During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "2005 iTest Problems/Problem 9"

(Created page with "==Problem== Find the area of the triangle with vertices of <math>(1,2)</math>, <math>(1,10)</math>, and <math>(5, 5)</math>. ==Solution 1== Since <math>(1,2)</math> and <math...")
 
(Solution 1)
Line 4: Line 4:
 
==Solution 1==
 
==Solution 1==
 
Since <math>(1,2)</math> and <math>(1,10)</math> are collinear, let that be the base. It therefore has length 8. The distance from <math>(5, 5)</math> to this line is the same as the distance from <math>(5,5)</math> to <math>(1,5)</math>, which is equal to 4. The area of this triangle is <math>\frac{8\cdot4}{2}=\boxed{16}</math>.
 
Since <math>(1,2)</math> and <math>(1,10)</math> are collinear, let that be the base. It therefore has length 8. The distance from <math>(5, 5)</math> to this line is the same as the distance from <math>(5,5)</math> to <math>(1,5)</math>, which is equal to 4. The area of this triangle is <math>\frac{8\cdot4}{2}=\boxed{16}</math>.
 +
 +
==See Also==
 +
{{iTest box|year=2005|num-b=8|num-a=10}}

Revision as of 16:34, 13 October 2025

Problem

Find the area of the triangle with vertices of $(1,2)$, $(1,10)$, and $(5, 5)$.

Solution 1

Since $(1,2)$ and $(1,10)$ are collinear, let that be the base. It therefore has length 8. The distance from $(5, 5)$ to this line is the same as the distance from $(5,5)$ to $(1,5)$, which is equal to 4. The area of this triangle is $\frac{8\cdot4}{2}=\boxed{16}$.

See Also

2005 iTest (Problems, Answer Key)
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 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 TB1 TB2 TB3 TB4