Question 5
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.
Let the vertices of quadrilateral be
A(−5, 7) , B(−4, −5)
C(−1, −6) , D(4, 5)
Joining AC
There are 2 triangles formed ABC & ACD
Hence,
Area of quadrilateral ABCD = Area of ∆ ABC + Area of ∆ ADC
Finding area ∆ ABC
Area of triangle ABC = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ]
Here
x1 = −5 , y1 = 7
x2 = −4 , y2 = −5
x3 = −1 , y3 = −6
Putting values
Area of triangle ABC = 1/2 [ −5(−5 –(−6)) + (−4)(−6 – 7) + (−1)(7 – (−5)) ]
= 1/2 [ −5(−5 + 6) – 4(−13) + (−1)(7 + 5)]
= 1/2 [ −5(1) – 4(−13) + (−1)(12)]
= 1/2 [−5 + 52 − 12]
= 𝟏/𝟐 [35] square units
Similarly,
Finding area ∆ ADC
Area of triangle ADC = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ]
Here
x1 = −5, y1 = 7
x2 = 4, y2 = 5
x3 = −1, y3 = −6
Area of triangle ADC = 1/2 [ −5(5 –(−6)) + 4(−6 – 7) + (−1)(7 – 5) ]
= 1/2 [ −5(5 + 6) + 4(−13) + (−1)(2)]
= 1/2 [ −5(11) + 4(−13) + (−1)(2)]
= 1/2 [ −55 − 52 – 2]
= 1/2 [ −109]
But area cannot be negative,
∴ Area of triangle ADC = 𝟏/𝟐 [ 109] square units
Area of quadrilateral ABCD
= Area of triangle ABC + Area of triangle ADC
= 1/2 [ 35 + 109]
= 1/2 [ 144]
= 72 square units

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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