Question

find the area of triangle with its vertices as (1,3),(4,2)and(-2,5) ​

Answer 1

Answer:

Formula is add so

13 +42+25=70

Answer 2

Answer:

Question :

Find the area of triangle with its vertices as (1,3),(4,2)and(-2,5).

$\begin{gathered}\end{gathered}$

Solution :

Here :

• ➛ Coordinate of A = (1, 3)
• ➛ Coordinate of B = (4, 2)
• ➛ Coordinate of C = (-2, 5)

To Calculate the area of the triangle with given three vertices the formula is :

$\small\star\; \pink{\underline{\boxed{\bf{Area_{\:(triangle)} = \dfrac{1}{2}\Bigg[x_1 \bigg(y_2 - y_3 \bigg) + x_2 \bigg(y_3 - y_1 \bigg) + x_3 \bigg(y_1 - y_2 \bigg)\Bigg]}}}}$

Where :

$\begin{gathered}\small\star \: \sf{\underline{\underline{\purple{Here}}}}\begin{cases}\sf{\quad x_1, \; y_1 = \bf{1, 3}}\\\\\sf{\quad x_2,\; y_2 = \bf{4, 2}}\\\\\sf{\quad x_3, \;y_3 = \bf{-2, 5}}\end{cases}\\\\\end{gathered}$

Now, substuting the all the given values in the formula :

${\implies{\small{\sf{Area_{\:(triangle)} = \dfrac{1}{2}\Bigg[x_1 \bigg(y_2 - y_3 \bigg) + x_2 \bigg(y_3 - y_1 \bigg) + x_3 \bigg(y_1 - y_2 \bigg)\Bigg]}}}}$

$\small{\implies{\sf{Area_{\:(triangle)} = \dfrac{1}{2}\Bigg[1 \bigg(2 - 5 \bigg) + 4\bigg(5 - 3 \bigg) + - 2 \bigg(3- 2 \bigg)\Bigg]}}}$

$\small{\implies{\sf{Area_{\:(triangle)} = \dfrac{1}{2}\Bigg[1 \times - 3 + 4 \times 2+ - 2 \times 1\Bigg]}}}$

$\small{\implies{\sf{Area_{\:(triangle)} = \dfrac{1}{2}\Bigg[ - 3 + 8 - 2 \Bigg]}}}$

$\small{\implies{\sf{Area_{\:(triangle)} = \dfrac{1}{2} \times 3}}}$

$\small{\implies{\sf{Area_{\:(triangle)} = \dfrac{1}{\cancel{2}} \times \cancel{3}}}}$

$\small{\implies{\sf{\red{Area_{\:(triangle)} = 1.5}}}}$

⌗ Hence, the area of triangle is 1.5.

$\begin{gathered}\end{gathered}$

Learn More :

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Triangle:-}\\ \\ \star\sf Triangle \:area = \dfrac{1}{2}\times b \times h\\ \\ \star\sf Triangle \: perimeter=a+b+c\\\\ \star\sf Scalene\:\triangle=\sqrt{s (s-a)(s-b)(s-c)}\\\\\star\sf Equilateral\: \triangle\:area = \dfrac{\sqrt{3}}{4}\times{side}^{2}\\\\\star\sf Equilateral \:\triangle\:perimeter = 3 \times side\\\\\star\sf Isosceles\: \triangle\:area= \dfrac{3}{4}\sqrt{{4b}^{2}-{a}^{2}}\\\\\star\sf Isosceles\:\triangle\:perimeter=a+2b\end{minipage}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

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