# Ex 7.3, 3

Last updated at Oct. 4, 2017 by Teachoo

Last updated at Oct. 4, 2017 by Teachoo

Transcript

Ex 7.3 , 3 Find the area of the triangle formed by joining the mid−points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle. Let the vertices of triangle be A(0, −1) , B(2, 1), C(0, 3) Let the mid−point of AB be P BC be Q AC be R Joining points P, Q , R We get ∆PQR We need to find area of ∆ABC & area of ∆PQR Finding area of ∆ ABC Area of triangle ABC = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] Here x1 = 0 , y1 = −1 x2 = 2 , y2 = 1 x3 = 0 , y3 = 3 Putting values Area of triangle ABC = 1/2 [ 0(1 − 3) + 2(3 – (−1)) + 0(−1 − 1) ] = 1/2 [ 0 + 2(3 + 1) + 0] = 1/2 [ 0 + 2(4) + 0] = 1/2 [ 8] = 4 Finding area of triangle PQR First we have to find coordinates of P, Q & R Since P is the mid−point of AB Coordinates of P = ((𝑥1 + 𝑥2)/2, (𝑦1 +𝑦2)/2) = ((0 + 2)/2, (−1 + 1)/2) = (2/2, 0/2) = (1, 0) Similarly, Q is the mid−point of BC Coordinates of Q = ((𝑥1 + 𝑥2)/2, (𝑦1 +𝑦2)/2) = ((2 + 0)/2, (1 + 3)/2) = (2/2, 4/2) = (1, 2) Similarly, R is the mid−point of AC Coordinates of R = ((𝑥1 + 𝑥2)/2, (𝑦1 +𝑦2)/2) = ((0+0)/2, (−1 + 3)/2) = (0/2, 2/2) = (0, 1) So, the coordinates are P(1, 0), Q(1, 2), R(0, 1) Now, we need to find out the area of ∆PQR Area of triangle PQR = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] Here x1 = 1 , y1 = 0 x2 = 1 , y2 = 2 x3 = 0 , y3 = 1 Putting values Area of triangle PQR = 1/2 [ 1(2 − 1) + 1(1 − 0) + 0(0 − 2) ] = 1/2 [ 1(1) + 1(1) + 0] = 1/2 [1 + 1] = 1/2 [2] = 1 Hence, the required ratio (𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑃𝑄𝑅)/(𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐶) = 1/4

Class 10

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.