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19

Part of 2015 AMC 10

Problems(2)

Section of Right Triangle

Source: 2015 AMC 10A #19

2/4/2015
The isosceles right triangle ABCABC has right angle at CC and area 12.512.5. The rays trisecting ∠ACB\angle{ACB} intersect ABAB at DD and EE. What is the area of △CDE\triangle{CDE}?
<spanclass=′latex−bold′>(A)</span>523<spanclass=′latex−bold′>(B)</span>503−754<spanclass=′latex−bold′>(C)</span>1538<spanclass=′latex−bold′>(D)</span>50−2532<spanclass=′latex−bold′>(E)</span>256<span class='latex-bold'>(A) </span>\frac{5\sqrt{2}}{3}\qquad<span class='latex-bold'>(B) </span>\frac{50\sqrt{3}-75}{4}\qquad<span class='latex-bold'>(C) </span>\frac{15\sqrt{3}}{8}\qquad<span class='latex-bold'>(D) </span>\frac{50-25\sqrt{3}}{2}\qquad<span class='latex-bold'>(E) </span>\frac{25}{6}
geometrytrigonometrysymmetrytrig identitiesLaw of SinesanglesAMC
A triangle

Source: 2015 AMC 10B/12B #19

2/26/2015
In △ABC\triangle{ABC}, ∠C=90∘\angle{C} = 90^{\circ} and AB=12AB = 12. Squares ABXYABXY and ACWZACWZ are constructed outside of the triangle. The points X,Y,ZX, Y, Z, and WW lie on a circle. What is the perimeter of the triangle?
<spanclass=′latex−bold′>(A)</span> 12+93<spanclass=′latex−bold′>(B)</span> 18+63<spanclass=′latex−bold′>(C)</span> 12+122<spanclass=′latex−bold′>(D)</span> 30<spanclass=′latex−bold′>(E)</span> 32 <span class='latex-bold'>(A)</span>\ 12+9\sqrt{3}\qquad<span class='latex-bold'>(B)</span>\ 18+6\sqrt{3}\qquad<span class='latex-bold'>(C)</span>\ 12+12\sqrt{2}\qquad<span class='latex-bold'>(D)</span>\ 30\qquad<span class='latex-bold'>(E)</span>\ 32
geometryperimeterAMC 10AMC