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trf05
 
 

Area of triangle on the right is twice area of the one left

by trf05 Fri Feb 29, 2008 1:59 pm

Image

I can't seem to get to the right answer.. Please help!
crisz
 
 

by crisz Mon Mar 03, 2008 2:06 am

here you have to use the ratio: s/S=h/H
blue_lotus
 
 

by blue_lotus Mon Mar 03, 2008 5:22 am

There are two triangles. The area of big triangle is twice that of smaller triangle.
The 2 triangles are similar triangle using A-A-A

The triangle can be any kind of triangle. As long as the angles x,y,z match in both triangle.
We can take value of x=y=z =60 . I mean let us consider this triangle as equilateral triangle.

For equilateral triangle , Area = root(3)/4 * side ^2

Now from the given data:
Area of small triangle = 1/2 Area of Large triangle

=> Root(3)/4 *s^2 = 1/2 [ root(3)/4 *S^2 ]
=> s^2 = 1/2 * S^2

Take root on both sides
=> s = S/root(2)

i.e S = root(2)* s
RonPurewal
Students
 
Posts: 19744
Joined: Tue Aug 14, 2007 8:23 am
 

by RonPurewal Mon Mar 03, 2008 6:16 am

the following is absolutely true:

blue_lotus Wrote:The triangle can be any kind of triangle.


so why not pick a triangle that's friendlier than an equilateral triangle (which introduces icky square roots that most people would rather not have to deal with)?

how about this:
the answer choices are all different, all the time (because they're multiplying by different numbers). therefore, you can pick ANY triangle you want (which means you can also just let 'little s' be 1).

so:

* let little s = 1
* let the triangles be 45-45-90 triangles with the left and bottom sides perpendicular
* area of small triangle = (1/2)(1)(1) = 1/2
* so area of big triangle = 1
* therefore, (1/2)(big S)(big S) = 1
* (big S) squared = 2
* big S = root 2

the equilateral triangle of course works, but it's much harder to deal with!