Question

The stupa of Sanchi is a great example of architecture in India. Its base part is
cylindrical in shape called Medhi. The dome of this stupa is called Anda, is
hemispherical in shape. It also contains a cubical shaped part called Harmika at the
top.

a. Find the lateral surface area of the Harmika, if the side of this part is 8 m
i 128 m2

ii 256 m2

iii 512 m2

iv 384 m2

b. The diameter and height of the Medhi are 42 m and 12 m respectively. If the
volume of a brick is 0.01 m3

, find the number of bricks used in the cylindrical

base
i 16,63,400 ii 1,66,320 iii 16,63,200 iv 17,20,180

c. The diameter of Anda is 42 m. Find the volume of the Anda.
i 19568 m3

ii 17545 m3

iii 18406 m3

iv 19404 m3

d. A priest is watching Chhatri at the top of the Stupa. If the height of the
Chhatri from the ground level is 60 m and the angle of elevation of the
Chhatri from the priest is 300

, find the distance of the priest from the base of

the Stupa.
i 60√3 m ii 20√3 m iii 40√3 m iv 60/√3 m
e. The radius of pradakshina path is 25 m. A boy walks 7 rounds on this path.
Find the distance covered by this boy.
i 1200 m ii 1100 m iii 1000 m iv 1500 m

Answer 1

Answer:

the stupa 63237 4363 the radius of earth is 25 a boy walks 7 rounds on this part III the distance covered by this boy 12000 aur 1000 beta aur beta find the distance of the priced from

Answer 2

Given,

Elements of stupa of Sanchi,

Medhi which is cylindrical in shape.

Anda which is hemispherical in shape.

Harmika which is cubical in shape.

To find,

a.) Lateral surface area of Harmika.

b.) Number of bricks used to build Medhi.

c.) Volume of Anda.

d.) Distance of priest from the base of stupa.

e.) Distance covered by the boy on pradakshina path.

Solution,

a.)

Given that side of the cube is, $a=8m$

And we know that,

Lateral surface of the cube,

$LSA=4side^{2}$

So,

$LSA=4a^{2}$

⇒$LSA=4(8)^{2}$

⇒$LSA=4(64)$

⇒$LSA=256m^{2}$

Therefore, lateral surface area of Harmika is $256m^{2}$. (Option ii)

b.)

Diameter of cylindrical Medhi, $d=42m$

⇒Radius, $r=21m$

Height of Medhi, $h=12m$.

So,

Volume of cylinder will be

⇒$V=\pi r^{2}h$

⇒$V=\frac{22}{7}$×$21$×$21$×$12$

⇒$V=22(63)(12)$

⇒$V=16632m^{3}$

Hence, volume of Medhi is $16632m^{3}$.

Now that we are given volume of brick,

$v=0.01m^{3}$

Number of bricks required will be,

⇒$N=\frac{V}{v}$

⇒$N=\frac{16632}{0.01}$

⇒$N=1,66,320$ bricks.

Therefore, number of bricks used to build Medhi was 1,66,320. (Option ii)

c.)

Diameter of hemispherical Anda,

$d=42m$.

Radius, $r=21m$

So, the volume of hemisphere will be,

⇒$V=\frac{2\pi r^{3}}{3}$

⇒$V=\frac{2(22)(21)^{3}}{3(7)}$

⇒$V=2(22)(21)(21)$

⇒$V=19404m^{3}$

Therefore, volume of Anda is $19404m^{3}$. (Option iv)

d.)

Height of chhatri from ground, $h=60m.$

Angle of elevation, $\alpha =30$°

In the triangle thus formed,

$tan\alpha =\frac{h}{x}$

Where, $x=$ distance between priest and the foot of stupa.

So,

⇒$tan30=\frac{60}{x}$
⇒$\frac{1}{\sqrt{3} }=\frac{60}{x}$

⇒$x=60\sqrt{3} m$

Therefore, distance between priest and foot of stupa is $60\sqrt{3}m$. (Option i)

e.)

Radius of pradakshina path, $r=25m$

So,

Circumference of path,

$C=2\pi r$

⇒$C=2(\frac{22}{7}) (25)$

⇒$C=\frac{1100}{7}m$

A boy walks 7 rounds.

So, total distance travelled by the boy,

⇒$s=7C$

⇒$s=7(\frac{1100}{7} )$

⇒$s=1100m$

Therefore, the boy walked $1100m$ distance. (Option ii)