Slide56.JPG

Slide57.JPG
Slide58.JPG
Slide59.JPG
Slide60.JPG


Transcript

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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.