Question

Find the area of the triangle formed by joining the mid - points of the sides of the triangle whose vertices are (0,1) , (2,1) and (0,3). Find the ratio of this area to the area of the triangle.

Answer 1

ar (triangle ABC) = 1 ÷ 2 [0 (1-3) + 2(3+1) + 0(-1-1) ] sq. units = 1 ÷ 2 [ 0 + 8 + 0 ] sq. units

= 8 ÷ 2
= 4 sq. units ----------- (i)

Also coordinates of mid - points of sides are D (1,2) , E (0,1) and F (1,0)

• ar (triangle DEF) = 1 ÷ 2 [ 1 (1-0) + 0 (0-2) + 1 (2-1) ] sq. units

ar (triangle DEF) = 1 ÷ 2 [ 1 + 0 + 1 ]

ar (triangle DEF) = 2 ÷ 2
= 1 sq. unit ------------(ii)

From (i) and (ii),

ar (ABC) : ar (DEF)

= 4 : 1 ans.

# Hope this helps.

Answer 2

Solution :-

D , E and F are midpoints of side AB , AC and BC respectively !!


• Coordinates of a mid point =

For x :- ( x1 + x2 )/2

For y :- ( y1 + y2 )/2

=> Coordinates of Point D :-

For x = ( 0 + 0 )/2 = 0/2 = 0

For y = ( 1 + 3 )/2 = 4/2 = 2

D ( 0 , 2 )

=> Coordinates of Point E

For x = ( 0 + 2 )/2 = 2/2 = 1

For y = ( 1 + 1 )/2 = 2/2 = 1

E ( 1 , 1 )

=> Coordinates of point F

For x = ( 0 + 2 )/2 = 2/2 = 1

For y = ( 3 + 1 )/2 = 4/2 = 2

F ( 1 , 2 )

• Area of Triangle :-
$= \frac{1}{2} (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)$

x1 = 0
x2 = 1
x3 = 1

y1 = 2
y2 = 1
y3 = 2

$area \: of \: def = \frac{1}{2} (0(1 - 2) + 1(2 - 2) + 1(2 - 1)$

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

$= \frac{1}{2 } {unit}^{2}$

• Area of ∆ABC

x1 = 0
x2 = 0
x3 = 2

y1 = 1
y2 = 3
y3 = 1


$area \: of \: abc = \frac{1}{2} (0(3 - 1) + 0(1 - 1) + 2(1 - 3))$


$= \frac{1}{2} | - 4|$

$= \frac{1}{2} \times 4$

$= 2 {unit}^{2}$


• Ratio of the area of DEF to the area of ∆ABC is :-

$\frac{area \: of \: def}{area \: of \: abc} = \frac{ \frac{1}{2} }{2}$


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


$= \frac{1}{4}$


area of ∆DEF : area of ∆ABC = 1 : 4