Question

If A(-7,5),B(-6,-7),C(-3,-8) and D(2,3) are the vertices of a quadrilateral ABCD then find the areaof the quadrilateral.

Answer 1

Here, we will use the formula Area of triangle=12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)].
Now,
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC

So,
Area of triangle ABC=12[(−3)(−7−(−8))+(−2)(−8−5)+(1)(5−(−7))]
Area of triangle ABC=12[(−3)(1)+(−2)(−13)+(1)(12)]
Area of triangle ABC=12[−3+26+12]
Area of triangle ABC=352                                                                                                ... (1)

Also,
Area of triangle ADC=12[(−3)(3−(−8))+(6)(−8−5)+(1)(5−3)]
Area of triangle ADC=12[(−3)(11)+(6)(−13)+(1)(2)]
Area of triangle ADC=12[−33−78+2]
Area of triangle ADC=1092                                                                                                ... (2)

Adding equation (1) and (2), we get
Area of quadrilateral ABCD=352+1092
Area of quadrilateral ABCD=1442
Area of quadrilateral ABCD=72

Therefore, area of quadrilateral ABCD is 72 square units.

Answer 2

Answer:

72 squareunits

Step-by-step explanation:

It is given that A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) are the vertices of a quadrilateral ABCD.

Area of the quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD

∴ Area of the quadrilateral ABCD = square units

Hence, the area of the given quadrilateral is 72 square units.