Check sibling questions

Ex 10.1, 1 - Draw a quadrilateral whose vertices (-4, 5)

Ex 10.1, 1 - Chapter 10 Class 11 Straight Lines - Part 2
Ex 10.1, 1 - Chapter 10 Class 11 Straight Lines - Part 3
Ex 10.1, 1 - Chapter 10 Class 11 Straight Lines - Part 4
Ex 10.1, 1 - Chapter 10 Class 11 Straight Lines - Part 5

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Transcript

Ex 10.1, 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 square units Hence area of quadrilateral is 60.5 sq. unit

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.