NCERT Solutions Ganita Prakash Chapter 6 Sharan’s house plan — In-text Questions
Book page 147 Updated on2026-09-05
Q1.
Now, find out the missing dimensions and area of Sharan’s home. Below is the plan. Some of the measurements are given. a. Find the missing measurements.
Answer
The plot is 42 ft wide (given). Work across the top row first.
Top row: 12 (bedroom) + Toilet + 18 (kitchen) + Utility = 42
Utility: area 70 = ? × 10 → Utility = 7 ft × 10 ft
So Toilet width = 42 − 12 − 18 − 7 = 5 ft, and its height matches the kitchen, 10 ft
Now the left column gives the height of the plot:
Master Bedroom 15 ft + Small Bedroom 10 ft = 25 ft
Hall height = 25 − 10 = 15 ft, and its width is 5 + 18 = 23 ft (as printed)
Entrance = 7 ft × 15 ft
Sharan's completed plan — 42 ft × 25 ft, with every room's dimensions and area filled in.
Here the rooms account for the plot exactly — Sharan's plan has no leftover passage.
Q3.
What are the dimensions of all the different rooms in Sharan’s house? Compare the areas and perimeters of Sharan’s house and Charan’s house.
Answer
The dimensions are listed in the table above. Now compare the two houses.
Charan's house
Sharan's house
Plot
35 ft × 30 ft
42 ft × 25 ft
Area
1050 sq ft
1050 sq ft
Perimeter
2 × (35 + 30) = 130 ft
2 × (42 + 25) = 134 ft
Conclusion: the two houses have exactly the same area (1050 sq ft), but Sharan's house has the longer perimeter — 134 ft against 130 ft, that is 4 ft more.
Why: Charan's plot (35 × 30) is closer to a square than Sharan's (42 × 25). For the same area, the more square-like the plot, the shorter its boundary. So Charan would need less compound wall — and pay less for it.
Math Talk: if you had to build a boundary wall at ₹500 per foot, Charan pays 130 × 500 = ₹65,000 and Sharan pays 134 × 500 = ₹67,000 — ₹2000 more for exactly the same floor space.