Question

If A(–5, 7), B(–4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

Answer 1

Answer:

Joining BD, there are two triangles.

Area of quad ABCD = Ar △ABD + Ar △BCD

Ar △ABD = ½| (-5)(-5 - 5) + (-4)(5 - 7) + (4)(7 + 5) |

= 53 sq units

Ar △BCD = ½| (-4)(-6 - 5) + (-1)(5 + 5) + (4)(-5 + 6) |

= 19 sq units

Hence, area of quad ABCD = 53 + 19 = 72 sq units.

Answer 2

Hey.....

Step-by-step explanation:

For triangle ABC

Area of triangle

= 1/2(x1(y22y3) + x2(y33y1) + x3(y11y2))

= 1/2((5((5 + 6))4((667))1(7 + 5))

= 1/2((5 + 52212)

= 1/2(35)

For triangle ACD

Area of triangle

= 1/2(x1(y22y3) + x2(y33y1) + x3(y11y2))

= 1/2((5((665))1(557) + 4(7 + 6))

= 1/2((55 + 2 + 52)

= 1/2(1)

Area of ABCD = Area of ABC + Area of ACD

= 18 sq units.

Hope this helps ☺️