Question

find the area of the triangle defined by the vertices (0,0),(3,0),(2,3)​

Answer 1

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Explanation:

Given :

Three vertices of a triangle

let Vertices be A(0,0),B(3,0) &C(2,3)

To Find :

Area of a triangle

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$\bold{\boxed{Area \: Of \: a \: Triangle = \frac{1}{2} [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]}}$

X1=0,X2=3,X3=2

Y1=0,Y2=0,Y3=3

Now ,finding area of triangle by putting this values in above formula:-

$= > \frac{1}{2} [0(0 - 3) + 3(3 - 0) + 2(0 - 0)]$

$= > \frac{1}{2} [0( - 3) + 9 + 0)]$

$= > \frac{1}{2} \times 9 = \frac{9}{2}$

Hence,Area of Triangle =$\bold{\red{\frac{9}{2}}}$sq.unit

Note : If area comes in negative value then we put mode because area can't be negative.

Answer 2

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

Given :

Three vertices of a triangle

let Vertices be A(0,0),B(3,0) &C(2,3)

To Find :

Area of a triangle

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$\bold{\boxed{Area \: Of \: a \: Triangle = \frac{1}{2} [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]}}$

X1=0,X2=3,X3=2

Y1=0,Y2=0,Y3=3

Now ,finding area of triangle by putting this values in above formula:-

$= > \frac{1}{2} [0(0 - 3) + 3(3 - 0) + 2(0 - 0)]$

$= > \frac{1}{2} [0( - 3) + 9 + 0)]$

$= > \frac{1}{2} \times 9 = \frac{9}{2}$

Hence,Area of Triangle =$\bold{\red{\frac{9}{2}}}$sq.unit

Note : If area comes in negative value then we put mode because area can't be negative.