CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard

Class 10
Solutions of Sample Papers for Class 10 Boards

## The diagrams show the plans for a sun room. It will be built onto the wall of a house. The four walls of the sunroom are square clear glass panels. The roof is made using

• Four clear glass panels, trapezium in shape, all the same size
• One tinted glass panel, half a regular octagon in shape

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Note: Check more Case Based Questions for - Coordinate Geometry (Chapter 7 Class 10)

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### Transcript

Question 17 Case Study Based-1 SUN ROOM The diagrams show the plans for a sun room. It will be built onto the wall of a house. The four walls of the sunroom are square clear glass panels. The roof is made using • Four clear glass panels, trapezium in shape, all the same size • One tinted glass panel, half a regular octagon in shape(a) Refer to Top View Find the mid-point of segment joining the points J (6, 17) & I (9, 16). (i) (33/2,15/2) (ii) (3/2,1/2) (iii) (15/2,33/2) (iv) (1/2,3/2) Mid point of line segment joining J (6, 17) & I (9, 16) is Mid point = ((6 + 9)/2 ,(17 + 16)/2 ) = (𝟏𝟓/𝟐 ,𝟑𝟑/𝟐 ) So, (iii) is correct (b) Refer to Top View The distance of the point P from the y-axis is (i) 4 (ii) 15 (iii) 19 (iv) 25 Distance of Point P from y-axis is 4 (c) Refer to Front View The distance between the points A and S is (i) 4 (ii) 8 (iii) 16 (iv) 20 Distance between the points A and S is 16 (d) Refer to Front View Find the co-ordinates of the point which divides the line segment joining the points A and B in the ratio 1:3 internally. (i) (8.5,2.0) (ii) (2.0,9.5) (iii) (3.0,7.5) (iv) (2.0,8.5)Here, Here, Point A(1, 8) and B (5, 11) Let Point X (x, y) divide AB in ratio 1:3 internally Therefore, Coordinates of X = ((1(5) + 3(1))/(1 + 3),(1(10) + 3(8))/(1 + 3)) = ((5 + 3)/4,(10 + 24)/4) = (8/4,34/4) = (2, 8.5) (e) Refer to Front View If a point (x, y) is equidistant from the Q(9, 8) and S(17, 8), then (i) x + y = 13 (ii) x – 13 = 0 (iii) y – 13 = 0 (iv )x – y = 13 Let required point be X (x, y) Now, Point X is equidistant from Q(9, 8) & S (17, 8) Hence, QX = SX √((𝑥 −9)2+(𝑦−8)2) = √((𝑥 −17)2+(𝑦−8)2) Squaring both sides (𝑥 −9)2+(𝑦−8)2=(𝑥 −17)2+(𝑦−8)2 (𝑥 −9)2=(𝑥 −17)2 𝑥^2+9^2−18𝑥=𝑥^2+17^2−34𝑥 34𝑥−18𝑥=17^2−9^2 16𝑥=(17−9)(17+9) 16𝑥=8 × 26 𝑥=(8 × 26)/16 𝑥=13 𝒙−𝟏𝟑=𝟎 So, (ii) is correct