Question

Given a quadrilateralABCD with vertices A(5, 3) , B(6, 4), C (− 3, − 2) and D( − 14, 7) Find the area of ABCD area =_______ sq. units.

Answer 1

Answer:

Step-by-step explanation:

first of all we divide it in 2 triangles

1 st ABC

AREA OF ABC = 1/2(A1(B2-C2) +B1(C2-A2) +C1(A2-B2) )

AREA OF ABC =1/2 ( 5(4-(-2)) + 6(-2-3) + (-3)(3-4) )

AREA OF ABC =1/2 (30-30+3)

AREA OF ABC =1/2 (3)

AREA OF ABC = 1.50

2  nd ADC

AREA OF ADC = 1/2(A1(D2-C2) +D1(C2-A2) +C1(A2-D2) )

AREA OF ABC =1/2 ( 5(7-(-2)) + (-14)(-2-3) + (-3)(3-(-14)) )

AREA OF ABC =1/2 (45+70-51)

AREA OF ABC =1/2 (64)

AREA OF ABC = 32

SO,

AREA OF QUADRILATERAL IS 33.50 sq. units

(1.50+32)

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