Difference between revisions of "1980 AHSME Problems/Problem 7"
J314andrews (talk | contribs) (Shrunk diagram because it was huge.) |
|||
(2 intermediate revisions by the same user not shown) | |||
Line 1: | Line 1: | ||
==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> | ||
+ | size(150); | ||
defaultpen(linewidth(0.7)+fontsize(10)); | defaultpen(linewidth(0.7)+fontsize(10)); | ||
real r=degrees((12,5)), s=degrees((3,4)); | real r=degrees((12,5)), s=degrees((3,4)); | ||
Line 24: | Line 25: | ||
== Solution == | == Solution == | ||
− | + | Draw <math>\overline{AC}</math>. By the Pythagorean Theorem, <math>AC = \sqrt{AB^2 + BC^2} = \sqrt{3^2+4^2} = 5</math>. Notice that <math>AD^2 = AC^2 + CD^2 = 169</math>, so by the converse of the Pythagorean Theorem, <math>\triangle ACD</math> is right. Thus 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}} |
Latest revision as of 17:27, 25 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 . By the Pythagorean Theorem,
. Notice that
, so by the converse of the Pythagorean Theorem,
is right. Thus 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.