Math, asked by adithyarebel48, 3 months ago

15. Find the area of the
Traingle whose vertices are
(5,4) (3,8) &(-2,10)​

Answers

Answered by javeedpasha3136
2

Step-by-step explanation:

Let A(5,4)=X¹,Y¹ B(3,8)=X²,Y² andC(-2,10)=X³,Y³

Area of triangle ABC=1/2(x¹(y²-y³)+x²(y³-y¹)+x³(y¹-y²))

1/2(5(8-10)+3(10-4)+(-2)(4-8))

1/2(5(-2)+3(6)-2(-4)

1/2(-10+18+8)

1/2(-10+26)

1/2(16)

8 square units

Answered by Anonymous
47

Given:-

Vertices of triangle are as follows:-

  • (5, 4)
  • (3, 8)
  • (-2, 10)

To Find:-

  • Area of the triangle.

Solution:-

We are given with three points (or vertices) of the triangle i.e. (5, 4), (3, 8) and (-2, 10)

From here we get the following:-

  • \tt{x_1 = 5}
  • \tt{x_2 = 3}
  • \tt{x_3 = -2}
  • \tt{y_1 = 4}
  • \tt{y_2 = 8}
  • \tt{y_3 = 10}

We already know:-

  • \dag{\boxed{\underline{\pink{\tt{Area\:of\:triangle = \dfrac{1}{2}[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]}}}}}

Putting all the values in the formula:-

 = \sf{\dfrac{1}{2}[5(8-10) + 3(10 - 4) + \{-2(4 - 8)\}]}

 = \sf{\dfrac{1}{2}[5 \times (-2) + 3 \times 6 - 2 \times (-4)]}

 = \sf{\dfrac{1}{2}[-10 + 18 + 8]}

 = \sf{\dfrac{1}{2}[26 - 10]}

 = \sf{\dfrac{1}{2} \times 16}

 = \sf{\dfrac{1}{\not{2}} \times \not{16}}

 = \sf{8}

The area of the triangle is 8 square units.

______________________________________

Learn More!!!

↦ What are x₁, x₂, x₃, y₁, y₂ and y₃?

  • x₁ is the abscissa of the first point given i.e. (5, 4)
  • x₂ is the abscissa of the second point given i.e. (3, 8)
  • x₃ is the abscissa of the third point given i.e. (-2, 10)
  • y₁ is the ordinate of the first point i.e. (5, 4)
  • y₂ is the ordinate of the second point i.e. (3, 8)
  • y₃ is the ordinate of the third point i.e. (5, 4)

NOTE:- The highlighted words represent abscissa and ordinate of the respective points.

______________________________________

Similar questions