Question 1132689

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<pre>
The side length of the larger square is (6+12) = 18 centimeters, so its area is  {{{18^2}}} = 324 cm^2.


Now, there are 4 congruent right angled triangles  ALD,  DKC, CBJ  and  BIA.


The area of the smaller square is 324 cm^2  <U>MINUS</U>  4 times the area of the triangle ALD.


Triangle ALD is similar to triangle ADH (they both are right angled triangles and have common acute angle DAL).


Triangle ALD has the hypotenuse  AD = 6+12 = 18 cm.

Triangle ADH has the hypotenuse  AH = {{{sqrt(18^2 + 6^2)}}} = {{{6*sqrt(3^2 + 1^2)}}} = {{{6*sqrt(10)}}}.


The ratio of the hypotenuses  |AH|/|AD| = {{{(6*sqrt(10))/18}}} = {{{sqrt(10)/3}}}  (the similarity coefficient).


Hence, the ratio of the areas of the triangles  ADH  and  ALD  is the square of the similarity coefficient, i.e.  {{{10/9}}}.


The area of the triangle  ADH  is  {{{(1/2)*18*6}}} = 54 cm^2.


Hence, the area of the triangle  ALD  is  {{{54*(9/10)}}} cm^2 = 48.6 cm^2.


Then the area of the smaller square is (as I explained it above)  324 - 4*48.6 = 129.6 cm^2= 129 6/10=129 3/5 cm^2 cm^2     <U>ANSWER</U>
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B) {{{129}}}{{{ 3/5 }}}