Question

Find area of triangle formed by joining the midpoints of sides of triangle whose vertices are A(2,1) B(4,3) C(2,5)

Answer 1

area of triangle ABC is 4 times of ar of DEF

Answer 2

Answer: 1 squnit.

Step-by-step explanation: In the attached figure, ABC is a triangle with vertices  A(2,1), B(4,3) and C(2,5). ΔDEF is formed bu joining the mid-points D, E and F of the sides AB, BC and CA respectively of ΔABC.

Co-ordinates of point D are given by $(\dfrac{2+4}{2},\dfrac{1+3}{2})=(3,2).$

Co-ordinates of point E are $(\dfrac{4+2}{2},\dfrac{3+5}{2})=(3,4).$

And, co-ordinates of point F are $(\dfrac{2+2}{2},\dfrac{5+1}{2})=(2,3).$

Therefore, area of ΔDEF is given as

$A = \dfrac{1}{2}(3(4-3)+3(3-2)+2(2-4))\\\\\Rightarrow A= \dfrac{1}{2}(3+3-4)\\\\\Rightarrow A= 1.$

Thus, the area of the triangles formed by joining the mid-points is $1~sq~unit.$