Question

The vertices of a A ABC are A
-5
-1
B
(3.-5), C(5, 2). Show that
the area of the A ABC is four times the area of the triangle formed by
joining the mid-points of the sides of the triangle ABC.
​

Answer 1

Answer:

Step-by-step explanation:

If A (x1,y1), B (x2,y2), C(x3,y3) are the vertices of ΔABC, then  

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

∴ Area of ΔABC = 1/2 | -5 (-5-2) + 3 (2 - (-1)) + 5(-1-(-5))|

                          = 1/2 | -5 (-7) + 3 (3) + 5 (4)

                           = 1/2 (35 + 9 + 20 )

                          = 1/2 (64)

                         =  32                 ...... ......(1)

now suppose, the midpoint of a side AB of ΔABC is D

∴ D (x1,y1) = D ( -5+3 / 2, -1-5 / 2) = D (-1,-3)

The midpoint of BC is E.

∴ E (x2,y2) = E ( 3+5/ 2, -5+2 / 2) = E (4,-1.5)

and the midpoint of AC is F.

∴ F (x3,y3) = F ( -5+5/ 2, -1+2 / 2) = F (0, 0.5)

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

∴ Area of ΔDEF= 1/2 | -1 (-1.5-0.5) + 4 (0.5 - (-3)) + 0 (-3-(-1.5))|

                          = 1/2 | -1 (-2) + 4 (3.5) + 0 (-1.5)

                          = 1/2 ( 2 + 14 + 0 )

                         = 1/2 (16)

                         =  8

If Area of  ΔABC is four times the area of the triangle formed by

joining the mid-points of the sides of the triangle ABC.

Area of  ΔABC = 4 x Area of  ΔDEF

                       = 4 x 8

                       = 32               ...... ......(2)

Thus, from result (1) and (2), we can say that the Area of  ΔABC is four times the area of ΔDEF (the triangle formed by  joining the mid-points of the sides of the triangle ABC.)