Difference between revisions of "1995 AJHSME Problems/Problem 24"
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<math>\text{(A)}\ 6.4 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 7.2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10</math> | <math>\text{(A)}\ 6.4 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 7.2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10</math> | ||
| − | ==Solution== | + | ==Solution 1== |
The answer is C=7.2 | The answer is C=7.2 | ||
Use pythagoras to figure it out. A 4-6 triangle and approx 7.2 | Use pythagoras to figure it out. A 4-6 triangle and approx 7.2 | ||
~Aarav22 | ~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== | ==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 12:16, 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 
![[asy] unitsize(12); pair A,B,C,D,P,Q,W,X,Y,Z; A = (0,0); B = (12,0); C = (20,6); D = (8,6); W = (18,0); X = (30,0); Y = (38,6); Z = (26,6); draw(A--B--C--D--cycle); draw(W--X--Y--Z--cycle); P = (8,0); Q = (758/25,6/25); dot(A); dot(B); dot(C); dot(D); dot(W); dot(X); dot(Y); dot(Z); dot(P); dot(Q); draw(A--B--C--D--cycle); draw(W--X--Y--Z--cycle); draw(D--P); draw(Z--Q); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",P,S); label("$A$",W,SW); label("$B$",X,S); label("$C$",Y,NE); label("$D$",Z,NW); label("$F$",Q,E); [/asy]](http://latex.artofproblemsolving.com/9/e/2/9e2bbb0a9b55e0060cb42c71e75562e282f6032e.png)
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 | ||
