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 given triangle.

Answer 1

Hint: In this type of question, we need to find the area of the triangle.
Also we have to use the formula for Mid- point formula, and use that for finding the coordinates of $P,Q,R$
After this we will find the area of the remaining triangles.
Finally, we will find the ratio of this area to the area of the given triangle.

Complete step-by-step answer:
It is given that the vertices of the triangle be $A(0,1),B(2,1)$ and $C(0,3)$.
First of all we plot the vertices of the triangle $ABC$ in $XY$-plane.
Since $P,Q,R$ are mid points of the corresponding edges $AB,BC,AC$.
Here the vertices of the triangle diagram as follows,

We need to find the area enclosed by the triangle $ABC$
Area of the $\Delta ABC = \dfrac{1}{2}\left[ {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right]$
Here, coordinates
${x_1} = 0,{y_1} = 1$
${x_2} = 2,{y_2} = 1$
${x_3} = 0,{y_3} = 3$
Area of the $\Delta ABC = \dfrac{1}{2}\left[ {0(1 - 3) + 2(3 - 1) + 0(1 - 1)} \right]$
On subtracting the bracketed terms we get,
  $ = \dfrac{1}{2}\left[ {0 + 2(2) + 0} \right]$
On adding the terms we get,
$ = \dfrac{1}{2}\left( 4 \right)$
Let us divided the terms we get,
$ = 2$Square units
Also it is given that $P$ be the midpoint of $AB$
Now we are going to find the coordinates for $P$ using the midpoint formula
$\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
Here, the vertices of the \[{\text{AB}}\],
${x_1} = 0,{y_1} = 1$
${x_2} = 2,{y_2} = 1$
Putting the above value in the formula we get,
$ = \left( {\dfrac{{0 + 2}}{2},\dfrac{{1 + 1}}{2}} \right)$
On simplification we get the coordinates of $P$,
$ = {\text{P}}\left( {1,1} \right)$
Similarly, Q is the midpoint of BC,
Here, the vertices of the \[{\text{BC}}\],
${x_2} = 2,{y_2} = 1$
${x_3} = 0,{y_3} = 3$
Putting the value, we get
$\left( {\dfrac{{2 + 0}}{2},\dfrac{{1 + 3}}{2}} \right)$
On adding the terms we get,
$ = \left( {\dfrac{2}{2},\dfrac{4}{2}} \right)$
Let us divided the terms we get the coordinates of ${\text{Q}}$ ,
$ = {\text{Q}}\left( {1,2} \right)$
Similarly, R is the midpoint of AC,
Here the vertices
${x_1} = 0,{y_1} = 1$
${x_3} = 0,{y_3} = 3$
Putting the values in the above formula we get,
$\left( {\dfrac{{0 + 0}}{2},\dfrac{{1 + 3}}{2}} \right)$
On adding the terms we get,
$ = \left( {\dfrac{0}{2},\dfrac{4}{2}} \right)$
Let us divided the terms we get the coordinates of ${\text{R}}$ ,
$ = {\text{R}}\left( {0,2} \right)$
Hence, the coordinates for $P(1,1),Q(1,2),R(0,2)$
Here the vertices of the triangle $PQR$ diagram as follows,

Now we need to find the area enclosed by the triangle $PQR$
Area of the $\Delta {\text{PQR}} = \dfrac{1}{2}\left[ {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right]$
Here, the Coordinates of \[{\text{PQR}}\]
${x_1} = 1,{y_1} = 1$
${x_2} = 1,{y_2} = 2$
${x_3} = 0,{y_3} = 2$
Putting the values in the formula,
Area of $\Delta {\text{PQR}} = \dfrac{1}{2}\left[ {1(2 - 2) + 1(2 - 1) + 0(1 - 2)} \right]$
On some simplification we get,
$ = \dfrac{1}{2}\left[ {0 + 1 + 0} \right]$
Let us add the terms and we get the answer,
$ = \dfrac{1}{2}$ Square units.
Hence the required ratio is
$\dfrac{{{\text{Area of}}\;\Delta PQR}}{{{\text{Area of}}\;\Delta ABC}} = \dfrac{{\dfrac{1}{2}}}{2} = \dfrac{1}{4}$ Sq. unit
$\therefore $The required ratio of the Area of the triangle ${\text{Area of}}\;\Delta PQR$ and ${\text{Area of}}\;\Delta ABC$ are $1:4$

Note: For performing the multiplication of fraction,
First we have to solve by simplifying the fraction, and then multiply the numerator with the numerator and the denominator with the denominator.