Question

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

Answer 1

Given :

In a △ABC another △DEF is inside the △ABC such that D,E and F are the midpoints of AB, BC and CA of △ABC respectively as shown in attached figure.

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Solution :

$\sf ar(\triangle ABC) \\ \\= \small \frac{1}{2} |0(1 - 3) + 2(3 - 2) + 0(2 - 1)| \\ \\ = \sf \frac{1}{2} |2| \\ \\ = \bf \: 1 \: sq \: unit$

Here,

D,E and F are the midpoints of AB, BC and CA of triangle respectively

$\small \sf Coordinate \: of \: D=( \frac{2 + 0}{2} \: , \: \frac{1 + 2}{2} ) \\ \\ = ( 1 \: , \: \frac{3}{2} )$

$\small \sf Coordinate \: of \: E=( \frac{2 + 0}{2} \: , \: \frac{1 + 3}{2} ) \\ \\ = (1,2)$

$\small\sf coordinates \: of \: F= (\frac{0 + 0}{2} \: , \: \frac{3 + 2}{2} ) \\ \\ = (0, \frac{5}{2} )$

Now,

$\sf ar(\triangle DEF)\\ \\ = \small \frac{1}{2} |1( 2 - \frac{5}{2}) + 1( \frac{5}{2} - \frac{3}{2} )+ 0( \frac{3}{2} - 2) | \\ \\ = \sf \frac{1}{2} | - \dfrac{1}{2} + 1 | \\ \\ = \sf \frac{1}{2} | \frac{1}{2} | \\ \\ = \bf \frac{1}{4} \: sq \: unit$

⇒ar(△DEF):ar(△ABC)=1:4

Answer 2

Given :-

vertices are (0,2) ,(2,1) and (0,3).

To Find :-

Find the ratio of this area

to the area of the given triangle.

Solution :-

By using mid point formula

$\sf (x,y) = \bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg)$

Let the triangle be ABC and points be DEF

For D

$\sf D = \bigg(\dfrac{0+2}{2},\dfrac{1+2}{2} \bigg)$

$\sf D = \bigg(\dfrac{2}{2},\dfrac{3}{2}\bigg)$

$\sf D = 1,\dfrac{3}{2}$

For E

$\sf E = \bigg(\dfrac{2+0}{2},\dfrac{1+3}{2}\bigg)$

$\sf E = \bigg(\dfrac{2}{2},\dfrac{4}{2}\bigg)$

$\sf E = 1,2$

For F

$\sf F = \bigg(\dfrac{0+0}{2},\dfrac{3+2}{2}\bigg)$

$\sf F = \bigg(\dfrac{0}{2},\dfrac{5}{2}\bigg)$

Finding area of triangle

$\sf Area_{\triangle} = \dfrac{1}{2}\bigg|x_1(y_2 - y_3) + x_2(y_3-y_1)+x_3 (y_1-y_2)\bigg|$

$\sf Area_{\triangle} = \dfrac{1}{2} \bigg|0(1-3) + 2(3-2) + 0(2-1)\bigg|$

$\sf Area_{\triangle} = \dfrac{1}{2}\bigg|0(-2) + 2(1) + 0(1)\bigg|$

$\sf Area_{\triangle} =\dfrac{1}{2}\times 2$

$\sf Area_{\triangle} = 1$

Now

$\sf Ratio = 1:4$