Difference between revisions of "1988 AHSME Problems/Problem 24"

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==Solution==
 
==Solution==
 
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Let the trapezium have diagonal legs of length <math>x</math> and a shorter base of length <math>y</math>. Drop altitudes from the endpoints of the shorter base to the longer base to form two right-angled triangles, which are congruent since the trapezium is isosceles. Thus using the base angle of <math>\arcsin(0.8)</math> gives the vertical side of these triangles as <math>0.8x</math> and the horizontal side as <math>0.6x</math>. Now notice that the sides of the trapezium can be seen as being made up of tangents to the circle, and thus using the fact that "the tangents from a point to a circle are equal in length" gives <math>2y + 0.6x + 0.6x = 2x</math>. Also, using the given length of the longer base tells us that <math>y + 0.6x + 0.6x = 16</math>. Solving these equations simultaneously gives <math>x=10</math> and <math>y=4</math>, so the height of the trapezium is <math>0.8 \times 10 = 8</math>. Thus the area is <math>\frac{1}{2}(4+16)(8) = 80</math>, which is <math>\boxed{\text{C}}</math>.
  
  

Latest revision as of 14:29, 27 February 2018

Problem

An isosceles trapezoid is circumscribed around a circle. The longer base of the trapezoid is $16$, and one of the base angles is $\arcsin(.8)$. Find the area of the trapezoid.

$\textbf{(A)}\ 72\qquad \textbf{(B)}\ 75\qquad \textbf{(C)}\ 80\qquad \textbf{(D)}\ 90\qquad \textbf{(E)}\ \text{not uniquely determined}$


Solution

Let the trapezium have diagonal legs of length $x$ and a shorter base of length $y$. Drop altitudes from the endpoints of the shorter base to the longer base to form two right-angled triangles, which are congruent since the trapezium is isosceles. Thus using the base angle of $\arcsin(0.8)$ gives the vertical side of these triangles as $0.8x$ and the horizontal side as $0.6x$. Now notice that the sides of the trapezium can be seen as being made up of tangents to the circle, and thus using the fact that "the tangents from a point to a circle are equal in length" gives $2y + 0.6x + 0.6x = 2x$. Also, using the given length of the longer base tells us that $y + 0.6x + 0.6x = 16$. Solving these equations simultaneously gives $x=10$ and $y=4$, so the height of the trapezium is $0.8 \times 10 = 8$. Thus the area is $\frac{1}{2}(4+16)(8) = 80$, which is $\boxed{\text{C}}$.


See also

1988 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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