NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

NCERT Solutions Class 9 Maths
Chapter – 13 (Surface Areas and Volumes)

The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 13 Surface Areas and Volumes Exercise 13.9 has been provided here to help the students in solving the questions from this exercise.

Chapter 13: Surface Areas and Volumes

Exercise – 13.9

1. A wooden bookshelf has external dimensions as follows: Height = 110cm, Depth = 25cm, Breadth = 85cm (see fig. 13.31). The thickness of the plank is 5cm everywhere. The external faces are to be polished, and the inner faces are to be painted. If the rate of polishing is 20 paise per cm² and the rate of painting is 10 paise per cm², find the total expenses required for polishing and painting the surface of the bookshelf.

Answer – The external dimensions of the bookshelf,
Breadth, B = 85 cm
Depth, D = 25 cm
Height, H = 110 cm
The thickness of the plank, t = 5 cm
Internal measures of the bookshelf is,
The breadth of each shelf, b = B – 2t
⇒ b = 85 cm – 2 × 5 cm = 75 cm
Depth of each shelf, d = D – t
⇒ d = 25 cm – 5 cm = 20 cm
Height of each shelf, h = H – 4t
= (110 cm – 4 × 5 cm) ÷ 3
= 90 cm / 3
= 30 cm
Now, Surface area to be polished = External 5 surfaces of the bookshelf + border of the shelf
= 2(B + H) D + BH + 2Ht + 4bt
= [2 × (85 cm + 110 cm) × 25 cm] + (85 cm × 110 cm) + (2 × 110 cm × 5 cm) + (4 × 75 cm × 5 cm)
= 9750 cm² + 9350 cm² + 1100 cm² + 1500 cm²= 21700 cm²Cost of polishing at the rate of 20 paise per cm² = 21700 cm² × (₹ 20/100) / cm²
= ₹ 4340
Surface area to be painted = Internal 5 surfaces of 3 shelves
= 3 [2(b + h)d + bh]
= 3 [2 × (75 cm + 30 cm) × 20 cm + (75 cm × 30 cm)]
= 3 [4200cm² + 2250 cm²]
= 3 × 6450 cm²= 19350 cm²Cost of painting at the rate of 10 paise per cm² = 19350 cm² × (₹ 10/100) / cm²
= ₹ 1935
Total expense required for polishing and painting = ₹ 4340 + ₹ 1935
= ₹ 6275
Thus, the total expense required for polishing and painting the surface of the bookshelf is ₹ 6275.

2. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in fig. 13.32. Eight such spheres are used forth is the purpose and are to be painted silver. Each support is a cylinder of radius 1.5cm and height 7cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm^2, and black paint costs 5 paise per cm².

Answer –
Diameter of the wooden sphere = 21 cm
The radius (R) of the wooden sphere = $\frac{21}{2}$ cm
Surface area for wooden sphere = 4πR²= 4 × $\frac{22}{7}$ × $\frac{21}{2}$ cm × $\frac{21}{2}$ cm
= 1386 cm²Since the support is a cylinder of radius, r = 1.5 cm
Area of the circular end of the cylinder = πr²= $\frac{22}{7}$ × 1.5 cm × 1.5 cm
= 7.07 cm²So, the area of each wooden sphere to be painted = 1386 cm² – 7.07 cm² = 1378.93 cm²Total area of the 8 spheres to be painted = 8 × 1378.93 cm² = 11031.44 cm²Cost of silver painting the wooden spheres at the rate of 25 paise per cm²= 11031.44 × ₹ (25/100)
= ₹ 2757.86
Now,
Radius of the cylinder, r = 1.5 cm
Height of the cylinder, h = 7 cm
Curve surface area of the cylinder = 2πrh
= 2 × $\frac{22}{7}$ × 1.5 cm × 7 cm
= 66 cm²CSA of 8 cylindrical support to be painted = 8 × 66 cm² = 528 cm²Cost of black painting the cylindrical support at 5 paise per cm²= 528 × ₹ (5/100)
= ₹ 26.40
Hence the cost of paint required = ₹ 2757.86 + ₹ 26.40
= ₹ 2784.26
Thus, the cost of paint required is ₹ 2784.26 (approx.)

3. The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?

Answer –
Let the radius of the sphere = r
Then its diameter, d= 2r
The Surface area of a sphere = 4πr²The Curved surface area of the sphere = 4πr²Now it is given in the question that the diameter of the sphere is decreased by 25% hence a new sphere is formed.
Therefore, the diameter of the new sphere can be written as:
= 2r – (25%) of (2r)
= 2r – $\frac{25}{100}$ × (2r)
= 2r – (r/2)
= 3r/2
Radius of the new sphere = $\frac{1}{2}$ × $\frac{3r}{2}$
= $\frac{3r}{4}$
Hence, curved surface area of the new sphere = 4π $\left ( \frac{3r}{4} \right )^2$
= 4π $\left ( \frac{9r^2}{16} \right )$

= $\frac{9\pi r^2}{4}$
Now, decrease in the original curved surface area = 4πr – $\frac{9\pi r^2}{4}$
= $\frac{16 \pi r^2 - 9\pi r^2}{4}$

= $\frac{7\pi r^2}{4}$
So, the percentage decrease in the curved surface area is,

= $\frac{7\pi r^2}{4}$ × $\frac{1}{4 \pi r^2}$ × 100%

= $\frac{7}{16}$ × 100%
= 43.75%
Therefore, the percentage decrease in the surface area of the sphere is 43.75% .

NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9