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>\ | + | 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 |
<asy> | <asy> | ||
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<math>\text{(A)} \ 32 \qquad \text{(B)} \ 36 \qquad \text{(C)} \ 39 \qquad \text{(D)} \ 42 \qquad \text{(E)} \ 48</math> | <math>\text{(A)} \ 32 \qquad \text{(B)} \ 36 \qquad \text{(C)} \ 39 \qquad \text{(D)} \ 42 \qquad \text{(E)} \ 48</math> | ||
+ | |||
+ | == Solution == | ||
+ | |||
+ | Connect C and A, and we have a 3-4-5 right triangle and 5-12-13 right triangle. The area of both is <math> \frac{3\cdot4}{2}+\frac{5\cdot12}{2}=36\Rightarrow\boxed{(B)} </math>. | ||
+ | |||
+ | == See also == | ||
+ | {{AHSME box|year=1980|num-b=6|num-a=8}} |
Revision as of 19:08, 31 March 2013
Problem
Sides and of convex polygon have lengths 3, 4, 12, and 13, respectively, and is a right angle. The area of the quadrilateral is
Solution
Connect C and A, and we have a 3-4-5 right triangle and 5-12-13 right triangle. The area of both 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 |