2009 DAT QR 40

[froggy94]

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Hey guys!

I had a question on #40 of the QR for the 2009 DAT-

The question is: Two sides of an isosceles triangle have length 10. If the area is 48, what is the longest possible third side?

A. 10
B. 12
C. 14
D. 16
E. 18

The answer was D.
I'm taking my DAT Wednesday so any help is much appreciated!!

 

[Lnguyen7]

A =1/2*base*height
so b*h= 2A = 2*48 = 96

By pythagorean theorem:
side^2 = (b/2)^2 + h^2
100 = (b/2)^2 + h^2

Now I think the quickest way to do this is by plugging in each of the base lengths and see which one satisfies both equations instead of solving them. The only problem with this approach is we tend to select the choice that seems to be correct without checking all of the choices. You can quickly eliminate A, C and E because those will not give an exact area of 48. B and D both give area of 48 - TRAP! Since the question asks for the longest possible side, only choice D is correct.

 

[froggy94]

That makes sense- thank you so much!

 

[DrNutmeg]

A =1/2*base*height
so b*h= 2A = 2*48 = 96

By pythagorean theorem:
side^2 = (b/2)^2 + h^2
100 = (b/2)^2 + h^2

Now I think the quickest way to do this is by plugging in each of the base lengths and see which one satisfies both equations instead of solving them. The only problem with this approach is we tend to select the choice that seems to be correct without checking all of the choices. You can quickly eliminate A, C and E because those will not give an exact area of 48. B and D both give area of 48 - TRAP! Since the question asks for the longest possible side, only choice D is correct.

you can only apply Pythagorean theorem to right triangles. the question doesn't say it's a right triangle.

 

[Lnguyen7]

you can only apply Pythagorean theorem to right triangles. the question doesn't say it's a right triangle.

If you draw an isosceles triangle with a height, you will see two right triangles. Hence side^2 = (b/2)^2 + h^2 and not side^2 = b^2 + h^2

 

[DrNutmeg]

ahh I see.