# Case Study Question 03

## Chapter 13: Surface Areas and Volumes

### Class 10

Read the following passage carefully and answer the following questions:

Visit to Sanchi Stupa

Ajay is a Class X student. His class teacher Mrs. Kiran arranged a historical trip to great Stupa of Sanchi. She explained that Stupa of Sanchi is great example of architecture in India. Its base part is cylindrical in shape. The dome of this stupa is hemispherical in shape, known as Anda. It also contains a cubical shape part called Hermika at the top. Path around Anda is known as Pradakshina Path. Question.1.
Find the lateral surface area of the Hermika, if the side of cubical part is 8 m.

(a) 128 m^{2}
(b) 256 m^{2}
(c) 512 m^{2}
(d) 1024 m^{2}

(b) : Lateral surface area of Hermika which is cubical in shape = 4 a^{2}
= 4 \times (8)^{2}
= 256 m^{2}

Question.2.
The diameter and height of the cylindrical base part are respectively 42 m and 12 m. If the volume of each brick used is 0.01 m^{3}, then find the number of bricks used to make the cylindrical base.

(a) 1663200
(b) 1580500
(c) 1765000
(d) 1865000

(a) : Diameter of cylindrical base = 42 m
So, Radius of cylindrical base (r) = 21 m
Height of cylindrical base (h) = 12 m
Therefore, Number of bricks used = = \frac{(\frac{22}{7}) \times 21\times 21\times 12}{0.01}
= 1663200

Question.3.
If the diameter of the Anda is 42 m, then the volume of the Anda is

(a) 17475 m^{3}
(b) 18605 m^{3}
(c) 19404 m^{3}
(d) 18650 m^{3}

(c) : Given, diameter of Anda which is hemispherical in shape = 42 m
⇒ Radius of Anda (r) = 21 m
Therefore, Volume of Anda = \frac{2}{3} \pi r^{3}
= \frac{2}{3} \times \frac{22}{7} \times 21 \times21 \times 21
=  44 \times 21 \times 21
= 19404 m^{3}

Question.4.
The radius of the Pradakshina path is 25 m. If Buddhist priest walks 14 rounds on this path, then find the distance covered by the priest.

(a) 1860 m
(b) 3600 m
(c) 2400 m
(d) 2200 m

(d) : Given, radius of Pradakshina Path (r) = 25 m
So, Perimeter of path = 2 \pi r
= (2 \times \frac{22}{7} \times 25) m
Therefore, Distance covered by priest = 14 \times 2 \times \frac{22}{7}\times 25
= 2200 m

Question.5.
The curved surface area of the Anda is

(a) 2856 m^{2}
(b) 2772 m^{2}
(c) 2473 m^{2}
(d) 2652 m^{2}

(b) : Since, radius of Anda (r) = 21 m
Therefore, Curved surface area of Anda = 2 \pi r^{2}
= 2 \times \frac{22}{7}\times 21 \times 21
= 2772 m^{2}

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