Question

Find the area of quadrilateral ABCD whose vertices are A (1, 0), B (5, 3), C (2, 7), D ( -2, 4).

Answer 1

Answer:

Area of the quadrilateral ABCD= 25 sq.units

Step-by-step explanation:

Given:-

A quadrilateral ABCD whose vertices are A (1, 0), B (5, 3), C (2, 7), D ( -2, 4).

To find:-

Find the area of quadrilateral ABCD whose vertices are A (1, 0), B (5, 3), C (2, 7), D ( -2, 4).

Solution:-

Given vertices of a quadrilateral ABCD are

A (1, 0), B (5, 3), C (2, 7), D ( -2, 4).

AC and BD are diagonals

The diagonal divides the quadrilateral ABCD into two triangles ∆ABC and ∆ADC

To find the area of a quadrilateral ABCD first we have to find the areas of two triangles ∆ABC and ∆ADC and adding them.

area ( ABCD) = area (∆ABC ) + area(∆ADC)

i)Area of ∆ABC:-

A (1, 0), B (5, 3), C (2, 7)

Let (x1, y1)=(1,0) => x1 = 1 and y1 = 0

Let (x2, y2)=(5,3) =>x2 =5 and y2=3

Let (x3, y3)=(2,7)=>x3=2 and y3=7

Area of a triangle whose vertices (x1, y1);(x2, y2) and (x3, y3) is

∆=(1/2) | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) | sq.units

=>Area (∆ABC)

= (1/2) | 1(3-7) + 5(7-0) + 2(0-3) |

=> (1/2) | 1(-4) + 5(7) +2(-3) |

=>(1/2) | -4 +35 -6 |

=>(1/2) | 35 -10 |

=> (1/2) | 25 |

=> (1/2)×25

=>25/2

=>12.5 sq.units

Area of ∆ABC = 12.5 sq.units-----(1)

ii) Area of ∆ADC:-

A(1, 0), C (2, 7), D ( -2, 4)

Let (x1, y1) = (1,0) =>x1 = 1 and y1 = 0

Let (x2,y2) = (-2,4)=>x2 = -2 and y2 = 4

Let (x3, y3) = (2,7) => x3 = 2 and y3 = 7

Area of a triangle whose vertices (x1, y1);(x2, y2) and (x3, y3) is

∆=(1/2) | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) | sq.units

=>Area (∆ADC)

=>(1/2) | 1(4-7) + (-2)(7-0) + 2(0-4) | sq.units

=>(1/2) | 1(-3) + (-2)(7) +2(-4) |

=>(1/2) | -3 -14 -8 |

=>(1/2) | -25 |

=>(1/2)(25)

=>25/2

=>12.5 sq.units

Area (∆ADC) = 12.5 sq.units -----(2)

Now,

Area of quadrilateral ABCD

=Area (∆ABC) + Area (∆ADC)

=>12.5 + 12.5

=>25 sq.units

Answer:-

Area of the quadrilateral ABCD whose vertices are A (1, 0), B (5, 3), C (2, 7), D ( -2, 4) = 25 sq.units

Used formula:-

Area of a triangle whose vertices (x1, y1);(x2, y2) and (x3, y3) is

∆=1/2 | x1(y2-y3)+x2(y3-y1)+x3(y1-y2) | sq.units