Surface Area of a Combination of Solids Questions in English

(B) $1$. The article consists of a cylinder and two cones at its ends.
$2$. Radius of cylinder $(r)$ = $30 \, cm$.
$3$. Height of each cone $(h_{cone})$ = $40 \, cm$.
$4$. Total height of the article = $180 \, cm$.
$5$. Height of the cylinder $(h_{cyl})$ = $180 - (40 + 40) = 100 \, cm$.
$6$. Slant height of the cone $(l)$ = $\sqrt{r^2 + h_{cone}^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, cm$.
$7$. Total Surface Area = (Curved Surface Area of Cylinder) + $2$ $\times$ (Curved Surface Area of Cone).
$8$. Total Surface Area = $(2 \pi r h_{cyl}) + 2 \times (\pi r l) = 2 \pi r (h_{cyl} + l)$.
$9$. Total Surface Area = $2 \times 3.14 \times 30 \times (100 + 50) = 188.4 \times 150 = 28260 \, cm^2$.

(C) The article consists of a cylinder and two hemispheres at its ends.
Given: Radius $r = 21 \, cm$,Total height $H = 92 \, cm$.
The height of the cylindrical part $h = H - 2r = 92 - 2(21) = 92 - 42 = 50 \, cm$.
The total surface area of the article = (Curved surface area of cylinder) + $2$ $\times$ (Curved surface area of hemisphere).
Total surface area = $2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$.
Substituting the values: $2 \times \frac{22}{7} \times 21 \times (50 + 2(21)) = 2 \times 22 \times 3 \times (50 + 42) = 132 \times 92 = 12,144 \, cm^2$.

(D) The radius $(r)$ of the hemispherical dome is $21\, m$.
The curved surface area $(CSA)$ of a hemisphere is given by the formula $2\pi r^2$.
$CSA = 2 \times \frac{22}{7} \times 21 \times 21$
$CSA = 2 \times 22 \times 3 \times 21$
$CSA = 44 \times 63 = 2772\, m^2$.
The cost of painting is ₹ $80$ per $m^2$.
Total cost = $CSA \times \text{Rate} = 2772 \times 80 = 2,21,760$.
Thus,the total cost of painting the outer surface is ₹ $2,21,760$.

(A) Given: Radius $(r)$ = $7\, cm$,Height $(h)$ = $24\, cm$.
First,find the slant height $(l)$ of the cone: $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\, cm$.
The surface area to be painted consists of the curved surface area of the cone and the curved surface area of the hemisphere.
Curved Surface Area of Cone = $\pi rl = \frac{22}{7} \times 7 \times 25 = 550\, cm^2$.
Curved Surface Area of Hemisphere = $2\pi r^2 = 2 \times \frac{22}{7} \times 7 \times 7 = 308\, cm^2$.
Total Surface Area = $550 + 308 = 858\, cm^2$.
Cost of painting = Total Surface Area $\times$ Rate = $858 \times 2 = ₹ 1716$.

(B) The article consists of a cylinder with two cones attached at its ends.
Radius of cylinder $(r)$ = $21 \, cm$.
Height of cylinder $(h)$ = $50 \, cm$.
Height of cone $(H)$ = $28 \, cm$.
Slant height of cone $(l)$ = $\sqrt{r^2 + H^2} = \sqrt{21^2 + 28^2} = \sqrt{441 + 784} = \sqrt{1225} = 35 \, cm$.
The total surface area of the article = (Curved surface area of cylinder) + $2 \times$ (Curved surface area of cone).
Curved surface area of cylinder = $2 \pi r h = 2 \times \frac{22}{7} \times 21 \times 50 = 44 \times 3 \times 50 = 6600 \, cm^2$.
Curved surface area of one cone = $\pi r l = \frac{22}{7} \times 21 \times 35 = 22 \times 3 \times 35 = 2310 \, cm^2$.
Total surface area = $6600 + 2 \times 2310 = 6600 + 4620 = 11220 \, cm^2$.

