Difference between revisions of "2013 AMC 10B Problems/Problem 16"
m (→Problem) |
(→Solution 3) |
||
Line 42: | Line 42: | ||
From the solution above, we can find that the lengths of the diagonals are <math>6</math> and <math>4.5</math>. Now, since the diagonals of AEDC are perpendicular, we use the area formula to find that the total area is <math>\frac{6\times4.5}{2} = \frac{27}{2} = 13.5 = \boxed{\textbf{(B)} 13.5}</math> | From the solution above, we can find that the lengths of the diagonals are <math>6</math> and <math>4.5</math>. Now, since the diagonals of AEDC are perpendicular, we use the area formula to find that the total area is <math>\frac{6\times4.5}{2} = \frac{27}{2} = 13.5 = \boxed{\textbf{(B)} 13.5}</math> | ||
+ | |||
+ | ===Solution 4=== | ||
+ | |||
+ | From the solutions above, we know that the sides CP and AP are 3 and 4 respectively because of the properties of medians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its simple computation really, nothing large). Then we can find the areas of the remaining triangles based on side and ratio length of the bases. | ||
== See also == | == See also == |
Revision as of 08:50, 27 September 2018
Contents
[hide]Problem
In triangle , medians and intersect at , , , and . What is the area of ?
Solution
Solution 1
Let us use mass points: Assign mass . Thus, because is the midpoint of , also has a mass of . Similarly, has a mass of . and each have a mass of because they are between and and and respectively. Note that the mass of is twice the mass of , so AP must be twice as long as . PD has length , so has length and has length . Similarly, is twice and , so and . Now note that triangle is a right triangle with the right angle . Since the diagonals of quadrilaterals , and , are perpendicular, the area of is
Solution 2
Note that triangle is a right triangle, and that the four angles (angles and ) that have point are all right angles. Using the fact that the centroid () divides each median in a ratio, and . Quadrilateral is now just four right triangles. The area is
Solution 3
From the solution above, we can find that the lengths of the diagonals are and . Now, since the diagonals of AEDC are perpendicular, we use the area formula to find that the total area is
Solution 4
From the solutions above, we know that the sides CP and AP are 3 and 4 respectively because of the properties of medians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its simple computation really, nothing large). Then we can find the areas of the remaining triangles based on side and ratio length of the bases.
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.