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Triangle 2 - DS

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by pemdas » Mon Dec 26, 2011 2:47 am
karthikpandian19 wrote:E is the midpoint of AC in right triangle ABC shown above. If the area of ΔABC is 24, what is the area of ΔBED?

Image
you have additional info on pic - i'm using it for this q.

AB*AC/2=24, AB=AC hence AC^2=48 and AC=4sqroot(3)
EC=2sqroot(3) and 2ED^2=EC^2 or 2ED^2=12, ED^2=6 and ED=sqroot(6)
BC=sqroot(48+48)=4sqroot(6)
BD=3sqroot(6) or BC-ED
BD*ED/2 is req. square = 3sqroot(6)sqroot(6)/2=9[spoiler]


ans 9[/spoiler]
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by LalaB » Mon Dec 26, 2011 8:31 am
pemdas wrote:
karthikpandian19 wrote:E is the midpoint of AC in right triangle ABC shown above. If the area of ΔABC is 24, what is the area of ΔBED?

Image
you have additional info on pic - i'm using it for this q.

AB*AC/2=24, AB=AC hence AC^2=48 3)
EC=2sqroot(3) and 2ED^2=EC^2 or 2ED^2=12, ED^2=6 6)
[/spoiler]
or from here we can go this way -

Sabc/Sedc=AC^2/ED^2

24/Sedc=48/6

Sedc=3

since E is a midpoint, then triangles ABE and BEC have the same area. so, Sbec=Sabc/2=24/2=12

Sebc=Sbde +Sedc
12=Sbde+3

Sbde=9
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by pemdas » Mon Dec 26, 2011 9:10 am
i didn't get
Sabc/Sedc=AC^2/ED^2
another way would be S(ABC)-S(EDC)-S(ABE), S(EDC)=ED^2/2=3 and AB=AC=4sqroot(3) so S(ABE)=(1/2)*4sqroot(3)*2sqroot(3)=12. 24-3-12=9
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by LalaB » Mon Dec 26, 2011 9:14 am
pemdas wrote:i didn't get Sabc/Sedc=AC^2/ED^2

in two similar triangles, the ratio of their areas is the square of the ratio of their sides
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by LalaB » Mon Dec 26, 2011 9:27 am
pemdas wrote:i didn't get
Sabc/Sedc=AC^2/ED^2
another way would be S(ABC)-S(EDC)-S(ABE), S(EDC)=ED^2/2=3 and AB=AC=4sqroot(3) so S(ABE)=(1/2)*4sqroot(3)*2sqroot(3)=12. 24-3-12=9
pemdas, why do u calculate Sabe so long?

isnt it better just Sabc/2=24/2=12 ?(because E is a midpoint)
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by rijul007 » Mon Dec 26, 2011 9:52 am
karthikpandian19 wrote:E is the midpoint of AC in right triangle ABC shown above. If the area of ΔABC is 24, what is the area of ΔBED?

Image
Angle C = Angle B = 45 degrees
Therefore, AB = AC = 2x

Area of ΔABC = 24
1/2 * 2x * 2x = 24
x = 12 = 2√3


E is the mid point of AC
AE = EC = x = 2√3
Area of ΔAEB = 1/2 * x * 2x = 12


In ΔEDC,
EC = 2√3
ED = DC = y
2y^2 = 12
y = √6
Area of ΔEDC = 1/2 * y * y = 3

Area of ΔBED = Area of ΔABC - (Area of ΔAEB + Area of ΔEDC)
= 24 - (12 + 3)
= 24 - 15
= 9
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by pemdas » Mon Dec 26, 2011 10:05 am
gotcha you, how you tested the triangles ABC and EDC are similar. Do you know proportion of their sides? Are the sides in ABC and EDC known without the above calculations? Also, for similar triangles the proportion of all sides must be in the same ratio. Is the proportion of sides of triangles ABC and EDC in the same ratio?
LalaB wrote:
pemdas wrote:i didn't get Sabc/Sedc=AC^2/ED^2

in two similar triangles, the ratio of their areas is the square of the ratio of their sides
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by LalaB » Mon Dec 26, 2011 10:09 am
pemdas wrote:gotcha you, how you tested the triangles ABC and EDC are similar. Do you know proportion of their sides? Are the sides in ABC and EDC known without the above calculations? Also, for similar triangles the proportion of all sides must be in the same ratio. Is the proportion of sides of triangles ABC and EDC in the same ratio?
LalaB wrote:
pemdas wrote:i didn't get Sabc/Sedc=AC^2/ED^2

in two similar triangles, the ratio of their areas is the square of the ratio of their sides
)))))

pemdas, with all ur questions, soon I will have doubts even about my name )))

two triangles have the same 45 degree angle, and EC and DC are on both triangles.
Last edited by LalaB on Mon Dec 26, 2011 10:34 am, edited 1 time in total.
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by rijul007 » Mon Dec 26, 2011 10:10 am
pemdas wrote:gotcha you, how you tested the triangles ABC and EDC are similar. Do you know proportion of their sides? Are the sides in ABC and EDC known without the above calculations? Also, for similar triangles the proportion of all sides must be in the same ratio. Is the proportion of sides of triangles ABC and EDC in the same ratio?
LalaB wrote:
pemdas wrote:i didn't get Sabc/Sedc=AC^2/ED^2

in two similar triangles, the ratio of their areas is the square of the ratio of their sides
Triangle ABC and Triangle DEC are similar
Both are Right Isosceles triangles
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by pemdas » Mon Dec 26, 2011 10:40 am
beautiful
yes two triangles discussed are similar, i overlooked of the similarity to be not only

Corresponding sides are all in the same proportion

BUT also

Corresponding angles are the congruent (same measure)

good point, and we could use this head-start, as Lala figured S/s= (Side/side)^2

my only doubt is that midpoint to the side AC (non-hypotenuse) and BE will divide ABC into equal triangles was seen that AB is the only height of two triangles ABE and BEC. Hence S(ABE)=S(BEC)
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by LalaB » Mon Dec 26, 2011 11:02 am
@pemdas-

may be it helps
found on the internet -

In geometry, a median of a triangle is a line segment joining a vertex to the midpoint of the opposing side.

Each median divides the area of the triangle in half

source-https://en.wikipedia.org/wiki/Median_%28geometry%29
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by pemdas » Mon Dec 26, 2011 11:32 am
we just proved it - the smaller triangles share two sides in equal proportions 1/2 and 1/2 of the whole side and their altitudes (heights) will be the same. Hence 1/2*(base*height) will give equal squares for each triangle for the bases are equal (midpoint divides into halves) and height is the same. With some shapes altitude (height) will be also the side for smaller triangles, then the altitude will be also a bisector of the larger triangle. We just proved it based on your recent posts and examples given
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by ronnie1985 » Mon Dec 26, 2011 7:59 pm
(1/2)*AC*AB = 24 and AC = AB = > AC = sq rt(48) = 4rt(3)
Ar(tr.BDE) = ar(tr. BEC)-ar(tr.DEC) = 12-ar(tr.DEC)
EC = 2rt(3)
Also EC = DEsq+DCsq = > DE = DC = sqrt(6)
Ar(trDEC) = .5*DC*DE = 3
Ar(tr. BED) = 12-3=9
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by karthikpandian19 » Tue Dec 27, 2011 9:44 pm
OA is 9
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