Question

$\huge \fbox \red{❥ Question}$


Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end, as shown in the figure. The height of the cylinder is 1.45 m, and its radius is 30 cm. Find the total surface area of the bird-bath. (Take π = 22/7)



(REQUIRED QUALITY ANSWER)​

Answer 1

Answer:

Question :-

Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end, as shown in the figure. The height of the cylinder is 1.45 m, and its radius is 30 cm. Find the total surface area of the bird-bath. (Take π = 22/7)

Given :-

• Height of cylinder = 1.45 m
• Radius = 30 cm
• π = 22/7

To Find :-

TSA

Solution :-

Let us assume r as common radius and h be height of hollow cylinder.

Now,

Conversion of unit

$\fbox{1.45 m \: = 1(100) + 45 = 145\: cm}$

Now,

$\sf \: TSA \: (bird \: path) = CSA(cylinder) + CSA(hemisphere)$

$\sf \: TSA = 2\pi rh + 2\pi \: {r}^{2}$

$\sf \: TSA \: = 2\pi \: (height + radius)$

$\sf \: TSA = 2 \times \dfrac{22}{7} \bigg(145 + 30 \bigg)$

$\sf \: TSA = 2 \times \dfrac{22}{7} \bigg(175 \bigg)$

$\sf \: TSA = \dfrac{44}{7} \times 175$

$\sf \: TSA = 44 \times 25$

$\frak \pink{ TSA \: = 1100 \: {cm}^{2} }$

Answer 2

Given :-

• Height of cylinder = 1.45 m

• Radius = 30 cm

• π = 22/7

To Find :-

• TSA

Solution :-

Let us assume r as common radius and h be height of hollow cylinder.

Now,

Conversion of unit

${\sf{1.45 = 1 \times 100 + 45 = 145}} \\ \\ \sf{Now,} \\ \\ \sf{T.S.A(Bird \: path) = C.S.A(Cylinder) + C.S.A(Himesphere)} \\ \\ \sf{T.SA = 2\pi{rh} + 2\pi {r}^{2} } \\ \\ \sf{T.S.A = 2\pi(Hieght + Radius)} \\ \\ \sf{T.S.A = 2 \times \frac{22}{7}\bigg\lgroup145 + 30\bigg\rgroup } \\ \\ \sf{T.S.A = 2 \times \frac{22}{7} \bigg\lgroup175\bigg\rgroup} \\ \\ \sf{T.S.A = \frac{44}{7} \times 175} \\ \\ \sf{T.S.A = 44 \times 25} \\ \\ {\underline{\boxed{\bf{T.S.A = 1100 {cm}^{2} }}}}$

Therefore, T.S.A = 1100cm².