Question

In Fig. 6, ABC is a triangle coordinates of whose vertex A are (0, -1). D and E respectively are the mid-points of the sides AB and AC and their coordinates are (1, 0) and (0, 1) respectively. If F is the mid-point of BC, find the areas of ΔABC and ΔDF

Answer 1

The answer is in the image.

Hope, it is helpful to you...

Answer 2

Solution :

Given In ∆ABC , D,F and E are the

midpoints of sides AB, BC and CA

respectively ,

Let A( x1,y1 ) = ( 0 , -1 )

D( x2 , y2 ) = ( 1 , 0 ),

E( x3 , y3 ) = ( 0 , 1 ),

Now ,

i ) Area of ∆ADE

= 1/2|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|

=1/2|0( 0-1) + 1( 1 +1 ) + 0( -1 - 0 ) |

= 1/2 | 2 |

= 1 sq units ---( 1 )

*************************************
We know that ,

In ∆ABC , D,F and E are the midpoints

of sides AB , BC and CA respectively ,

then ∆ABC is divided into four

congruent triangles , when the three

midpoints are joined to each other .

( ∆DEF is called medial triangle )

****************************************

ii ) area of ∆ABC = 4 × ∆ADE

= 4 × 1

= 4 square units

iii ) area of ∆DEF = area of ∆ADE

= 1 square units.

•••••