Question

The vertices of ∆ ABC are A (-5,-1),B(3,-5),C(5,2) show that the area of the triangle ABC is 4 times the area of the triangle formed by joining the mid points of side of ∆ABC

Answer 1

Answer:

Step-by-step explanation:

Area=1/2[x1(y2-y3) +x2(y3-y1) +x3(y1-y2)]

=1/2[-5(-7) + 3(3) + 5(3)]

=1/2[-35+9+15]

=1/2[-24]

Area=12

AB=100

BC=52

AB-BC=48

Area =12×4

=48

AB-BC=Area

48=48

Answer 2

Answer:

Given: vertices of ΔABC = A( -5 , -1 ) , B( 3 , -5 ) & C( 5 , 2 )

To Prove: area of ΔABC = 4 times are of triangle formed by joining mid points of ΔABC

Let say the triangle formed by joining the vertices is ΔPQR

Formulas use are the following,

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

$Coordinates\:of\:Mid-Point=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Now,

$Area\:of\:\Delta\,ABC=\frac{1}{2}\left|-5(-5-2)+3(2+1)+5(-1+5)\right|$

                               $=\frac{1}{2}\left|-5(-7)+3(3)+5(4)\right|$

                               $=\frac{1}{2}\left|35+9+20\right|$

                               $=\frac{1}{2}\times64$

                               $=32\:unit^2$

Mid point of AB = P , Mid Point of CB = Q & Mid Point of AC = R

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

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

                          = $(-1,-3)$

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

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

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

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

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

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

$Area\:of\:\Delta\,PQR=\frac{1}{2}\left|-1(\frac{-3}{2}-\frac{1}{2})+4(\frac{1}{2}+3)+0(-3+\frac{-3}{2})\right|$

                               $=\frac{1}{2}\left|-1(\frac{-4}{2})+4(\frac{1+6}{2})+0\right|$

                               $=\frac{1}{2}\left|-1(-2)+2(7)+0\right|$

                               $=\frac{1}{2}\left|2+14+0\right|$

                               $=8\:unit^2$

Therefore, ar ΔABC = 4 × ΔPQR

Hence Proved,