Difference between revisions of "1964 AHSME Problems/Problem 22"
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The entire quadrilateral <math>ABEF</math> has area <math>\frac{1}{4} + \frac{1}{6} = \frac{5}{12}</math>. This is <math>5</math> times larger than the area of <math>\triangle DFE</math>, so the ratio is <math>1:5</math>, or <math>\boxed{\textbf{(C)}}</math>. | The entire quadrilateral <math>ABEF</math> has area <math>\frac{1}{4} + \frac{1}{6} = \frac{5}{12}</math>. This is <math>5</math> times larger than the area of <math>\triangle DFE</math>, so the ratio is <math>1:5</math>, or <math>\boxed{\textbf{(C)}}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \frac{A_{DFE}}{A_{ADE}} &= \frac{\frac{1}{3}DA}{DA} = \frac{1}{3} \ | ||
+ | A_{DFE} &= \frac{1}{3}A_{ADE} \ | ||
+ | A_{ADE} &= \frac{1}{2}A_{ADB} = \frac{1}{4}A_{ABCD} \ | ||
+ | \implies A_{DFE} &= \frac{1}{3} \cdot \frac{1}{4} A_{ABCD} = \frac{1}{12} A_{ABCD} \ | ||
+ | A_{ABEF} &= A_{ABD} - A_{DFE} \ | ||
+ | &= \frac{1}{2}A_{ABCD} - \frac{1}{12}A_{ABCD} \ | ||
+ | &= \frac{5}{12}A_{ABCD} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Therefore, <math>\frac{A_{DFE}}{A_{ABEF}} = \frac{\frac{1}{12}}{\frac{5}{12}} = \frac{1}{5}</math>, giving us the answer <math>\boxed{\textbf{(C)}}</math>. -nullptr07 | ||
==See Also== | ==See Also== |
Latest revision as of 22:28, 29 June 2023
Contents
[hide]Problem
Given parallelogram with
the midpoint of diagonal
. Point
is connected to a point
in
so that
. What is the ratio of the area of
to the area of quadrilateral
?
Solution
If it works for a parallelogram , it should also work for a unit square, with
. We are given that
is the midpoint of
, so
. If
is on
, then
. We note that
and
, so
means
, or
, and hence
.
We note that has a base
that is
and an altitude from
to
that is
. Therefore,
.
Quadrilateral can be split into
and
. The first triangle is
of the unit square cut diagonally, so
. The second triangle has base
that is
and height
to
that is
. Therefore,
.
The entire quadrilateral has area
. This is
times larger than the area of
, so the ratio is
, or
.
Solution 2
Therefore,
, giving us the answer
. -nullptr07
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
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