Question

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.

Answer 1

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

Answer 2

$\begin{gathered} \sf△ABC \\= \small \frac{1}{2} |0(1 - 3) + 2(3 - 2) + 0(2 - 1)| \\ \\ = \sf \frac{1}{2} |2| \\ \\ = \sf \: 1 \: sq \: unit\end{gathered}$

Here,

• D,E and F are the midpoints of AB, BC and CA of triangle respectively

$\begin{gathered}\small \sf Coordinate \: of \: D=( \frac{2 + 0}{2} \: , \: \frac{1 + 2}{2} ) \\ \\ = ( 1 \: , \: \frac{3}{2} )\end{gathered}$

$\begin{gathered}\small \sf Coordinate \: of \: E=( \frac{2 + 0}{2} \: , \: \frac{1 + 3}{2} ) \\ \\ = (1,2)\end{gathered}$

$\begin{gathered} \small\sf coordinates \: of \: F= (\frac{0 + 0}{2} \: , \: \frac{3 + 2}{2} ) \\ \\ = (0, \frac{5}{2} )\end{gathered}$

Now,

$\begin{gathered} \sf ar(△DEF)\\= \small \frac{1}{2} |1( 2 - \frac{5}{2}) + 1( \frac{5}{2} - \frac{3}{2} )+ 0( \frac{3}{2} - 2) | \\ \\ = \sf \frac{1}{2} | - \dfrac{1}{2} + 1 | \\ \\ = \sf \frac{1}{2} | \frac{1}{2} | \\ \\ = \sf \frac{1}{4} \: sq \: unit\end{gathered}$

⇒ar(△DEF):ar(△ABC)=1:4