Question

find the area of triangle whose vertices are (0,5) (8,0) and (2,0)
urgent plz​

Answer 1

$\green{\bigstar}\underline{\underline{\textsf{\textbf{ Given\::- }}}}$

• Vertices of triangle = (0,5) , (8,0) and (2,0).

$\green{\bigstar}\underline{\underline{\textsf{\textbf{ To\:Find\::- }}}}$

• Area of triangle = ?

$\green{\bigstar}\underline{\underline{\textsf{\textbf{ Step\:by\:step\:explanation\::- }}}}$

• To find area of triangle, using formula of area of triangle, i.e, $\blue{\bigstar}\tiny{\underline{\boxed{\bf{\orange{\dfrac{1}{2}|x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})|\:}}}}}$

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$\green{\bigstar}\underline{\underline{\textsf{\textbf{ Known\:values }}}}\begin{cases} & \sf{x_{1} = \bf{0}\:,\:\sf{y_{1}} = \bf{5}} \\ \\ & \sf{x_{2} = \bf{8}\:,\:\sf{y_{2}} = \bf{0}} \\ \\ & \sf{x_{3} = \bf{2}\:,\:\sf{y_{3}} = \bf{0}} \end{cases}$

$\green{\bigstar}\underline{\underline{\textsf{\textbf{ Putting\:all\:known\:values\::- }}}}$

$\dashrightarrow\qquad\sf Area_{(triangle)} = \dfrac{1}{2}|0(0 - 0) + 8(0 - 5) + 2(5 - 0)|$

$\dashrightarrow\qquad\sf Area_{(triangle)} = \dfrac{1}{2}|0(0) + 8(-5) + 2(5)|$

$\dashrightarrow\qquad\sf Area_{(triangle)} = \dfrac{1}{2}|0 + (-40) + 10|$

$\dashrightarrow\qquad\sf Area_{(triangle)} = \dfrac{1}{2}|-40 + 10|$

$\dashrightarrow\qquad\sf Area_{(triangle)} = \dfrac{1}{2}|-30|$

$\dashrightarrow\qquad\sf Area_{(triangle)} = \dfrac{1}{2}\:\times\:30$

$\dashrightarrow\qquad\sf Area_{(triangle)} = \dfrac{1}{\cancel{2}}\:\times\:\cancel{30}$

$\dashrightarrow\qquad{\boxed{\frak{\pink {Area_{(triangle)} = 15\:sq.units}}}}\:\purple\bigstar$

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$\small\therefore\:{\underline{\sf{Hence,\:area\:of\:triangle\:=\:\bf{15\:sq.units}\:\sf{respectively.}}}}$

$\green{\bigstar}\underline{\underline{\textsf{\textbf{ More\:to\:know\::- }}}}$

$\boxed{\begin{array}{c}\bullet\: \underline{\bf{Distance \: formula\::}} \\ \\ \quad\sf\sqrt{ \bigg(x_{2} - x_{1}\bigg)^2 + \bigg(y_{2} - y_{1}\bigg)^2 } \quad \\ \\ \bullet\: \underline{\bf{Section \: formula\::}} \\ \\ \quad\sf \bigg(\dfrac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}\:,\:\dfrac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}}\bigg)\quad \\ \\ \end{array}}$

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