Difference between revisions of "1980 AHSME Problems/Problem 7"
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==Problem== | ==Problem== | ||
− | Sides <math>AB,BC,CD</math> and <math>DA</math> of convex polygon <math>ABCD</math> have lengths 3, 4, 12, and 13, respectively, and <math>\angle CBA</math> is a right angle. The area of the quadrilateral is | + | Sides <math>AB</math>, <math>BC</math>, <math>CD</math> and <math>DA</math> of convex polygon <math>ABCD</math> have lengths <math>3</math>, <math>4</math>, <math>12</math>, and <math>13</math>, respectively, and <math>\angle CBA</math> is a right angle. The area of the quadrilateral is |
<asy> | <asy> | ||
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== Solution == | == Solution == | ||
− | + | Draw <math>\overline{AC}</math>. Then <math>\triangle ABC</math> is a <math>3-4-5</math> right triangle and <math>\triangle ACD</math> is a <math>5-12-13</math> right triangle. Therefore, the area of <math>ABCD</math> is <math> \frac{3\cdot4}{2}+\frac{5\cdot12}{2}=\boxed{(\textbf{B})\ 36} </math>. | |
== See also == | == See also == | ||
{{AHSME box|year=1980|num-b=6|num-a=8}} | {{AHSME box|year=1980|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:36, 14 July 2025
Problem
Sides ,
,
and
of convex polygon
have lengths
,
,
, and
, respectively, and
is a right angle. The area of the quadrilateral is
Solution
Draw . Then
is a
right triangle and
is a
right triangle. Therefore, the area of
is
.
See also
1980 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.