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

Solution:

Let (x1,y1)=A(-5,-1),

(x2,y2)=B(3,-5),

(x3,y3)=C(5,2)

Area of ∆ABC

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

=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)

ii)Let D,E and F are midpoints

of sides AB, BC and CA respectively,

D = (-1,-3),

E = (4,-3/2)

and

F = (0,1/2)

Area of ∆DEF

= 1/2|(-1)[-3/2-1/2]+4[1/2-(-3)]+0[-3-(-3/2)]|

= 1/2| (-1)(-2)+4×(7/2) |

= 1/2|2+14|

= 16/2

= 8 -----(2)

From (1) and (2) , we conclude

that

Area of ∆ABC = 4×Area ∆DEF

••••

Answer 2

Answer:

Area of Δ ABC = 4 Area of ΔMNP

Step-by-step explanation:

If a triangle is formed by joining mid points of triangle ABC

Let say mid points are M , N , P

ΔAMP , Δ BMN & Δ CPN will be similar to Δ ABC

Sides of these triangle will be half of Δ ABC

Area of each ΔAMP , Δ BMN & Δ CPN = Area of Δ ABC/(2²)

Area of ΔAMP = Δ BMN = Δ CPN = Area of Δ ABC/4

Area of ΔMNP = Area of Δ ABC - ( area of ΔAMP + Δ BMN  + Δ CPN )

= Area of Δ ABC - 3 Area of Δ ABC/4

= Area of Δ ABC/4

Area of Δ ABC = 4 Area of ΔMNP

Hence the 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