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:

Area of Δ ABC = 4 * Area of Δ DEF

Step-by-step explanation:

Formula used:

Area of the triangle formed by the points

$(x_1, y_1),(x_2, y_2)and(x_3, y_3)\:=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$The\:midpoint\:of \:the \:line \:joining (x_1,y_1)\: and \:(x_2,y_2) \:is\:(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Given points are

A(-5, -1), B(3, -5), C(5,2)

Area of Δ ABC

$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$=\frac{1}{2}[-5(-5-2)+3(2+1)+5(-1+5)]$

$=\frac{1}{2}[-5(-7)+3(3)+5(4)]$

$=\frac{1}{2}[35+9+20]$

$=\frac{1}{2}[64]$

$=32$square units............(1)

Midpoint of AB is D

$=(\frac{-5+3}{2},\frac{-1-5}{2})$

$=(\frac{-2}{2},\frac{-6}{2})$

$=(-1,-3)$

Midpoint of BC is E

$=(\frac{3+5}{2},\frac{-5+2}{2})$

$=(\frac{8}{2},\frac{-3}{2})$

$=(4,\frac{-3}{2})$

Midpoint of AC is F

$=(\frac{-5+5}{2},\frac{-1+2}{2})$

$=(\frac{0}{2},\frac{1}{2})$

$=(0,\frac{1}{2})$

$D(-1,-3),E(4,\frac{-3}{2}),F(0,\frac{1}{2})$

Area of Δ DEF

$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$=\frac{1}{2}[-1(\frac{-3}{2}-\frac{1}{2})+4(\frac{1}{2}-(-3))+0(-3-\frac{-3}{2})]$

$=\frac{1}{2}[-1(-2)+4(\frac{1+6}{2})+0]$

$=\frac{1}{2}[2+4(\frac{7}{2})+0]$

$=\frac{1}{2}[2+2(7)+0]$

$=\frac{1}{2}[16]$

$=8$ square units.............(2)

From (1) and (2)

Area of Δ ABC = 4 * Area of Δ DEF

Answer 2

Answer:

Step-by-step explanation:

A(-5, -1), B(3, -5), C(5, 2)

AB = √( (3-(-5))² + (-5 -(-1))² )  = √80 = 4√5

BC = √53

CA = √109

Let Say D , E & F are mid points  of AB , BC & CA

D =  (-1 , -3)  ,  mid point of AB

E = (4 , -3/2)  , mid point of BC

F = (0 , 1/2)   , mid point of  CA

DE = √109 / 2   = CA/2    =>  CA/DC  = 2

EF = √20 = 2√5 = 4√5/2  = AB/2    => AB/EF  = 2

DF = √53/ 2 = BC/2   => BC/DF  = 2

AB/EF = BC/DF = CA/DC = 2

=> ΔABC ≅ ΔEFD  

ΔEFD  is triangle formed by joining the mid-points of the sides of the triangle ABC

Ratio of Area of Similar triangles = (Ratio of sides of similar triangles)²

=> Area of ΔABC / Area of ΔEFD = (2)²

=> Area of ΔABC / Area of ΔEFD  = 4

=> Area of ΔABC = 4 * Area of ΔEFD