Question

In a triangle ABC, vertices A, B and C are (0, 0), (0, 2) and (3, 0) respectively. The area of triangle ABC is​

Answer 1

Answer:

3

Step-by-step explanation:

=1/2 |x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|

=1/2 |0(2-0)+0(0-0)+3(0-2)|

=1/2 |0+0+3(-2)|

=1/2 |-6|

=1/2 (6)

=3 square units

Answer 2

$\bf \underline{ \underline{\maltese\:Given} }$

$\sf Vertices \: of \: the \: \triangle \:ABC\: are :$

$\sf\implies(0, 0)$

$\sf\implies (0, 2)$

$\sf\implies(3, 0)$

$\bf \underline{ \underline{\maltese\:To \: find } }$

$\sf \implies Area \: of \: \triangle \: ABC = \: ?$

$\bf \underline{ \underline{\maltese\:Solution } }$

$\bf Area \: of \: triangle =$

$\underline{\boxed{ \sf \dfrac{1}{2}\ \times\ \bigg[x_1\bigg(y_2\ -\ y_3\bigg)\ +\ x_2\bigg(y_3\ -\ y_1\bigg)\ +\ x_3\bigg(y_1\ -\ y_2\bigg)\bigg]}}$

$\bf \underline{Where},$

$\implies \sf{x_1 =0 }$

$\implies \sf{x_2 = 0}$

$\implies \sf{x_3 =3 }$

$\implies \sf{y_1 =0 }$

$\implies \sf{y_2 = 2}$

$\implies \sf{y_3 =0 }$

$\bf \underline{Now},$

$\sf Substitute \: the \: values,$

$\sf \dfrac{1}{2}\ \times\ \bigg| \: 0\bigg(2\ -\ 0\bigg)\ +\ 0\bigg(0\ -\ 0\bigg)\ +\ 3\bigg(0\ -\ 2\bigg) \: \bigg|$

$\sf \dfrac{1}{2}\ \times\ \bigg| \:0(2)\ +\ 0(0)\ +\ 3(- 2)\: \bigg|$

$\sf \dfrac{1}{2}\ \times\ \bigg| \:0\ +\ 0\ +\ (- 6)\: \bigg|$

$\sf \dfrac{1}{2}\ \times\ \bigg| \:0\ +\ 0\ - 6\: \bigg|$

$\sf \dfrac{1}{2}\ \times\ \bigg| \: - 6\: \bigg|$

$\sf \dfrac{1}{2}\ \times 6 = \cancel{\dfrac{ 6}{2}} = 3$

$\underline{ \boxed{ \red{ \bf So, area \: of \: triangle \: is} \bf \: \: 3 \: \: \red{ square \: units. }}}$