# math URGENT!!

the midpoints of the sides of a triangle are (1,1),(4,3),and (3,5). find the area of the triangle. please answer and show how u did it step by step please!!!!

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3. 👁 46
1. It can be shown quite easily that the triangle created by joining the midpoints of a given triangle is similar to the original, and exactly 1/4 of its area.

There are several methods to find the area of a triangle if you know the coordinates of the vertices.
Use the method you learned to find the area of the small triangle, then multiply by 4

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posted by Reiny
2. how do i find the area of the little triangle?
Reiny,

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posted by dan
3. 1. label the points A(1,1) , B(4,3) , C(3,5)
Find the length of AB, let that be our base
AB = √( (4-1)^2 + (3-1)^2 ) = √13
now we need the height, that is, the distance from C to AB
equation for AB:
slope AB = 2/3
equation for AB: 2x - 3y + C = 0
(1,1) lies on it
2 -3 + C = 0 ---> C = 1
equation for AB : 2x - 3y + 1 = 0
the distance form some point (a,b) to the line Ax + By + C = 0 is
D = |Aa + Bb + C|/√(A^2 + B^2)

so our point outside the line is (3,5)
height = |2(3) -3(5) + 1|/√(4+9) = 8/√13

Area of small triangle = (1/2) base x height
= (1/2)(√13)(8/√13) = 4

2. Heron's Formula
Area of triangle = √(s(s-a)(s-b)(s-c) , where a, b, and c are the length, and s = (1/2) the perimeter.

a = BC = √5 = 2.236..
b = AC = √20 = 4.472..
c = AB = √13 = 3.60555. , we found that above
( store these in calculator's memory to keep accuracy)

s = (1/2)(above sum) = 5.156877...
s-a = 2.9208..
s-b = .68474...
s-c = 1.55132..

Area = √(5.156877(2.9208..)(.68474.)(1.5132..)
= √15.9999999999..
= 4 , Wow!

3. Using the cosine law and the above sides, you can find one of the angles.
then Area = (1/2) ab sinC, where C is the angle contained by a and b

4. EASIEST WAY:
list the ordered pairs in a column, repeating the 1st point you wrote down

1 1
4 3
3 5
1 1

Area = (1/2)|sum of downproducts - sum of upproducts|
= (1/2)|3+20+3 - (4+9+5)|
= (1/2)|26 - 18|
= 4

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posted by Reiny
4. so is that the area of big triangle or small?

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posted by dan

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