Find the area of the triangle formed by joining the midpoints of the triangle whose vertices are [0,-1] ,[2,1]and [0,3].Find the ratio of this area to the area of the given triangle.
Answers
Step-by-step explanation:
Let A(0,-1) , B(2,1) ,C(0,3) are vertices of the
Triangle.
D , E , F are midpoints of BC , CA and AB.
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The midpoint of the line segment joining
the points (x1,y1) and (x2 , y2 ) is P( x , y ).
x = ( x1 + x2 )/2 ;
y = ( y1 + y2 )/2
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Now ,
i ) mid point of B(2,1) , C(0,3) is D( x , y )
x = ( 2 + 0 )/2 = 1
y = ( 1 + 3 )/2 = 2
D = ( 1 , 2 )
Similarly ,
ii ) mid point of C( 0,3) , A(0,-1) = E( 0,1)
iii ) mid point of A(0,-1), B(2,1) = F(1,0)
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The area of the triangle formed by the
vertices ( x1,y1 ), ( x2, y2 ) , ( x3 , y3 ) is
1/2|x1 ( y2 - y3 )+x2( y3 - y1 ) +x3( y2 - y1) |
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iv ) Area of the triangle A( 0, -1), B( 2 , 1 )
C( 0 ,3 ) is
1/2|0( 1 - 3 ) + 2( 3 + 1 ) + 0 ( -1-1 ) |
= 1/2 | 8 |
= 4 sq units
v ) Area of the triangle D( 1,2 ) , E( 0 , 1 ),
and F( 1 , 0 ) is
1/2 | 1( 1 - 0 ) + 0( 0 - 2 ) + 1( 2 - 1 ) |
= 1/2 | 2 |
= 1 sq units
vi )
ratio = ( area ∆ABC )/( area ∆DEF )
= 4/1
= 4 : 1
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Answer:
Find the ratio of this area to the area of the given triangle. There is no any given triangle