Question

The students of class 10th of a school undertake to work for the campaign 'Say no to plastic' in the city. They took the map of the city and formed co-ordinate plane on it to divide their areas. Group A took the region covered between the co-ordinates(1,1) (-3,2) (-2,2) and (1,-3) taken in order. Find the area of the region covered by Group A

Answer 1

Answer:

The area of the region covered is 13.5 square units.

Step-by-step explanation:

Given:

Group A took the region covered between the co-ordinates(1,1) (-3,2) (-2,2) and (1,-3) taken in order.

To Find:

The area bounded within the region

Solution:

Let the points be known as

P(-3,2) , Q(1,1), R (1,-3) and S(-2,-2)

The region covered by group A is the region PORS

To get the area covered we need to divide the region into two parts,

Which is, Triangle's PQS and QRS

Thus, the required area will be

Area of the region = Area of triangle PQR + Area of triangle PRS

Area of the region = $\frac{1}{2}| -3(1+2) + 1(-2-2) -2(2-1) | + \frac{1}{2} |1(-3 + 2)+1(-2-1) + (-2)(1+3)|$

$\frac{1}{2}| -9-4-2 | + \frac{1}{2} |-1-3-8|$

$\frac{1}{2}| -15 | + \frac{1}{2} |-12|$

$= \frac{15}{2} +\frac{12}{2}$

Area of the region =  13.5 sq units.

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Answer 2

Answer:

The territory included is 13.5 square units in size.

step-by-step explanation:

Region's Area  Region of Area A. Assume that the smoother curves just on finite interval [a, b] is represented by the non - negative real

function $y = f(x)$ provided. the region enclosed by the x-axis, the vertical lines $x = a$ and $x = b$, and the curve of $f(x)$

The area between the coordinates $(1,1) (-3,2) (-2,2)$ and $(1,-3)$ was chosen by Group A.

The region contained the area.

The points should be defined as

$P(-3,2), Q(1,1), R(1,-3), and S(-2,-2)$

Group A's coverage area is known as PORS.

Divide the space into two sections so that it may be covered.

which are the PQS and QRS of Triangle.

Consequently, the needed space will be

Size of the area = PQR triangle's area + Triangle PRS area

$=\frac{1}{2} |-3(1+2)+1(-2-2)-2(2-1)|+\frac{1}{2} |1(-3+2)+1(-2-1)+(-2)(1+3)|$

$=\frac{1}{2} |-9-4-2|+\frac{1}{2} |-1-3-8|$

$=\frac{1}{2} |-15|+\frac{1}{2} |-12|$

$=\frac{15}{2} +\frac{12}{2}$

$= 13.5 units$

Area of the region $= 13.5 units$

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