GSEB Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.9

Question 1. A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (See figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner surfaces 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: First, we will calculate the surface area to be polished.
Surface area for polishing \( = [(110 \times 85) + 2(110 \times 25) + 2(85 \times 25) + 2(110 \times 5) + 4(75 \times 5)] \)
\( = (9350 + 5500 + 4250 + 1100 + 1500) \) cm² \( = 21700 \) cm²
Expenses for polishing at 20 paise per cm² \( = 21700 \times 20 \) paise
\( = \frac{21700 \times 20}{100} \) Rs \( = 4340 \) Rs

Next, we will find the surface area to be painted.
Surface area for painting \( = [2(20 \times 90) + 6(75 \times 20) + (75 \times 90)] \)
\( = (3600 + 9000 + 6750) \) cm² \( = 19350 \) cm²
Expenses for painting at 10 paise per cm² \( = 19350 \times 10 \) paise
\( = \frac{19350 \times 10}{100} \) Rs \( = 1935 \) Rs

Finally, we compute the total expenses needed for polishing and painting the bookshelf surface.
Total expenses \( = \) Rs \( 4340 + \) Rs \( 1935 = \) Rs \( 6275 \)
In simple words: First, calculate the area for polishing and its cost. Then, calculate the area for painting and its cost. Add both costs together to get the total money needed.

Exam Tip: Remember to differentiate between external and internal surfaces and account for the plank thickness when calculating dimensions for painting versus polishing areas.

Question 2. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in figure. Eight such spheres are used for this purpose and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm².
Answer: For a wooden sphere:
Diameter \( = 21 \) cm
Radius \( (r) = \frac{21}{2} \) cm
Surface area of a wooden sphere to be painted \( = 4\pi r^2 - \pi (1.5)^2 \)
\( = 4 \times \frac{22}{7} \times (\frac{21}{2})^2 - \frac{22}{7} (1.5)^2 \)
\( = 1386 - 7.07 = 1378.93 \) cm² (approx.)
Surface area of eight wooden spheres \( = 1378.93 \times 8 = 11031.44 \) cm²
Cost of painting silver at 25 paise per cm² \( = 11031.44 \times 25 \) paise
\( = \frac{11031.44 \times 25}{100} \) Rs \( = 2757.86 \) Rs

For a cylindrical support:
Radius \( (r) = 1.5 \) cm
Height \( (h) = 7 \) cm
Surface area of a cylindrical support \( = 2\pi rh \)
\( = 2 \times \frac{22}{7} \times 1.5 \times 7 = 66 \) cm²
Surface area of eight cylindrical supports \( = 66 \times 8 = 528 \) cm²
Cost of painting black at 5 paise per cm² \( = 528 \times 5 \) paise
\( = \frac{528 \times 5}{100} \) Rs \( = 26.40 \) Rs

Total cost of paint required \( = (2757.86 + 26.40) \)
\( = 2784.26 \) Rs (approx.)
In simple words: Calculate the silver paint cost for all spheres by subtracting the area covered by cylinders. Then, calculate the black paint cost for all cylindrical supports. Finally, add these two costs together to get the total painting expense.

Exam Tip: Always remember to subtract the area where two shapes meet (like the sphere and cylinder base) if that part is not to be painted. Also, convert paise to rupees at the end for the final cost.

Question 3. The diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?
Answer: Let the initial diameter of the sphere be \( r \) cm. Then, the initial radius of the sphere will be \( \frac{r}{2} \) cm.
The curved surface area of the original sphere is \( 4\pi (\frac{r}{2})^2 = \pi r^2 \) cm².

If the diameter is decreased by 25%, the new diameter becomes:
\( r - \frac{25}{100}r = r - \frac{1}{4}r = \frac{3r}{4} \) cm.
The radius of the new sphere will be \( \frac{1}{2}(\frac{3r}{4}) = \frac{3r}{8} \) cm.

The new curved surface area of the sphere is \( 4\pi (\frac{3r}{8})^2 = 4\pi \frac{9r^2}{64} = \frac{9\pi r^2}{16} \) cm².

The decrease in the original curved surface area is:
\( \pi r^2 - \frac{9\pi r^2}{16} \)
\( = \frac{16\pi r^2 - 9\pi r^2}{16} \)
\( = \frac{7\pi r^2}{16} \)

The percentage of decrease in the original curved surface area is calculated as:
\( = \frac{\frac{7\pi r^2}{16}}{\pi r^2} \times 100 \% \)
\( = \frac{7}{16} \times 100 \% \)
\( = \frac{700}{16} \% \)
\( = \frac{175}{4} \% \)
\( = 43\frac{3}{4} \% \) or \( 43.75\% \)
Therefore, the original curved surface area decreases by \( 43.75\% \).
In simple words: When a sphere's diameter gets 25% smaller, its curved surface area also gets smaller. The total reduction in the surface area is 43.75%.

Exam Tip: Remember that surface area is proportional to the square of the radius or diameter. A percentage change in diameter will result in a squared percentage change in the area, so don't just multiply by 25%.