Difference between revisions of "1988 AHSME Problems/Problem 24"
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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 13:29, 27 February 2018
Problem
An isosceles trapezoid is circumscribed around a circle. The longer base of the trapezoid is 
,
and one of the base angles is 
. Find the area of the trapezoid.
Solution
Let the trapezium have diagonal legs of length 
and a shorter base of length 
. 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 
gives the vertical side of these triangles as 
and the horizontal side as 
. 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 
. Also, using the given length of the longer base tells us that 
. Solving these equations simultaneously gives 
and 
, so the height of the trapezium is 
. Thus the area is 
, which is 
.
See also
| 1988 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
