Question

the area of quadrilateral whose vertices are (-1,6),(-3,-9),(5,-8),(3,9) is :
a)94unit^2

b)96unit^2

c)169unit^2

d)none

Answer 1

Given : Vertices of quadrilateral (-1,6), (-3,-9), (5,-8), (3,9).

To find: Area of the quadrilateral = ?

Solution:

• Let the points be A (-1,6), B (-3,-9), C (5,-8), D (3,9).

• Now consider two triangles ABC and ADC. So sum of these triangles will result in area of quadrilateral.

• Area of triangle ABC,

              x1(y2-y3) + x2(y3-y1) + x3(y1-y2) / 2

• Putting values in this formula we get:

              -1(-1) + (-3)(-14) + 5(15) / 2

              1 + 42 + 75 / 2

              118 / 2

              59 sq units.

• Similarly for triangle ADC, we have:

             -1(17) + 3(-14) + 5(-3) / 2

             -17 -42 - 15 / 2

             -74 / 2

             -37 sq units

• We will ignore negative sign as area is always positive.

• So area of quadrilateral is : 59 + 37 = 96 sq cm.

Answer:

           So the area of quadrilateral is 96 sq cm.

Answer 2

answer is is 96 square unit hope it helps