Difference between revisions of "1992 AHSME Problems/Problem 24"
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− | Assume that the height of parallelogram <math>ABFG</math> | + | Assume that the height of parallelogram <math>ABFG</math> with respect to base <math>AB</math> is <math>x</math>. Then, the area of parallelogram <math>ABFG</math> is <math>AB * x</math>. The area of triangle <math>EFG</math> is <math>\frac{AB * x}{2}</math>, which is half of the area of parallelogram <math>ABFG</math>. |
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− | Thus, <math>[ | + | Thus, <math>[EFHG] = [EFG] + [FGH] = 1/2[ABFG] + 1/2[GFCD] = 1/2[ABCD] = 1/2(10) = \boxed{5}</math> |
== See also == | == See also == |
Latest revision as of 19:17, 4 September 2022
Contents
Problem
Let be a parallelogram of area
with
and
. Locate
and
on segments
and
, respectively, with
. Let the line through
parallel to
intersect
at
. The area of quadrilateral
is
Solution 1
Use vectors. Place an origin at
, with
. We know that
, and also
, and now we can find the area of
by dividing it into two triangles and using cross-products (the expressions simplify using the fact that the cross-product distributes over addition, it is anticommutative, and a vector crossed with itself gives zero).
Solution 2
We note that is a parallelogram because
and
. Using the same reasoning,
is also a parallelogram.
Assume that the height of parallelogram with respect to base
is
. Then, the area of parallelogram
is
. The area of triangle
is
, which is half of the area of parallelogram
.
Likewise, the area of triangle is half the area of parallelogram
.
Thus,
See also
1992 AHSME (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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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