RS Aggarwal Solutions Class 9 Maths Chapter 15: Being the first higher class in the students' lives and giving some weight to the academics of class 10, class 9 is crucial. For more advanced study in class 10, the majority of the 9th class chapters are extended.
Because maths is the hardest subject, it takes more practice than the other subjects combined. The topic of Chapter 15 Maths Class 9 RS Aggarwal is broad and encompasses several key ideas for the students. The RS Aggarwal solutions for the mensuration, volume, and surface area of solids in class 9 serve as a foundation for the extensive mensuration syllabus for class 10.RS Aggarwal Solutions Class 9 Maths Chapter 15 PDF
Question 1.
Solution:
(i) length =12cm, breadth = 8 cm and height = 4.5 cm ∴ Volume of cuboid = l x b x h = (12 x 8 x 4.5) cm3= 432 cm3 ∴ Lateral surface area of a cuboid = 2(l + b) x h = [2(12 + 8) x 4.5] cm2 = (2 x 20 x 4.5) cm2 = 180 cm2 ∴ Total surface area cuboid = 2(lb +b h+ l h) = 2(12 x 8 + 8 x 4.5 + 12 x 4.5) cm2 = 2(96 +36 +54) cm2 = (2 x186) cm2 = 372 cm2 (ii) Length 26 m, breadth =14 m and height =6.5 m ∴ Volume of a cuboid = l x b x h = (26 x 14 x 6.5) m3 = 2366 m3 ∴ Lateral surface area of a cuboid =2 (l + b) x h = [2(26+14) x 6.5] m2 = (2 x 40 x 6.5) m2 = 520 m2 ∴ Total surface area = 2(lb+ bh + lh) = 2(26 x 14+14 x6.5 +26 x6.5) = 2 (364+91+169) m2 = (2 x 624) m2= 1248 m2 (iii) Length = 15 m, breadth = 6m and height = 5 dm = 0.5 m ∴ Volume of a cuboid = l x b x h = (15 x 6 x 0.5) m3=45 m3 ∴ Lateral surface area = 2(l + b) x h = [2(15 + 6) x 0.5] m2 = (2 x 21×0.5) m2=21 m2 ∴ Total surface area =2(lb+ bh + lh) = 2(15 x 6 +6 x 0.5+ 15 x 0.5) m2 = 2(90+3+7.5) m2 = (2 x 100.5) m2 =201 m2Question 2.
Solution:
Length of Cistern = 8 m Breadth of Cistern = 6 m And Height (depth) of the Cistern =2.5 m ∴ Capacity of the Cistern = Volume of cistern ∴ Volume of Cistern = (l x b x h) = (8 x 6 x2.5) m3 =120 m3 Area of the iron sheet required = Total surface area of the system. ∴ Total surface area = 2(lb +bh +lh) = 2(8 x 6 + 6×2.5+ 2.5×8) m2 = 2(48 + 15 + 20) m2 = (2 x 83) m2=166 m2Question 3.
Solution:
Length of a room =9m, Breadth of a room = 8m And height of the room = is 6.5 m ∴ Area of 4 walls = Lateral surface area = 2 (l+ b) x h = [2 (9+8) x 6.5] m2 = (2 x 17 x 6.5) m2 =221 m2 ∴ Area not be whitewashed = (area of 1 door) + (area of 2 windows) = (2 x 1.5) m2 + (2 x 1.5 x 1) m2 = 3m2 + 3m2 =6m2 ∴ Area to be whitewashed = (221-6) m2 =215 m2 ∴ The cost of whitewashing the walls at the rate of Rs.6.40 per Square meter = Rs. (6.40 x 215) = Rs. 1376CBSE Class 9 Maths Syllabus | CBSE Class 9 Science Syllabus |
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