Question

ABC is a triangle coordinates of whose vertex are (0,-1) D and E are the mid points of the sides AB and AC and their coordinates are (1,0) and (0,1) . If F is the mid point of side BC find the areas of triangle ABC and triangle DEF.​

Answer 1

$254 - 1222 \\ \\ \binom{e \beta \gamma \gamma \gamma \gamma \gamma e \cot( \tan( \cos( \sin( {552.. - 12 \div \div \div \div \div 555552255 \times }^{?} ) ) ) ) }{?}$

a+b square

Answer 2

Answer:

Area of ΔABC is 4 units² and Area of ∆DEF is 1 units²

Step-by-step explanation:

Given In ∆ABC

D , F and E are the midpoints of sides AB, BC and CA respectively.

Coordinate of vertex are the folowing:

$A(x_1,y_1)=(0,-1)$

$D(x_2,y_2)=(1,0)$

$E(x_3,y_3)=(0,1)$

Now, using formula of area of triangle using coordinate geometry

Area of ∆ADE= $\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

                       =  $\frac{1}{2}\left|0(0-1)+1(1+1)+0(-1-0)\right|$

                       = $\frac{1}{2}\times2$

                       = 1 units²

Now we use a result which states that Mid point of sides of a triangle divides it into 4 congruent triangle when the three mid points are joined together.

Here, 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

So, Area of ∆ABC = 4 × ar ( ∆ADE )

                              = 4 × 1

                              = 4 units²

Also, area of ∆DEF = area of ∆ADE = 1 units²

Therefore, Area of ΔABC is 4 units² and Area of ∆DEF is 1 units²