Difference between revisions of "1979 AHSME Problems/Problem 1"
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Since the dimensions of <math>DEFG</math> are half of the dimensions of <math>ABCD</math>, the area of <math>DEFG</math> is <math>\dfrac{1}{2}\cdot\dfrac{1}{2}</math> of <math>ABCD</math>, so the area of <math>ABCD</math> is <math>\dfrac{1}{4}\cdot72 = \boxed{\textbf{(D) } 18}</math>. | Since the dimensions of <math>DEFG</math> are half of the dimensions of <math>ABCD</math>, the area of <math>DEFG</math> is <math>\dfrac{1}{2}\cdot\dfrac{1}{2}</math> of <math>ABCD</math>, so the area of <math>ABCD</math> is <math>\dfrac{1}{4}\cdot72 = \boxed{\textbf{(D) } 18}</math>. | ||
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+ | == See also == | ||
+ | {{AHSME box|year=1979|before=First question|num-a=2}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 11:48, 5 January 2017
Problem 1
If rectangle ABCD has area 72 square meters and E and G are the midpoints of sides AD and CD, respectively, then the area of rectangle DEFG in square meters is
Solution
Solution by e_power_pi_times_i
Since the dimensions of are half of the dimensions of , the area of is of , so the area of is .
See also
1979 AHSME (Problems • Answer Key • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
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 |
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