Difference between revisions of "2002 AMC 8 Problems/Problem 20"
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==Problem== | ==Problem== | ||
− | The area of triangle <math>XYZ</math> is 8 square inches. Points <math>A</math> and <math>B</math> are midpoints of congruent segments <math> \overline{XY} </math> and <math> \overline{XZ} </math>. Altitude <math> \overline{XC} </math> bisects <math> \overline{YZ} </math>. | + | The area of triangle <math>XYZ</math> is 8 square inches. Points <math>A</math> and <math>B</math> are midpoints of congruent segments <math>\overline{XY}</math> and <math>\overline{XZ}</math>. Altitude <math>\overline{XC}</math> bisects <math>\overline{YZ}</math>. The area (in square inches) of the shaded region is |
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==Solution 1== | ==Solution 1== | ||
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We know the area of triangle <math>XYZ</math> is <math>8</math> square inches. The area of a triangle can also be represented as <math>\frac{bh}{2}</math> or in this problem <math>\frac{XC\cdot YZ}{2}</math>. By solving, we have <cmath>\frac{XC\cdot YZ}{2} = 8,</cmath> <cmath>XC\cdot YZ = 16.</cmath> | We know the area of triangle <math>XYZ</math> is <math>8</math> square inches. The area of a triangle can also be represented as <math>\frac{bh}{2}</math> or in this problem <math>\frac{XC\cdot YZ}{2}</math>. By solving, we have <cmath>\frac{XC\cdot YZ}{2} = 8,</cmath> <cmath>XC\cdot YZ = 16.</cmath> | ||
With SAS congruence, triangles <math>XCY</math> and <math>XCZ</math> are congruent. Hence, triangle <math>XCY = \frac{8}{2} = 4</math>. (Let's say point <math>D</math> is the intersection between line segments <math>XC</math> and <math>AB</math>.) We can find the area of the trapezoid <math>ADCY</math> by subtracting the area of triangle <math>XAD</math> from <math>4</math>. | With SAS congruence, triangles <math>XCY</math> and <math>XCZ</math> are congruent. Hence, triangle <math>XCY = \frac{8}{2} = 4</math>. (Let's say point <math>D</math> is the intersection between line segments <math>XC</math> and <math>AB</math>.) We can find the area of the trapezoid <math>ADCY</math> by subtracting the area of triangle <math>XAD</math> from <math>4</math>. | ||
− | We find the area of triangle <math>XAD</math> by the <math>\frac{bh}{2}</math> formula- <math>\frac{XD\cdot AD}{2} = \frac{\frac{XC}{2}\cdot AD}{2}</math>. <math>AD</math> is <math>\frac{1}{4}</math> of <math>YZ</math>. The area of <math>XAD</math> is <cmath>\frac{\frac{XC}{2}\cdot \frac{YZ}{4}}{2} = \frac{16}{16} = 1</cmath>. | + | We find the area of triangle <math>XAD</math> by the <math>\frac{bh}{2}</math> formula- <math>\frac{XD\cdot AD}{2} = \frac{\frac{XC}{2}\cdot AD}{2}</math>. <math>AD</math> is <math>\frac{1}{4}</math> of <math>YZ</math> from solution 1. The area of <math>XAD</math> is <cmath>\frac{\frac{XC}{2}\cdot \frac{YZ}{4}}{2} = \frac{16}{16} = 1</cmath>. |
Therefore, the area of the shaded area- trapezoid <math>ADCY</math> has area <math>4-1 = \boxed{\text{(D)}\ 3}</math>. | Therefore, the area of the shaded area- trapezoid <math>ADCY</math> has area <math>4-1 = \boxed{\text{(D)}\ 3}</math>. | ||
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- sarah07 | - sarah07 | ||
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+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=zwy5U5IQi88 ~David | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=19|num-a=21}} | {{AMC8 box|year=2002|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:36, 26 June 2024
Contents
Problem
The area of triangle is 8 square inches. Points
and
are midpoints of congruent segments
and
. Altitude
bisects
. The area (in square inches) of the shaded region is
Solution 1
We know the area of triangle is
square inches. The area of a triangle can also be represented as
or in this problem
. By solving, we have
With SAS congruence, triangles and
are congruent. Hence, triangle
. (Let's say point
is the intersection between line segments
and
.) We can find the area of the trapezoid
by subtracting the area of triangle
from
.
We find the area of triangle by the
formula-
.
is
of
from solution 1. The area of
is
.
Therefore, the area of the shaded area- trapezoid has area
.
- sarah07
Video Solution
https://www.youtube.com/watch?v=zwy5U5IQi88 ~David
See Also
2002 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.