# Ex 10.1, 1 - Chapter 10 Class 11 Straight Lines

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex10.1, 1 Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7),(5, – 5) and (– 4, –2). Also, find its area Let points be A (–4, 5), B (0, 7), C (5, –5), D (–4, –2) Area of quadrilateral ABCD = Area of Δ ABD + Area of ΔBCD Finding Area of Δ ABD & Area of ΔBCD separately Calculating area of ∆ ABD Here x1 = –4 , y1 = 5 x2 = 0 , y2 = 7 x3 = –4 , y3 = –2 Area of ∆ABD = 1/2 |𝑥_1 (𝑦_2 − 𝑦_3 ) + 𝑥_2 (𝑦_3− 𝑦_1 ) + 𝑥_3 (𝑦_1 − 𝑦_3)| = 1/2 | − 4(7 − ( − 2)) + 0( − 2 − 5) + ( − 4)(5 − 7)| = 1/2 | − 4(7 + 2) + 0 − 4)( − 2)| = 1/2 | − 4(9) + 8| = 1/2 | − 28| = 1/2 × 28 = 14 Area of ΔBCD Here x1 = 0 y1 = 7 x2 = 5 y2 = –5 x3 = –4 y3 = –2 Area of ΔBCD = 1/2 |x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y3)| = 1/2 |0( − 5 − ( − 2)) + 5 ( − 2 − 7) − 4 (7 − ( − 5))| = 1/2 |0 + 5 ( − 9) − 4 (7 + 5)| = 1/2 |0 − 45 − 4 (12)| = 1/2 | − 45 − 48| = 1/2 | − 93| = 93/2 Now, Area of quadrilateral ABCD = Area of Δ ABD + Area of ΔBCD = 14 + 93/2 = (14 × 2 + 93)/2 = (28 + 93)/2 = 121/2 = 60.5 sq. units Hence area of quadrilateral is 60.5 sq. unit

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.