Difference between revisions of "1992 AHSME Problems/Problem 24"
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\text{(E) } 6</math> | \text{(E) } 6</math> | ||
− | == Solution == | + | == Solution 1 == |
− | <math>\fbox{C}</math> Use vectors. Place an origin at <math>A</math>, with <math>B = p, D = q, C = p + q</math>. We know that <math>\|p \times q\|=10</math>, and also <math>E=\frac{2}{3}p, F=p+\frac{2}{5}q, G = \frac{2}{5}q</math>, and now we can find the area of <math>EFHG</math> by dividing it into two triangles and using cross-products. | + | <math>\fbox{C}</math> Use vectors. Place an origin at <math>A</math>, with <math>B = p, D = q, C = p + q</math>. We know that <math>\|p \times q\|=10</math>, and also <math>E=\frac{2}{3}p, F=p+\frac{2}{5}q, G = \frac{2}{5}q</math>, and now we can find the area of <math>EFHG</math> 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). |
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+ | == Solution 2 == | ||
+ | We note that <math>ABFG</math> is a parallelogram because <math>AG = BF = 2</math> and <math>AG \parallel BF</math>. Using the same reasoning, <math>GFCD</math> is also a parallelogram. | ||
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+ | 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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+ | Likewise, the area of triangle <math>FGH</math> is half the area of parallelogram <math>GFCD</math>. | ||
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+ | Thus, <math>[EFHG] = [EFG] + [FGH] = 1/2[ABFG] + 1/2[GFCD] = 1/2[ABCD] = 1/2(10) = \boxed{5}</math> | ||
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+ | == Solution 3 == | ||
+ | After drawing and labeling your diagram, connect line <math>GF</math> and notice that there are now two triangles <math>EGF</math> and <math>GFH</math>. <math>ABFG</math> and <math>GFCD</math> are both parallelograms with a triangle inscribed in them with one of the sides as bases, so the combined area of the triangles, which forms <math>EFHG</math>, is just the area over <math>2</math>, so the answer is just <math>\frac{10}{2}=\boxed{5}</math> | ||
== See also == | == See also == |
Latest revision as of 15:49, 18 February 2025
Contents
[hide]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,
Solution 3
After drawing and labeling your diagram, connect line and notice that there are now two triangles
and
.
and
are both parallelograms with a triangle inscribed in them with one of the sides as bases, so the combined area of the triangles, which forms
, is just the area over
, so the answer is just
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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