Surface Areas and Volumes: 9 Math

NCERT Exercise 13.9 (Optional): 9th math

Surface Areas And Volumes Class nine Math NCERT Exercise 13.9 (optional)  Question (1) A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (see the 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.

Solution

Given, External height of the bookshelf = 110 cm

External depth of the bookshelf = 25 cm

And, External breadth of the bookshelf = 85 cm

And, thickness of the plank = 5 cm

Rate of polishing = 20 paise per cm²

And, rate of painting = 10 paise per cm²

And, external faces are to be polished and inner faces are to be painted

Thus, total expenses for painting and polishing of the given bookshelf = ?

Solution

Calculation of internal Surface Area of Compartment

Here, height of the book shelf = 110 cm

And, thickness of the plank = 5 cm

And, here it is clear after the observation of figure given, that there are three partition in the book shelf.

And total number of plank = 4

Thus, total thickness of planks = thickness of 1 plank × 4

= 5 cm × 4 = 20 cm

Thus, the total internal height of the bookshelf

= 110 cm – 20 cm = 90 cm

And, there are three partitions in the book shelf

Thus, height of each partition = 90/3 = 30 cm

And breadth of each partition = total breadth – (2 × thickness)

= 85 cm – (2 × 5 cm)

= 85 cm – 10 cm = 75 cm

Thus, breadth of each partition = 75 cm

And, depth of each partition = total depth – thickness of plank

= 25 cm – 5 cm = 20 cm

Now, internal curved surface area of one partition

= 2[(breadth × depth) + (height × depth)] + (height × breadth)

= 2 [ (75 cm × 20 cm) + (30 cm × 20 cm) ] + (30 cm × 75 cm )

= 2 (1500 cm² + 600 cm²) + 2250 cm²

= 2 × 2100 cm² + 2250 cm²

= 4200 cm² + 2250 cm²

= 6450 cm²

Thus, internal surface area of one compartment = 6450 cm²

And, here total number of compartment in the given book shelf = 3

Thus, total surface area of all the three compartments = 3 × 6450 cm³

= 19350 cm²

Calculation of cost of painting of internal area of compartment

Now, as given, rate of painting of internal surface of bookshelf = 10 paise/cm²

Now, since cost of painting internal surface of 1 m² = 10 paise

Therefore, cost of painting of 19350 cm²

= 19350 × 10 = 193500 paise

= Rs1935.00

Thus, cost of painting of internal surface of given bookshelf = Rs 1935.00

Calculation of outer surface Area of book shelf

As given, height of book shelf = 110 cm

Breadth of book shelf = 85 cm

And, depth of book shelf = 25 cm

Thus, total outer surface area of bookshelf

= 2[(height × depth) + (breadth × depth) + (height × breadth)]

= 2[(110 cm × 25 cm) + (85 cm × 25 cm) + (110 cm × 85 cm)]

= 2(2750 cm² + 2125 cm² + 9350 cm²)

= 2 × 14225 cm²

= 28450 cm²

Thus, total outer surface area of bookshelf = 28450 cm²

Now, since there are three compartments in book shelf and there is no front doors for compartment. This means outer surface of bookshelf is to be painted excluding the front faces of compartments.

Now, area of front face of one compartment = height of front face × breadth of front face

= 30 cm × 75 cm

= 2250 cm²

Thus, area of front face of 3 compartments = 3 × 2250 cm²

Thus, total area of all front face of all the 3 compartments = 6750 cm²

Thus, total outer area to be painted

= total outer area of bookshelf – area of front faces of all the 3 compartments

= 28450 cm² – 6750 cm²

= 21700 cm²

Thus, total outer area of book shelf to be painted = 21700 cm²

Calculation of polishing of outer surface including planks of given bookshelf

Since, cost of polishing of 1 cm² = 20 paise

Therefore, cost of polishing of 21700 cm² = 20 × 21700

= 4354000 paise = Rs 4340.00

Now, total cost of polishing and painting

= Total cost of painting + total cost of polishing

= Rs 1935.00 + Rs 4340.00

= Rs 6275.00

Thus, total cost of painting and polishing of given bookshelf = Rs 6275.00 Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.9 (optional)  Question (2) The front compound was 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.

Solution

Given, Diameter of wooden sphere = 21 cm

Therefore, radius of the sphere = 21/2 = 10.5 cm

Height of the cylindrical support pipe = 7 cm

And, radius of cylindrical support pipe = 1.5 cm

Number of wooden spheres to be decorated = 8

Rate of silver paint for spheres = 25 paise per cm²

And, rate of black paint which for cylindrical support = 5 paise per cm²

Therefore, total cost of paint = ?

