With D1 at (9, 0) and R1 at (11.5, 0), the width is 2.5 units (50 cm if 1 unit = 20 cm). This is quite narrow and would not easily accommodate a wheelchair.
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?
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The width is calculated by subtracting the x-coordinates: 11.5 minus 9 equals 2.5 units. If we assume a standard scale where 1 unit equals 20 cm, the width is 50 cm. This is generally considered uncomfortable for a main room door. Specifically, a standard wheelchair requires a minimum width of about 80 to 90 cm to pass through easily. Therefore, a person in a wheelchair would not be able to enter easily.
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Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-1/