Chapter 1
Orienting Yourself: The Use of Coordinates
CBSE Class 9
Mathematics Solutions
Educart Mathematics class Class 9 NCERT Exemplar cover
Question:

In the given figure, Reiaan's room is shown with points OABCOABC marking its corners. The xx- and yy-axes are marked in the figure. Point OO is the origin.

Question image

(A) If D₁R₁ represents the door to Reiaan's room, how far is the door from the left wall (the y-axis) of the room? How far is the door from the x-axis?

(B) What are the coordinates of D₁?

(C) If R₁ is the point (11.5, 0), how wide is the door? Do you think this is a comfortable width for the room door? If a person in a wheelchair wants to enter the room, will he/she be able to do so easily?

(D) If B₁ (0, 1.5) and B₂ (0, 4) represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door? [Page No. 4]

Answer: Verified

(A) From the figure, D1D_1 lies on the bottom wall (the xx-axis) at x=8x = 8.

So,

Distance of the door from the yy-axis (left wall) = 8 ft (the xx-coordinate of D1D_1).

Distance of the door from the xx-axis = 0 ft (the door is on the xx-axis itself).

(B) D1D_1 is on the xx-axis, i.e., y=0y = 0.

From the figure, its xx-coordinate 8.

Thus, D1=(8,0)D_1 = (8, 0).

(C) Both D1D_1 and R1R_1 lie on the xx-axis, so the door width is the difference of xx-coordinates.

Width of door = (xx-coordinate of R1R_1) - (xx-coordinate of D1D_1) = 11.58=3.511.5 - 8 = 3.5 ft.

A 3.5 ft wide door is quite comfortable, as standard residential doors are usually about 3 ft wide. A wheelchair typically requires a minimum clear width of about 32 inches, though 36 inches is recommended. Since 3.5 ft is well above this minimum width, a person in a wheelchair can enter easily.

(D) Both B1B_1 and B2B_2 lie on the yy-axis (x=0x = 0). The door width is the difference of yy-coordinates.

Bathroom door width = 41.5=2.54 - 1.5 = 2.5 ft.

The bathroom door (2.5 ft) is narrower than the room door (3.5 ft).

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