The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular lawn in the plot .The students are to sow the 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 triangle PQR if C is the origin?

Also calculate the areas of the triangles in these cases. What do you observe?

The Class X students of a secondary school in Krishinagar have been allotted  a rectangular plot of land for their gardening activity. Sapling of Gulmohar  are planted on the boundary at a

Asked by Pragya Singh | 1 year ago |  113

1 Answer

Solution :-

(i) Taking A as origin, coordinates of the vertices P, Q and R are,

From figure: P = (4, 6), Q = (3, 2), R (6, 5)

Here AD is the x-axis and AB is the y-axis.

 

(ii) Taking C as origin,

Coordinates of vertices P, Q and R are ( 12, 2), (13, 6) and (10, 3) respectively.

Here CB is the x-axis and CD is the y-axis.

Find the area of triangles:

Area of triangle PQR in case of origin A:

Using formula: Area of a triangle 

\( \dfrac{1}{2}\) × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

\( \dfrac{1}{2}\) [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)]

\( \dfrac{1}{2}\) (- 12 – 3 + 24 )

\( \dfrac{9}{2}\) sq unit

 

(ii) Area of triangle PQR in case of origin C:

Area of a triangle = \( \dfrac{1}{2}\) × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

\( \dfrac{1}{2}\) [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)]

\( \dfrac{1}{2}\) ( 36 + 13 – 40)

\( \dfrac{9}{2}\) sq unit

This implies, Area of triangle PQR at origin A = Area of triangle PQR at origin C

Area is same in both case because triangle remains the same no matter which point is considered as origin.

Answered by Abhisek | 1 year ago

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