Difference between revisions of "1995 AJHSME Problems/Problem 24"
Jonohampter (talk | contribs) (→Solution: (C) or 7.2.) |
Jonohampter (talk | contribs) (→Solution 2) |
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==Solution 2== | ==Solution 2== | ||
− | Use the area of parallelogram ABCD, which is 36, to find the altitude DF by using 36=\frac | + | Use the area of parallelogram ABCD, which is 36, to find the altitude DF by using 36=\frac{1}{2}BC*DF. We find BC by using Pythagorean theorem on triangle DEA which is a 6,8,10 triangle. We get that DA must be 10, and therefore BC must be 10. After this, we use 36=\frac(1)(2)10*DF, which you can solve to get 7.2=DF and therefore, the answer is (C) or 7.2. |
==See Also== | ==See Also== | ||
{{AJHSME box|year=1995|num-b=23|num-a=25}} | {{AJHSME box|year=1995|num-b=23|num-a=25}} |
Revision as of 12:17, 22 December 2024
Contents
Problem
In parallelogram , is the altitude to the base and is the altitude to the base . [Note: Both pictures represent the same parallelogram.] If , , and , then
Solution 1
The answer is C=7.2 Use pythagoras to figure it out. A 4-6 triangle and approx 7.2
~Aarav22
Solution 2
Use the area of parallelogram ABCD, which is 36, to find the altitude DF by using 36=\frac{1}{2}BC*DF. We find BC by using Pythagorean theorem on triangle DEA which is a 6,8,10 triangle. We get that DA must be 10, and therefore BC must be 10. After this, we use 36=\frac(1)(2)10*DF, which you can solve to get 7.2=DF and therefore, the answer is (C) or 7.2.
See Also
1995 AJHSME (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 | ||
All AJHSME/AMC 8 Problems and Solutions |