Question

Find the area of the triangle formed by joining the midpoint of the sides of triangle whose vertices A(2,3)B(4,-4)C(2,6)
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Answer 1

Given :

Vertices of triangle are : A(2, 3), B(4, -4) and C(2, 6)

Find :

Area of triangle.

Solution :

Continuing the Attachment..

We have coordinates :

→ D = (3, -1/2)

→ E = (3, 1)

→ F = (2, 9/2)

We know that..

$\bold{Area\:of\:triangle\:=\: 1/2[x_{1}( y_{2} - y_{3}) \: + \: x_{2}( y_{3} - y_{1}) \: + \: x_{3}( y_{1} - y_{2}) ]}$

Area of ΔABC :

We have..

• $x_{1}$ = 2
• $x_{2}$ = 4
• $x_{3}$ = 2

• $y_{1}$ = 3
• $y_{2}$ = -4
• $y_{3}$ = 6

Substitute the known values in above formula

$\Rightarrow\: 1/2[2( - 4 \: - \: 6) \: + \: 4( 6 \: - \: 3) \: + \: 2( 3 \: - \: (-4))]$

$\Rightarrow\:1/2[2(-10)\:+\:4(3)\:+\:2(3\:+\:4)]$

$\Rightarrow\:1/2[2(-10)\:+\:4(3)\:+\:2(7)]$

$\Rightarrow\:1/2[-\:20\:+\:12\:+\:14]$

$\Rightarrow\:1/2 \:\times\:6$

$\implies\:+\:3$

∴ Area of ΔABC = 3 sq. units

Now,

Area of ΔDEF :

We have..

• $x_{1}$ = 3
• $x_{2}$ = 3
• $x_{3}$ = 2

• $y_{1}$ = -1/2
• $y_{2}$ = 1
• $y_{3}$ = 9/2

$\Rightarrow\: 1/2[3( 1\: - \: 9/2) \: + \: 3( 9/2 \: - \: (-1/2)) \: + \: 2( -1/2 \: - \: 1)]$

$\Rightarrow\:1/2[3(-7/2)\:+\:3(5)\:+\:2(-3/2)$

$\Rightarrow\:1/2[-\:21/2\:+\:15\:-\:3$

$\Rightarrow\:1/2[(-\:21\:+\:30\:-\:6)/2]$

$\Rightarrow:1/2\:\times\:3/2$

$\implies\:3/4$ _____ (eq 2)

∴ Area of ΔDEF = 3/4 sq. units

If we have to find ratio.. then ratio :

→ Area of ΔDEF/Area of ΔABC = (3/4)/3

→ Area of ΔDEF/Area of ΔABC = 1/4

∴ Ratio is 1:4

But according to question we have to find the area of the triangle formed by joining the mid points.

And after joining mid points, we have ΔDEF.

✯ Answer :

Area of ΔDEF = 3/4 sq. units.

Answer 2

ANSWER:-

Given:

Find the area of the ∆ formed by joining the mid-points of the sides of triangle whose vertices A(2,3), B(4,-4) &C(2,6).

Solution:

Let the vertices of the ∆ABC be;

• A(2,3)
• B(4,-4)
• C(2,6)

Therefore,

Let D,E,F be the mid-points of the sides of this triangle.

Co-ordinates of D, E and F are given by;

$D= ( \frac{2 + 4}{2} \: \frac{3 + ( - 4)}{2}) = \frac{6}{2} , - \frac{1}{2} \\ \\ = > D= 3 , - \frac{1}{2} \\ \\ E = ( \frac{2 + 2}{2} , \frac{6 - 3}{2} ) = \frac{4}{2} , \frac{3}{2} \\ \\ = > E = 2 , \frac{3}{2} \\ \\ F = ( \frac{4 + 2}{2} , \frac{ - 4 + 6}{2} ) = \frac{6}{2} , \frac{2}{2} \\ \\ = > F= 3 ,1$

Let the vertices of ∆DEF;

• D(3, -1/2)
• E(2, 3/2)
• F(3 , 1)

Now,

We know that area of triangle is:

$= > \frac{1}{2} [x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2)]$

Area of ∆ABC,

$= > \frac{1}{2} [2( - 4 - 6) + 4(6 - 3) + 2(3 - ( - 4)]\\ \\ = > \frac{1}{2}[2 ( - 10) + 4(3) + 2(7)]\\ \\ = > \frac{1}{2} \times - 20 + 12 + 14 \\ \\ = > \frac{1}{2} \times 6 \\ \\ = > 3 sq. \: units$

Now,

Area of ∆DEF

$= > \frac{1}{2} [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) ]\\ \\ = > \frac{1}{2} [3( \frac{3}{2} - 1) + 2(1 - ( - \frac{1}{2} ) + 3( - \frac{1}{2} - \frac{3}{2} )] \\ \\ = > \frac{1}{2} [3( \frac{3 - 2}{2} ) + 2( \frac{2 + 1}{2} ) + 3( \frac{ - 1 - 3}{2} ) ]\\ \\ = > \frac{ 1}{2} [3( \frac{1}{2} ) + 2( \frac{3}{2} ) + 3( - \frac{4}{2} )] \\ \\ = > \frac{1}{2} ( \frac{3}{2} + \frac{6}{2} + ( \frac{ - 12}{2} )) \\ \\ = > \frac{1}{2} \times \frac{3 + 6 - 12}{2} \\ \\ = > \frac{1}{2} \times \frac{ - 3}{2} \\ \\ = > - \frac{3}{4} sq. \: units$

Therefore,

Required ratio (3: -3/4)

Hope it helps ☺️