Question

Question No. 19

Find the area of the triangle whose coordinates of vertices are A(6,3), B(-3,5) and (4,-2).

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Answer 1

Answer:

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Answer 2

$\large{\red{\bold{\underline{Given:}}}}$

$\sf \: Coordinates \: of \: triangle \: ABC \: are : \: A(6, 3), \\ \\ \sf \: B(-3,5) \: and \: C(4,-2).$

$\large{\green{\bold{\underline{To \: Calculate:}}}}$

$\sf \: Area \: of \: the \: triangle$

$\large{\blue{\bold{\underline{Formula \: Used:}}}}$

$\sf \: Area \: of \: triangle = \frac{1}{2} \times x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})$

$\large{\pink{\bold{\underline{Step \: by \: step \: Solution:}}}}$

$\sf \: Let \: us \: consider \: the \: respective \: coordinates \: as : \\ \\ A(6, 3) \rightarrow \: \sf \: (x_{1}, y_{1}) \\ \\ B(-3,5) \rightarrow \: \sf \: (x_{2}, y_{2}) \\ \\ C(4,-2) \rightarrow \: \sf \: (x_{3}, y_{3})$

$\large{\red{\bold{\underline{Calculation:}}}}$

$\sf \rightarrow \: Area \: of \: triangle = \frac{1}{2} \times 6(5 - ( - 2)) - 3( - 2 - 3) + 4(3 - 5) \\ \\ \rightarrow \sf \: Area \: of \: triangle = \frac{1}{2} \times( 6 \times 7) +(3 \times 5) - (4 \times 2) \\ \\ \rightarrow \sf \: Area \: of \: triangle = \frac{1}{2}\times( 42 + 15 - 8) \\ \\ \rightarrow \sf \: Area \: of \: triangle = \frac{1}{2} \times 49 \\ \\ \rightarrow \sf \: Area \: of \: triangle = 24.5$

$\large{\orange{\bold{\underline{Therefore:}}}}$

$\underline{ \sf \: The \: area \: of \: triangle \: is \: 24.5 \: Square \: units.}$