Find the area of the triangle formed by joining the midpoints of the sides of
the triangle whose vertices are (0,2) ,(2,1) and (0,3). Find the ratio of this area
to the area of the given triangle.
Answers
Step-by-step explanation:
Given :-
The triangle whose vertices are (0,2) ,(2,1) and (0,3).
To find :-
I) Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are (0,2) ,(2,1) and (0,3).
ii)Find the ratio of this area to the area of the given triangle.
Solution :-
Given vertices of a triangle are (0,2),(2,1),(0,3)
Finding Area of ∆ABC:-
Let (x1, y1) = A(0,2) => x1 = 0 and y1 = 2
Let (x2, y2) = B(2,1) => x2 = 2 and y2 = 1
Let (x3, y3) = C(0,3) => x3 = 0 and y3 = 3
We know that
Area of a triangle formed by the vertices (x1, y1) , (x2, y2) and (x3, y3) is
∆=(1/2) | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) | sq.units
On Substituting these values in the above formula
=> ∆ = (1/2) | 0(1-3)+2(3-2)+0(2-1) |
=> ∆ = (1/2) | 0(-2)+2(1)+0(1) |
=> ∆ = (1/2) | 0-2+0 |
=> ∆ = (1/2) |-2 |
=> ∆ = (1/2)(2)
=> ∆ = 2/2
=> ∆ = 1 sq.unit
Area of the triangle ABC = 1 sq.unit ----------(1)
Let the mid points of AB,BC and CA are D,E and F respectively
Finding mid point of AB :-
Let (x1, y1) = A(0,2) => x1 = 0 and y1 = 2
Let (x2, y2) = B(2,1) => x2 = 2 and y2 = 1
We know that
The mid point of the linesegment joining the points (x1, y1) and (x2, y2) is ({x1+x2}/2 , {y1+y2}/2)
=> Mid point of AB = ({0+2}/2 , {2+1}/2)
=> D = (2/2,3/2)
=> D = (1,3/2)
Finding mid point of BC :-
Let (x1, y1) = B(2,1) => x1 = 2 and y1 = 1
Let (x2, y2) = C(0,3) => x2 = 0 and y2 = 3
We know that
The mid point of the linesegment joining the points (x1, y1) and (x2, y2) is ({x1+x2}/2 , {y1+y2}/2)
=> Mid point of BC = ({2+0}/2 , {1+3}/2)
=> E= (2/2,4/2)
=> E = (1,2)
Finding mid point of AC :-
Let (x1, y1) = A(0,2) => x1 = 0 and y1 = 2
Let (x2, y2) = C(0,3) => x2 = 0 and y2 = 3
We know that
The mid point of the linesegment joining the points (x1, y1) and (x2, y2) is ({x1+x2}/2 , {y1+y2}/2)
=> Mid point of AC = ({0+0}/2 , {2+3}/2)
=> F= (0/2,5/2)
=> F = (0,5/2)
Finding area of ∆DEF :-
Let (x1, y1) = D(1,3/2) => x1 = 1 and y1 = 3/2
Let (x2, y2) = E(1,2)=> x2 = 1 and y2 = 2
Let (x3, y3) =F(0,5/2)=> x3 = 0 and y3 = 5/2
We know that
Area of a triangle formed by the vertices (x1, y1) , (x2, y2) and (x3, y3) is
∆=(1/2) | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) | sq.units
On Substituting these values in the above formula
=> ∆ = (1/2) | 1(2-(5/2))+1(5/2)-(3/2))+0(3/2)-2) |
=> ∆ = (1/2) | 1(4-5)/2 +(5-3)/2 +0|
=> ∆ = (1/2) |(-1/2)+(2/2)+0 |
=> ∆ = (1/2) |-1/2+1|
=> ∆=(1/2)(1/2)
=> ∆ = 1/4
=> ∆ = 1/4 sq.unit
Area of the triangle DEF = 1/4 sq.units ----------(2)
Ratio of area ∆DEF to area of ∆ABC
=> 1/4:1
=> 1:4
Answer :-
1)The area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are (0,2) ,(2,1) and (0,3) is 1/4 sq.units
ii)The ratio of this area to the area of the given triangle is 1 :4
Used formulae:-
- The mid point of the linesegment joining the points (x1, y1) and (x2, y2) is ({x1+x2}/2 , {y1+y2}/2)
- Area of a triangle formed by the vertices (x1, y1) , (x2, y2) and (x3, y3) is
- ∆=(1/2) | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) | sq.units
Here,
- D,E and F are the midpoints of AB, BC and CA of triangle respectively
Now,