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Ex 7.4, 5 The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in the Fig. The students are to sow seeds of flowering plants on the remaining area of the plot. (i) Taking A as origin, find the coordinates of the vertices of the triangle. If A is the origin, AD and AB are coordinate axes. our graph will look like this The vertices of the ∆PQR are P (4, 6), Q (3, 2) and R (6, 5) Ex 7.4, 5 (ii) What will be the coordinates of the vertices of Δ PQR if C is the origin? Also calculate the areas of the triangles in these cases. What do you observe? If C is the origin, CD and CB are the coordinate axes. Our graph would be like The vertices of the ∆PQR are P (−12, –2), Q (−13, −6) and R (−10, −3) Calculating the area Area of ∆PQR with vertices P (4, 6), Q (3, 2) & R (6, 5) Here, 𝑥_1=4 , 〖 𝑦〗_1=6 𝑥_2=3 , 𝑦_2=2 𝑥_3=6 , 𝑦_3=5 ar (∆PQR) = 1/2 [𝑥_1 (𝑦_2−𝑦_3 )+ 𝑥_2 (𝑦_3−𝑦_1 )+𝑥_3 (𝑦_1−𝑦_2 )] = 1/2 [4 (2−5)+3 (5−6) + 6 (6−2)] Area of ∆PQR with vertices P (−12, –2), Q (−13, −6) & R (−10, −3) Here, 𝑥_1=−12 , 𝑦_1=−2 𝑥_2=−13 , 𝑦_2=−6 𝑥_3=−10 , 𝑦_3=−3 ar (∆PQR) = 1/2 [𝑥_1 (𝑦_2−𝑦_3 )+ 𝑥_2 (𝑦_3−𝑦_1 )+𝑥_3 (𝑦_1−𝑦_2 )] = 1/2 [(−12) (−6+3)+(−13)(−3+2)+−10(−2+6)] = 1/2 [4 (−3)+3 (−1)+6 (4)] = 1/2 [−12−3+24] = 1/2 (9) = 𝟗/𝟐 sq. units = 1/2 [(−12) (−3)+(−13)(−1)+(−10)(4)] = 1/2 [36+13−40] = 1/2 (9) = 𝟗/𝟐 sq. units = 1/2 [(−12) (−3)+(−13)(−1)+(−10)(4)] = 1/2 [36+13−40] = 1/2 (9) = 𝟗/𝟐 sq. units We observe, that the areas are same in both the cases.

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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.