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# Geometry

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Director
Joined: 03 Sep 2006
Posts: 885
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Geometry [#permalink]  10 Dec 2006, 00:19
I can't draw the figure here. ( nevertheless figures are not drawn to scale is given )

There are two triangles shown in a figure. Their corresponding angles are all equal. Lengths of sides are different. One has the base given as s and other one has the base S. Area of the triangle with base S is twice the area of triangle with base s. Then in terms os s, S=

why?
Intern
Joined: 17 Nov 2006
Posts: 19
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Kudos [?]: 0 [0], given: 0

Re: Geometry [#permalink]  10 Dec 2006, 00:28
LM wrote:
I can't draw the figure here. ( nevertheless figures are not drawn to scale is given )

There are two triangles shown in a figure. Their corresponding angles are all equal. Lengths of sides are different. One has the base given as s and other one has the base S. Area of the triangle with base S is twice the area of triangle with base s. Then in terms os s, S=

why?

The area of the bigger one will be square of the ratio,
AREA/area=(S/s)^2=2
S/s=root2
Senior Manager
Joined: 05 Oct 2006
Posts: 268
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Kudos [?]: 4 [0], given: 0

Re: Geometry [#permalink]  10 Dec 2006, 00:38
let L and l be the corresponding heights in 2 traingles.
Let x and y be the corresponding legths of side in 2 triangles.

since the area of one is double.

1/2 SL = 2 * 1/2 sl
so, S=2sl/L.........eq 1

from similar triangle,

s/S= y/x
also y/x = l/L
hence l/L = s/S

putting this in eq 1

S=2*s*s/S

hence S = s*sqrt2

LM wrote:
I can't draw the figure here. ( nevertheless figures are not drawn to scale is given )

There are two triangles shown in a figure. Their corresponding angles are all equal. Lengths of sides are different. One has the base given as s and other one has the base S. Area of the triangle with base S is twice the area of triangle with base s. Then in terms os s, S=

why?
Re: Geometry   [#permalink] 10 Dec 2006, 00:38
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