Question

The Class $X$ students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmoharare planted on the boundary at a distance of $1$ m from each other. There is a triangular grassy lawn in the plot as shown in the Fig. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking $A$ as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of $\Delta PQR$ if $C$ is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?

Answer 1

(i) If we take $A$ as origin then $AD$ is $X$-axis and $AB$ is the $Y$-axis Then the coordinates of the vertices of $△PQR$ is

$P=(4,6),Q=(3,2),R=(6,5)$

(ii) If we take $C$ as origin then $CB$ is $X$-axis and $CD$ is the $Y$-axis Then the coordinates of the vertices of $△PQR$ is

$P=(12,2),Q=(13,6),R=(10,3)$

Area of the triangle =$\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

First case-

Area of $△PQR=\frac{1}{2}\left[4\right(2-5)+3(5-6)+6(6-2\left)\right]$

$=\frac{1}{2}(4\times -3+3\times -1+6\times 4)$

$=\frac{1}{2}(-12-3+24)=\frac{9}{2}$Sq.unit

Second case-

Area of $△PQR=\frac{1}{2}\left[12\right(6-3)+13(3-2)+10(2-6\left)\right]$

$=\frac{1}{2}(12\times 3+13\times 1+10\times -4)$

$=\frac{1}{2}(36+13-40)=\frac{9}{2}$Sq.unit

Hence we observed that the area of the triangle in both case is equal.