Difference between revisions of "1995 AJHSME Problems/Problem 24"

(Solution: (C) or 7.2.)
(Solution 2)
Line 36: Line 36:
  
 
==Solution 2==
 
==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.
+
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

Problem

In parallelogram $ABCD$, $\overline{DE}$ is the altitude to the base $\overline{AB}$ and $\overline{DF}$ is the altitude to the base $\overline{BC}$. [Note: Both pictures represent the same parallelogram.] If $DC=12$, $EB=4$, and $DE=6$, then $DF=$

[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]

$\text{(A)}\ 6.4 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 7.2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$

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 (ProblemsAnswer KeyResources)
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