Question

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Answer 1

$Hey\:!!..$

The answer goes here....

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》To find :

Area of the triangle formed by joining the midpoints and the ratio of this area to the area of the given triangle.

》Given :

Vertices of triangle = $A(0,-1)$ ; $B(2,1)$ , $C(0,3)$

》Solution :

Given vertices of triangle are $A(0,-1)$ ; $B(2,1)$ , $C(0,3)$

Now, let the midpoint of AB be P, BC be Q, CA be R.

Construction : Join P, Q, R

Joining the points we get $\triangle PQR$

Now, according to question we need to find the area of $\triangle PQR$ & $\triangle ABC$

So, area of $\triangle ABC$ -

In $\triangle ABC$ ,

$x_{1}=0$ & $y_{1}=-1$

$x_{2}=2$ & $y_{2}=1$

$x_{3}=0$ & $y_{3}=3$

Area of $\triangle$ = $\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

⟹ $\frac{1}{2}[0(1-3)+2(3-(-1))+0(-1-1)]$

⟹ $\frac{1}{2}[0+2(3+1)+0]$

⟹ $\frac{1}{2}(8)$

⟹ $4$

Now, area of $\triangle PQR$ -

In order to find the area of $\triangle PQR$ we first need to find the coordinates of P, Q and R.

Since, P is midpoint of AB. Therefore coordinates of P -

⟹ $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

⟹ $(\frac{0+2}{2}, \frac {-1+1}{2})$

⟹ $(\frac{2}{2}, \frac{0}{2})$

⟹ $(1,0)$

Since, Q is midpoint of BC. Therefore coordinates of Q -

⟹ $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

⟹ $(\frac{2+0}{2}, \frac {1+3}{2})$

⟹ $(\frac{2}{2}, \frac{4}{2})$

⟹ $(1,2)$

Since, R is midpoint of AC. Therefore coordinates of R -

⟹ $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

⟹ $(\frac{0+0}{2}, \frac{-1+3}{2})$

⟹ $(\frac{0}{2}, \frac{2}{2})$

⟹ $(0,1)$

So, the coordinates are $P(1,0)$ , $Q(1,2)$ , $R(0,1)$

Now,

$x_{1}=1$ & $y_{1}=0$

$x_{2}=1$ & $y_{2}=2$

$x_{3}=0$ & $y_{3}=1$

Area of $\triangle$ = $\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

⟹ $\frac{1}{2}[1(2-1)+1(1-0)+0(0-2)]$

⟹ $\frac{1}{2}[1(1)+1(1)+0]$

⟹ $\frac{1}{2}(2)$

⟹ $1$

Now, the required ratio -

$\frac{area\:of\:\triangle\:PQR}{area\:of\:\triangle\:PQR}$ = $\frac{1}{4}$

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Thanks !!..

Answer 2

Answer:

1:4.

Step-by-step explanation:

Let the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).

Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by

D = (0+2/2 , -1+1/2) = (1,0)

E = (0+0/2 , -3-1/2) = (0,1)

F = (2+0/2 , 1+3/2) = (1,2)

Area of a triangle = 1/2 {x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)}

Area of ΔDEF = 1/2 {1(2-1) + 1(1-0) + 0(0-2)}

                       = 1/2 (1+1) = 1 square units

Area of ΔABC = 1/2 [0(1-3) + 2{3-(-1)} + 0(-1-1)]

                        = 1/2 {8} = 4 square units

Therefore, the required ratio is 1:4.