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# PS - Area of shaded region

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14 Oct 2005, 22:04
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See Attached... took me > 3mins to solve ... shorter method?
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PS - Area.gif [ 2.68 KiB | Viewed 1136 times ]

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14 Oct 2005, 22:56
BD= sqrt ( 20^2+15^2) = 25
BD* CE = 20*15=300 ---->CE = 12
BE= sqrt ( 20^2-12^2) = 16
The area is : 12*16/2= 96

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14 Oct 2005, 23:24
96
laxieqv is probably quicker, this was how I did it

DB = sqrt(400+225) = 25
Area (BCD) = 05*20*15 = 05*25*CE => CE = 12
Tri BEC ~ Tri CED (note order of vertices)
=> BE/CE = BC/CD =20/15
=> BE = 20*12/15 = 16

Area = 1/2*16*12 = 96

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14 Oct 2005, 23:52
Another method: (It took me little less than 2 mins)

1) Find the area of the triangle BCD = 1/2 * b * h = 1/2 * 20 * 15 = 150

2) Find the area of the triangle CED.

hypotenuse is 15 (Remember the ratio is 3:4:5 - hence the sides should be 9:12:15)

area = 1/2 * 12 * 9 = 54

3) Area of shaded region = area BCD - area CED = 150 - 54 = 96

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15 Oct 2005, 06:57
thanks guys... Keep forgetting about similar triangles!!

And yes, OA is 96.
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15 Oct 2005, 08:58
wow...i just estimated it to 96...

well I figured (1) the entire area has to be less than half of that of rectangle...i.e 150

then I said, how many right angle triangles can I cut in a rectangle...looks like 5, so I divide 300/5=60, so thats the area of each right angle triangle
then the area of the shaded area has to be 150-50=90 or very close to it...96 it is...took less than 40 seconds to do this...

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15 Oct 2005, 09:00
I dont think you can assume its 3:4:5 triangle'

what if its is 5:12:13?

sudhagar wrote:
Another method: (It took me little less than 2 mins)

1) Find the area of the triangle BCD = 1/2 * b * h = 1/2 * 20 * 15 = 150

2) Find the area of the triangle CED.

hypotenuse is 15 (Remember the ratio is 3:4:5 - hence the sides should be 9:12:15)

area = 1/2 * 12 * 9 = 54

3) Area of shaded region = area BCD - area CED = 150 - 54 = 96

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15 Oct 2005, 09:13
fresinha12 wrote:
wow...i just estimated it to 96...

well I figured (1) the entire area has to be less than half of that of rectangle...i.e 150

then I said, how many right angle triangles can I cut in a rectangle...looks like 5, so I divide 300/5=60, so thats the area of each right angle triangle
then the area of the shaded area has to be 150-50=90 or very close to it...96 it is...took less than 40 seconds to do this...

Can you demonstrate how? I mean the bold part ....thank you ...just want to know how this very interesting solution works

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15 Oct 2005, 09:36
First of it should be 6 triangles...

so lets see..

if you look at the trinagle with one side equal to 15 (i.e. the prependicular) then you roughly sketch it on a piece of paper and you will see there are 6 such trinagles that you can cut...

each triangle would have an area roughly=50

so then you get 150 (from half the area of the rectangle)-50 its about 100.

laxieqv wrote:
fresinha12 wrote:
wow...i just estimated it to 96...

well I figured (1) the entire area has to be less than half of that of rectangle...i.e 150

then I said, how many right angle triangles can I cut in a rectangle...looks like 5, so I divide 300/5=60, so thats the area of each right angle triangle
then the area of the shaded area has to be 150-50=90 or very close to it...96 it is...took less than 40 seconds to do this...

Can you demonstrate how? I mean the bold part ....thank you ...just want to know how this very interesting solution works

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15 Oct 2005, 10:41
vikramm wrote:
See Attached... took me > 3mins to solve ... shorter method?

took me 30 seconds

notice the 9:12:15 triangle

15^2 and 20^2 = 625

hype is 25

base is 25 - 9 = 16

16*12 = 192/2 = 96

knowing your 3:4:5s, 5:12:13s, and 8:15:17s saves a heck of a lot time! (geek)

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16 Oct 2005, 09:21
bigtooth81 wrote:
I got C

Bigtooth your avatar is quite sultry.

I think you should replace it with this one which I think is more politically correct:

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16 Oct 2005, 09:55
or this

(better match )

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16 Oct 2005, 09:59
gsr wrote:
or this

(better match )

No. Yours is way too raunchy GSR. Your emoticons are French kissing.

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16 Oct 2005, 19:55
In my opinion knowing the 3:4:5 triangle is the best approach. The first step you notice that the big triangle is 5*(3:4:5), the second step you know the small triangle is also a 3:4:5 with the long side =15. You can figure out h=12 in like 30 seconds.
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16 Oct 2005, 20:06
HongHu wrote:
In my opinion knowing the 3:4:5 triangle is the best approach. The first step you notice that the big triangle is 5*(3:4:5), the second step you know the small triangle is also a 3:4:5 with the long side =15. You can figure out h=12 in like 30 seconds.

I also agree that this is the best approach ^_^

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16 Oct 2005, 20:06
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