Question

Find area of triangle formed by joining the midpoints of sides of triangle whose vertices are(0;-1) (2;1) (0;3) also find the ratio of areas of two triangles

Answer 1

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

Let the midpoint of AB be P
BC be Q
AC be R

Now let us join the points P,Q,R

we get triangle PQR

Area of Triangle ABC:
Area of Triangle ABC =1/2 [x1(y2-y3) +x2 (y3-y1]+x3(y1-y2)]

here
x1=0, y1= -1
x2=2, y2=1
x3=0, y3=3
Substituting the values we get
=1/2[0(1-3)+2(3-(-1)+0(-1-1)]
=1/2 [0+2(3+1)+0]=1/2[8]=4

Finding the area of Triangle PQR:

coordinates of P=(x1+x2/2 , y1+y2/2)
=(0+2/2, -1+1/2)
=(1,0)

similarly
 Coordinates of Q=(2+0/2, 1+3/2)=(1,2)

Coordinates of R:=(0+0/2, -1+3/2)=(0,1)

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

x1=1,y1=0
x2=1, y2=2
x3=0,y3=1                                    
Area of triangle PQR
=1/2[1[2-1) +1(1-0)+0(0-2)]
=1/2[1x1 +1x1+0]=1/2(1+1]
=2/2=1

Hence the required  ratio is:

Area of ΔPQR/Area of ΔABC =1/4

Answer 2

Answer:

It may Help you...

plzz follow me...

#RaghavPandey...