Difference between revisions of "1997 AHSME Problems/Problem 9"
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The other triangle has area <math>1</math>, and the whole square has area <math>4</math>. As above, we find the area of the quadrilateral by subtracting the two triangles, and we get <math>\frac{11}{5}</math>, which is <math>\boxed{C}</math>. | The other triangle has area <math>1</math>, and the whole square has area <math>4</math>. As above, we find the area of the quadrilateral by subtracting the two triangles, and we get <math>\frac{11}{5}</math>, which is <math>\boxed{C}</math>. | ||
+ | |||
+ | == See also == | ||
+ | {{AHSME box|year=1997|num-b=8|num-a=10}} |
Revision as of 09:30, 9 August 2011
Contents
Problem 9
In the figure, is a
square,
is the midpoint of
, and
is on
. If
is perpendicular to
, then the area of quadrilateral
is
Solution 1
Since and
, we have
.
From those two equations, we find that and
Now that we have and
, we can find the area of the bottom triangle
:
The area of left triangle is
The area of the square is .
Thus, the area of the remaining quadrilateral is , and the answer is
Solution 2
Place the square on a coordinate grid so that and
. Line
is
. Line
goes through
and has slope
, so it must be
The intersection of the two lines is , and
thus has coordinates
. The altitude from
to
thus has length
, so the area of the triangle
is
.
The other triangle has area , and the whole square has area
. As above, we find the area of the quadrilateral by subtracting the two triangles, and we get
, which is
.
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 |