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mbaMission
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Equilateral Triangle Question 07 Jun 2009, 00:30
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Pls help.



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Equilateral Triangle Question 07 Jun 2009, 02:58
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mbaMission
Pls help.
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D

The larger area =36square root3
the smaller area=9square root3

So the area of the border=27square root 3
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Equilateral Triangle Question 07 Jun 2009, 03:06
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How did you get the length of bigger triangle?

sondenso
mbaMission
Pls help.

D

The larger area =36square root3
the smaller area=9square root3

So the area of the border=27square root 3
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Equilateral Triangle Question 07 Jun 2009, 03:51
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mbaMission
Pls help.

D

The larger area =36square root3
the smaller area=9square root3

So the area of the border=27square root 3
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How to calculate the larger area?
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bigoyal
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Equilateral Triangle Question 07 Jun 2009, 04:10
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Attachment:
solution.jpg
solution.jpg [ 12.81 KiB | Viewed 2208 times ]

refer to the attachment.

For smaler triangle, s = 18 / 3 =6
Hieght = √(6^2 + 3^2) = 3√3
Area = 1/2 * 6 * 3√3 = 9√3

For larger traingle, given DQ = √3
Angle QBD = 30 degree (since QB will bisect angle CBA)
Angle QDA = 90 degree, and hence angle DQB = 60

sides of 30-60-90 triangle are always in ratio of 1:√3:2
since DQ = √3
=> DB = √3 * √3 = 3
and QB = 2√3 = AP

Hence, base BC = 6 + 3 + 3 = 12
Height = 3√3 + √3 + 2√3 = 6√3
Area = 1/2 * 12 * 6√3 = 36√3

Diff = 27√3

OA is D.

Please let me know if you need more explanation
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Equilateral Triangle Question 07 Jun 2009, 04:39
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bigoyal
Attachment:
solution.jpg

refer to the attachment.

For smaler triangle, s = 18 / 3 =6
Hieght = √(6^2 + 3^2) = 3√3
Area = 1/2 * 6 * 3√3 = 9√3

For larger traingle, given DQ = √3
Angle QBD = 30 degree (since QB will bisect angle CBA)
Angle QDA = 90 degree, and hence angle DQB = 60

sides of 30-60-90 triangle are always in ratio of 1:√3:2
since DQ = √3
=> DB = √3 * √3 = 3
and QB = 2√3 = AP

Hence, base BC = 6 + 3 + 3 = 12
Height = 3√3 + √3 + 2√3 = 6√3
Area = 1/2 * 12 * 6√3 = 36√3

Diff = 27√3

OA is D.

Please let me know if you need more explanation
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gr8 job ! thanks mate :wink:
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Equilateral Triangle Question 07 Jun 2009, 22:50
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I think the height of the smaller traingle should be 5.
By using Pythagoras theorem....

(Height)^2 + (Base)^2 = (Hypot..)^2
h^2 + 3^2 = 6^2
or H=5.
How is it 3sqrt(3) ???????????????
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Equilateral Triangle Question 07 Jun 2009, 23:36
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mdfrahim
I think the height of the smaller traingle should be 5.
By using Pythagoras theorem....

(Height)^2 + (Base)^2 = (Hypot..)^2
h^2 + 3^2 = 6^2
or H=5.
How is it 3sqrt(3) ???????????????
Show more

Altitude of equilateral triangle is (sqr rt 3) / 2 X (length of side)
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Equilateral Triangle Question 08 Jun 2009, 02:23
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mdfrahim
I think the height of the smaller traingle should be 5.
By using Pythagoras theorem....

(Height)^2 + (Base)^2 = (Hypot..)^2
h^2 + 3^2 = 6^2
or H=5.
How is it 3sqrt(3) ???????????????
Show more

You are calculating it incorrectly.
h^2 + 3^2 = 6^2
or, h^2 = 36 - 9 = 27
or, h = √27 = 3√3



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