EXERCISE 13.9
Question 1:
A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (see the given figure). The thickness of the plank is 5 cm 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:
External height (l) of book self = 85 cm
External breadth (b) of book self = 25 cm
External height (h) of book self = 110 cm
External surface area of shelf while leaving out the front face of the shelf
= lh + 2 (lb + bh)
= [85 × 110 + 2 (85 × 25 + 25 × 110)] cm2
= (9350 + 9750) cm2
= 19100 cm2
Area of front face = [85 × 110 − 75 × 100 + 2 (75 × 5)] cm2
= 1850 + 750 cm2
= 2600 cm2
Area to be polished = (19100 + 2600) cm2 = 21700 cm2
Cost of polishing 1 cm2 area = Rs 0.20
Cost of polishing 21700 cm2 area Rs (21700 × 0.20) = Rs 4340
It can be observed that length (l), breadth (b), and height (h) of each row of the book shelf is 75 cm, 20 cm, and 30 cm respectively.
Area to be painted in 1 row = 2 (l + h) b + lh
= [2 (75 + 30) × 20 + 75 × 30] cm2
= (4200 + 2250) cm2
= 6450 cm2
Area to be painted in 3 rows = (3 × 6450) cm2 = 19350 cm2
Cost of painting 1 cm2 area = Rs 0.10
Cost of painting 19350 cm2 area = Rs (19350 × 0.1)
= Rs 1935
Total expense required for polishing and painting = Rs (4340 + 1935)
= Rs 6275
Therefore, it will cost Rs 6275 for polishing and painting the surface of the bookshelf.
Page No 237:
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 the given 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:
Radius (r) of wooden sphere $=\left(\frac{21}{2}\right) \mathrm{cm}$=10.5cm
Surface area of wooden sphere = 4πr2
$=\left[4 \times \frac{22}{7} \times(10.5)^{2}\right] \mathrm{cm}^{2}=1386 \mathrm{cm}^{2}$
Radius (r1) of the circular end of cylindrical support = 1.5 cm
Height (h) of cylindrical support = 7 cm
CSA of cylindrical support = 2πrh
$=\left[2 \times \frac{22}{7} \times(1.5) \times 7\right] \mathrm{cm}^{2}$
=66cm2
Area of the circular end of cylindrical support = πr2
$=\left[\frac{22}{7} \times(1.5)^{2}\right] \mathrm{cm}^{2}$
= 7.07 cm2
Area to be painted silver = [8 × (1386 − 7.07)] cm2
= (8 × 1378.93) cm2 = 11031.44 cm2
Cost for painting with silver colour = Rs (11031.44 × 0.25) = Rs 2757.86
Area to be painted black = (8 × 66) cm2 = 528 cm2
Cost for painting with black colour = Rs (528 × 0.05) = Rs 26.40
Total cost in painting = Rs (2757.86 + 26.40)
= Rs 2784.26
Therefore, it will cost Rs 2784.26 in painting in such a way.
Question 3:
The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?Answer:
Let the diameter of the sphere be d.Radius (r1) of sphere $=\frac{d}{2}$
New radius $\left(\mathrm{r}_{2}\right)$ of sphere $=\frac{d}{2}\left(1-\frac{25}{100}\right)=\frac{3}{8} d$
CSA (S1) of sphere $=4 \pi r_{1}^{2}$
$=4 \pi\left(\frac{d}{2}\right)^{2}=\pi d^{2}$
CSA (S2) of sphere when radius is decreased $=4 \pi r_{2}^{2}$
$=4 \pi\left(\frac{3 d}{8}\right)^{2}=\frac{9}{16} \pi d^{2}$
Decrease in surface area of sphere = S1 − S2
$=\pi d^{2}-\frac{9}{16} \pi d^{2}$
$=\frac{7}{16} \pi d^{2}$
Percentage decrease in surface area of sphere$=\frac{S_{1}-S_{2}}{S_{1}} \times 100$
$=\frac{7 \pi d^{2}}{16 \pi d^{2}} \times 100$
$=\frac{700}{16}$
=43.75%
0 comments:
Post a Comment