Question

Find the area of ∆ ABC with vertices A(5,3), B(4,5) and C(3,1).​

Answer 1

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$\text{\large\underline{\red{Given:-}}}$

• Find the area of triangle ABC with vertices A(5,3) ,B(4,5) and C(3,1) .

$\text{\large\underline{\purple{To find:-}}}$

• ➡ The area of triangle =?

Formula :-

$\begin{gathered}\red\bigstar\sf{ Area \: of \: \triangle \: ABC =\dfrac{1}{2} \times A=\left[\begin{array}{ccc}x_{1}&y_{1}&z_{1}\\x_{2}&y_{2}&z_{2}\\x_{3}&y_{3}&z_{3}\end{array}\right]}\\\end{gathered}$

$\text{\large\underline{\orange{Solution:-}}}$

• (x1,y1)=(5=3)
• $\begin{gathered}\\ \sf {(x_{2} , y_{2} )=(4,5)}\end{gathered}$
• $\begin{gathered}\\ \sf {(x_{3} , y_{3} ) =(3,1)}\end{gathered}$

Substitute the vertices values in this formula:-

$\begin{gathered}\\ \implies\large\sf{\: Area \: of \: triangle =\dfrac{x_{1} (y_{2} -y_{3})+x_{2}(y_{3} -y_{1})+x_{3}(y_{1} -y{2})}{2}}\end{gathered}$

$\begin{gathered}\\ \implies\large\sf{Area \: of \: triangle=\dfrac{5(5-1)+4(1-3)+3(3-5)}{2}}\end{gathered}$

$\begin{gathered}\\ \implies\large\sf{ Area \: of \: triangle =\dfrac{5(4)+4(-2)+3(-2)}{2}}\end{gathered}$

$\begin{gathered}\\ \implies\large\sf{Area \: of \: triangle =\dfrac{20-8-6}{2}}\end{gathered}$

$\begin{gathered}\\ \implies\large\sf{Area \: of \: triangle =\dfrac{20-15}{2}}\end{gathered}$

$\begin{gathered}\\ \implies\large\sf{ Area \: of \: triangle ABC =\dfrac{5}{2}}\end{gathered}$

$\begin{gathered}\\ \implies\large\sf{ Area \: of \: triangle ABC =2.5 sq.units}\end{gathered}$

$\red\bigstar\boxed{\large\sf{Area \: of \: \triangle \: ABC =2.5sq.units}}$

____________________________

Additional Information :-

• $\begin{gathered}\\ \large\sf{(1) \: Area \: of \: triangle =\dfrac{b \times h}{2}}\end{gathered}$

• $\begin{gathered}\\ \large\sf{(2) \: Area \: of \: parallelogram =B \times h}\end{gathered}$

• $\begin{gathered}\\ \large\sf{(3) \: Area \: of \: rectangle = L \times w}\end{gathered}$

• $\begin{gathered}\\ \large\sf{(4) \: Area \: of \: square =L^2}\end{gathered}$

• $\begin{gathered}\\ \large\sf{(5) \: Area \: of \: circle =\pi r^2}\end{gathered}$

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Answer 2

Answer:

Question

• Find the area of ∆ ABC with vertices A(5,3), B(4,5) and C(3,1).

$\bf\underbrace\red{Answer:}$

Given :-

• A ∆ ABC with vertices A(5,3), B(4,5) and C(3,1).

To find :

• Area of triangle.

Formula used :-

Let us consider a ∆ ABC having vertices

$\bf \:A(x_1, y_1) , B(x_2, y_2) ,C(x_3, y_3)\:$

then, area of triangle ABC is given by

$\bf \ Area =\dfrac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Solution:-

$\bf \:Here, x_1= 5, x_2= 4, x_3=3, y_1=3,y_2=5,y_3=1$

Area of triangle =

$\bf \ Area =\dfrac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$\bf ⟹\ Area =\dfrac{1}{2} [5(5 - 1) + 4(1 - 3) + 3(3 - 5)]$

$\bf ⟹\ Area =\dfrac{1}{2} [5 \times 4 - 4 \times 2 - 3 \times 2]$

$\bf ⟹\ Area =\dfrac{1}{2} [20 - 8 - 6]$

$\bf ⟹\ Area =\dfrac{1}{2} \times 6 = 3 \: sq. \: units$