In order to conduct Sports Day activities in your School, lines have been drawn with chalk powder at a distance of 1 m each, in a rectangular shaped ground ABCD, 100 flowerpots have been placed at a distance of 1 m from each other along AD, as shown in given figure below. Niharika runs 1/4 th the distance AD on the 2nd line
and posts a green flag. Preet runs 1/5 th distance AD on the eighth line and posts a red flag

This question is inspired from  Ex 7.2, 3 - Chapter 7 Class 10 - Coordinate Geometry

In order to conduct Sports Day - Teachoo.jpg

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Question 1

Find the position of green flag
(a) (2, 25)
(b) (2, 0.25)
(c) (25, 2)
(d) (0, –25)

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Question 2

Find the position of red flag
(a) (8, 0)
(b) (20, 8)
(c) (8, 20)
(d) (8, 0.2)

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Question 3

What is the distance between both the flags?
(a) √41
(b) √11
(c) √61 
(d) √51

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Question 4

If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
(a) (5, 22.5)
(b) (10, 22)
(c) (2, 8.5)
(d) (2.5, 20)

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Question 5

If Joy has to post a flag at one-fourth distance from green flag, in the line segment joining the green and red flags, then where should he post his flag?
(a) (3.5, 24)
(b) (0.5, 12.5)
(c) (2.25, 8.5)
(d) (25, 20)

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  1. Chapter 7 Class 10 Coordinate Geometry (Term 1)
  2. Serial order wise

Transcript

Question In order to conduct Sports Day activities in your School, lines have been drawn with chalk powder at a distance of 1 m each, in a rectangular shaped ground ABCD, 100 flowerpots have been placed at a distance of 1 m from each other along AD, as shown in given figure below. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5 th distance AD on the eighth line and posts a red flag Given that there are 100 flowers between A & D at 1m distance each Niharika runs 1/4th of the distance AD on 2nd line So, Niharika’s x−cordinate = 2 Niharika’ y – cordinate = 1/4 × 100 = 25 ∴ Coordinates of Niharika = G (2, 25) Also, Preet runs 1/5th of the distance AD on 2nd line So, Preet’s x−ordinate = 8 Preet’s y – ordinate = 1/5 × 100 = 20 ∴ Coordinates of Preet = R (8, 20) Putting values in formula NP = √(( 8 −2)2+(20 −25)2) = √((6)2+(−5)2) = √((6)2+(5)2) = √(36+25) = √61 Hence, The distance between both flags = √𝟔𝟏 metres Question 1 Find the position of green flag (a) (2, 25) (b) (2, 0.25) (c) (25, 2) (d) (0, –25) Position of green flag = Point G = (2, 25) So, the correct answer is (a) Question 2 Find the position of red flag (a) (8, 0) (b) (20, 8) (c) (8, 20) (d) (8, 0.2) Position of green flag = Point R = (8, 20) So, the correct answer is (c) Question 3 What is the distance between both the flags? (a) √41 (b) √11 (c) √61 (d) √51 Distance between both flags = RG = √(( 𝟖 −𝟐)𝟐+(𝟐𝟎 −𝟐𝟓)𝟐) = √((6)2+(−5)2) = √((6)2+(5)2) = √(36+25) = √𝟔𝟏 m So, the correct answer is (c) Question 4 If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag? (a) (5, 22.5) (b) (10, 22) (c) (2, 8.5) (d) (2.5, 20) Since Rashmi had to post a blue flag exactly halfway between the line segment joining the two flags It will be the mid-point of RG Thus, our diagram looks like Here, x = (𝑥1 + 𝑥2)/2 and y = (𝑦1 + 𝑦2)/2 Substituting the value in the formula x = (𝒙𝟏 + 𝒙𝟐)/𝟐 x = (2 + 8)/2 x = 10/2 x = 5 y = (𝒚𝟏 + 𝒚𝟐)/𝟐 y = (25 + 20)/2 y = 45/2 y = 22.5 ∴ Rashmi (x, y) = (5, 22.5) So, the correct answer is (a) Question 5 If Joy has to post a flag at one-fourth distance from green flag, in the line segment joining the green and red flags, then where should he post his flag? (a) (3.5, 24) (b) (0.5, 12.5) (c) (2.25, 8.5) (d) (25, 20) Now, Joy had to post a flag at one-fourth distance from green flag, in the line segment joining the green and red flags Thus, Distance between Green Flag & Joy = 1/4 × Distance between Green and Red Flag GJ = 𝟏/𝟒 × GR GJ = 1/4 × (GJ + JR) GJ = 1/4 × GJ + 1/4 × JR GJ − 1/4 × GJ = 1/4 × JR 3/4 × GJ = 1/4 × JR 𝑮𝑱/𝑱𝑹=𝟏/𝟑 Thus, Joy divides GR in the ratio 1:3 Finding x x = (𝑚1 𝑥2 + 𝑚2 𝑥1)/(𝑚1 + 𝑚2) Where, m1 = 1, m2 = 3 x1 = 2, x2 = 8 Putting values x = (1 × 8 + 3 × 2)/(1 + 3) x = (8. + 6)/4 x = 14/4 x = 3.5 Finding y y = (𝑚1 𝑦2 + 𝑚2 𝑦1)/(𝑚1 + 𝑚2) Where, m1 = 1, m2 = 3 y1 = 25, y2 = 20 Putting values y = (1 × 20 + 3 × 25)/(1 + 3) y = (20 + 75 )/4 y = (95 )/4 y = 23.75 Thus, Required Point = (3.5, 23.75)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.