14)
(147-87)(h/14) = 60
(60/14)h = 60
h = 14
So the North area is (12/14)14 = 12.
Excellent work!
Similar to MarieAnne below:
If 'y = altitude of AEB', then 'y/h = 2/7' , as AEB~CED, ratios of sides and altitudes =2:5.
'2[Area(ADE)] = 2[Area(ADB) - Area(AEB)]', so
'2(30) = 6h - (2/7)6h',
'h = 10(7/5) = 14', so 'y = 4')
'Area(AED) = (1/2) (6)(4) = 12'
The diagram is fairly accurate, the only answer that is close when you compare it to 30square units is 12 square units.
△AEB=30*6/15=12
This is the shortest solution I've seen so far. Excellent work!
Sir ,I am new here
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without pen: ruclips.net/video/ayHskqZmphM/видео.html
Let's draw 2 additional rhombuses. See the picture:
drive.google.com/file/d/1T3zRV2xiSz9nAxqrNVahUjB1ucr_otBL/view?usp=share_link