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

Made by

Davneet Singh

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.

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