Question

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

Answer 1

Given:

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

To find :

➡ The area of triangle =?

Formula :

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

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

Solution:

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

Substitute the vertices values in this formula :

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

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

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

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

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

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

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

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

____________________________

Additional Information :

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

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

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

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

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

Answer 2

Bhaiya nhi Saiyan aur agr bura lage to lgna v chaiye.