The two sides of a right triangle are 10 square root 2

Question

The two sides of a right triangle are 10 \sqrt{2}  and 5 \sqrt{2} . Which of the following can be the value of the third side?

I. 5 \sqrt{2} 
II. II. 5 \sqrt{6} 
III. III. 5 \sqrt{10}

A) II only 
B) III only 
C) I and II only 
D) II and III only 
E) I, II, and III

Salsanousi · Answer

Square root of 2 is approx. 1.414

10* 1.414 = 14.14 (1)
5 * 1.414 = 7.07 (2)

The third side is (1) - (2) < x < (1) + (2)

5 sqrt 2 < x < 15 sqrt 2

Sqrt 6 is slightly greater than sqrt 4

Sqrt 10 is slightly greater than sqrt 9

II and III

Answe choice D

Posted from my mobile device

KanishkM · Answer

Difference of two sides < Third side < Sum of two sides 
5√2 < Third side < 15√2 (1)

taking a square of (1)
50 <  (Third side)^2  < 450

Answer is D

fskilnik · Answer

?\,\,\,:\,\,\,3{\rm{rd}}\,\,{\rm{side}}\,\,\,\,\left( {\,{\rm{length}}\,} \right)

\left. \matrix{
 {\left( {10\sqrt 2 } \right)^2} = {\left( {5\sqrt 2 } \right)^2} + {x^2}\,\,\, \Rightarrow \,\,\,\,200 = 50 + {x^2}\,\,\,\,\mathop \Rightarrow \limits^{x\, > \,0} \,\,\,\,x = 5\sqrt 6 \,\,\,\,\, \hfill \cr 
 {y^2} = {\left( {5\sqrt 2 } \right)^2} + {\left( {10\sqrt 2 } \right)^2}\,\,\, \Rightarrow \,\,\,\,{y^2} = 250\,\,\,\,\mathop \Rightarrow \limits^{y\, > \,0} \,\,\,\,y = 5\sqrt {10} \,\, \hfill \cr} \right\}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\,\,:\,\,\,5\sqrt 6 \,\,{\rm{or}}\,\,5\sqrt {10} \,\,\,\, \Rightarrow \,\,\,\,\left( D \right)

Important: in the case of right triangles, the  a^2=b^2+c^2  equality guarantees that the (positive) numbers obtained will be lengths of a viable triangle!
(In short: there is no need to check whether each side is less than the sum of the other two, for instance.)

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

bumpbot · Answer

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The two sides of a right triangle are 10 square root 2

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The two sides of a right triangle are 10 square root 2

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