Calculation of Surface area of Spheres

As given, radius of spheres = 10.5 cm

Now, We know that, Surface Area of a Shpere = 4 ℼ r²

Thus, surface area of given wooden spheres

= 4 × `22/7` × (10.5 cm)²

= 4 × `22/7` × 10.5 cm × 10.5 cm

= 4 × 22 × 1.5 cm × 10.5 cm

= 88 × 15.75 cm²

= 1386 cm²

Thus, surface area of one wooden sphere = 1386 cm²

Here, since, wooden spheres area installed on given cylindrical mount, thus, surface area of a sphere is painted silver excluding the area covered by cylindrical sphere.

Now, since as per question, the radius of cylindrical mount = 1.5 cm

And, we know that, area of a circle = ℼ r²

Thus, approximate area of sphere excluded while painting

= ℼ × (1.5 cm)²

= `22/7` × 2.25 cm²

= 7.071 cm²

Thus, approximate area of one wooden sphere to be painted silver

= Area of one sphere – Area excluded due to overlap of cylindrical mount

= 1386 cm² – 7.071 cm²

= 1378.929 cm²

≃ 1379 cm²

Thus, total surface area of one sphere to be painted = 1379 cm²

Calculation of silver paint for one sphere

Now, since cost silver paint for 1 cm² = 25 paise = Rs 0.25

Thus, cost silver paint for 1379 cm² = Rs 0.25 × 1379

= 344.75

Thus, cost of silver paint for one sphere = Rs 344.75

Calculation of Curved Surface Area of Cylindrical Mount

Here, height of cylindrical mount = 7 cm

And, radius of cylindrical mount = 1.5 cm

Now, we know that Curved Surface Area of a Cylinder

= 2 ℼ r h

Thus, curved surface area of one given cylindrical mount

= 2 × `22/7` × 1.5 cm × 7 cm

= 2 × 22 × 1.5 cm × 1 cm

= 44 × 1.5 cm²

= 66 cm²

Thus, curved surface area to be painted black of one cylindrical mount = 66 cm²

Calculation of black paint for one cylindrical mount

Now, since cost of black paint for 1 cm² = 5 paise

Thus, cost of black paint for 66 cm² = 5 × 66 paise

= 330 paise = Rs 3.30

Calculation of total cost of painting

Now total cost of painting of one spherical mount

= Cost of silver paint for one sphere + cost of black paint for one cylindrical mount

= Rs 344.75 + Rs 3.30

= Rs 348.05

Thus, cost of all given 8 similar wooden sphere along with cylindrical mount

= cost of painting for one spherical mount × 8

= Rs 348.05 × 8

= Rs 2784.40

Thus, total cost of painting for all the given balls along with mount

≃ Rs 2784.40 Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.9 (optional)  Question (3) The diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?

Solution

Given, decrease in diameter of a sphere = 25%

Therefore, percent of decrease in curved surface area of given sphere = ?

As per question, here diameter of a sphere is decreased by 25%, thus radius will also decrease by 25%.

Now, let radius of the given sphere 1 (R) = 100 m

Therefore, radius of sphere 2 after decrease of 25 % (r) = 75 m

Now, We know that, Curved Surface area of a sphere

= 4 ℼ r²

Thus, Curved surface area of sphere 1

= 4 `22/7` × (100 m)²

= 4 × `22/7` 10000 m²

= 125717.285 m²

Now, curved surface area of sphere 2

= 4 × `22/7` (75 m)²

= 4 × `22/7` × 5625 m²

= 70714.285 m²

Now, decrease in curved surface area because of decrease in radius by 25%

= Curved surface area of sphere 1 – curved surface area of sphere 2

= 125717.285 m² – 70714.285 m²

= 55003 m²

Calculation of decrease in curved surface area of given sphere

Now, in 125717.285 m², decrease = 55003

Thus, in 100, decrease `=(55003 m^2)/(125717.285 m^2)xx100`

= 43.75%

Thus, decrease in curved surface area of a sphere due to 25% decrease in radius

= 43.75% Answer

Shortcut Method

Let, radius of sphere 1 = 100 m

Therefore, radius after decrease of 25% of sphere 2 = 75 m

Now, ratio of radii of sphere 1 and sphere 2 = 100:75

`=100/75 = (25xx4 )/(25xx3)=4/3`

Now, we know that, ratio of curved surface areas two spheres = square of their radii

`=(4/3)^2 = 16/9`

Thus, here curved surface area of sphere 1 = 16

And, curved surface area of sphere 2 = 9

And, decrease in curved surface area = 16 – 9 = 7

Now, since in 16 decrease = 7

Therefore, in 1 decrease = 7/16

Therefore, in 100 decrease `=7/16xx100 = 43.75%`

Thus, decrease in curved surface area of a sphere due to 25% decrease in radius

= 43.75% Answer

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