(C) Given: Radius $(r)$ = $3.5\, m$,Height $(h)$ = $12\, m$.
First,find the slant height $(l)$ of the cone using the formula $l = \sqrt{r^2 + h^2}$.
$l = \sqrt{(3.5)^2 + (12)^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\, m$.
The curved surface area $(CSA)$ of the cone is given by $\pi rl$.
$CSA = \frac{22}{7} \times 3.5 \times 12.5 = 22 \times 0.5 \times 12.5 = 11 \times 12.5 = 137.5\, m^2$.
The cost of painting is at the rate of ₹ $30/m^2$.
Total cost = $137.5 \times 30 = ₹ 4125$.

(D) The article consists of a cylinder and two hemispheres at its ends.
Given: Radius of cylinder $(r)$ = $20 \, cm$,Total height $(H)$ = $80 \, cm$.
The height of the cylindrical part $(h)$ = $H - 2r = 80 - 2(20) = 80 - 40 = 40 \, cm$.
The total surface area of the article = (Curved surface area of cylinder) + $2$ $\times$ (Curved surface area of hemisphere).
$= 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$.
$= 2 \times 3.14 \times 20 \times (40 + 2 \times 20)$.
$= 125.6 \times (40 + 40) = 125.6 \times 80 = 10048 \, cm^2$.

(A) The article consists of a cylinder,a hemisphere,and a cone.
Radius of cylinder $(r)$ = $28 \, cm$,Height of cylinder $(h_1)$ = $54.5 \, cm$.
Radius of hemisphere $(r)$ = $28 \, cm$.
Radius of cone $(r)$ = $28 \, cm$,Height of cone $(h_2)$ = $21 \, cm$.
Slant height of cone $(l)$ = $\sqrt{r^2 + h_2^2} = \sqrt{28^2 + 21^2} = \sqrt{784 + 441} = \sqrt{1225} = 35 \, cm$.
The total surface area is the sum of the curved surface area of the cylinder,the curved surface area of the hemisphere,and the curved surface area of the cone.
Total Surface Area = $2\pi rh_1 + 2\pi r^2 + \pi rl$.
Total Surface Area = $\pi r (2h_1 + 2r + l) = \frac{22}{7} \times 28 \times (2 \times 54.5 + 2 \times 28 + 35)$.
Total Surface Area = $22 \times 4 \times (109 + 56 + 35) = 88 \times 200 = 17600 \, cm^2$.

(B) $1$. Identify the dimensions: Radius $(r)$ = $5 \, cm$,Height of each cone $(h_{cone})$ = $12 \, cm$,Total height $(H)$ = $41 \, cm$.
$2$. Calculate the height of the cylinder $(h_{cyl})$: $h_{cyl} = H - 2 \times h_{cone} = 41 - 2(12) = 41 - 24 = 17 \, cm$.
$3$. Calculate the slant height $(l)$ of the cone: $l = \sqrt{r^2 + h_{cone}^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm$.
$4$. The total surface area of the article consists of the curved surface area of the cylinder and the curved surface areas of the two cones.
$5$. Total Surface Area = $2 \pi r h_{cyl} + 2 \times (\pi r l) = 2 \pi r (h_{cyl} + l)$.
$6$. Substitute the values: $2 \times 3.14 \times 5 \times (17 + 13) = 31.4 \times 30 = 942 \, cm^2$.

(C) The article consists of a cylinder and a cone joined together.
Given: Radius of cylinder and cone $(r) = 6 \, cm$,Height of cone $(h_{cone}) = 8 \, cm$,Total height of article $= 20 \, cm$.
Height of cylinder $(h_{cyl}) = 20 \, cm - 8 \, cm = 12 \, cm$.
Slant height of cone $(l) = \sqrt{r^2 + h_{cone}^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, cm$.
Total Surface Area = (Curved Surface Area of Cylinder) + (Curved Surface Area of Cone) + (Area of base of cylinder).
Total Surface Area = $(2 \pi r h_{cyl}) + (\pi r l) + (\pi r^2)$.
Total Surface Area = $\pi r (2 h_{cyl} + l + r) = 3.14 \times 6 \times (2 \times 12 + 10 + 6)$.
Total Surface Area = $18.84 \times (24 + 10 + 6) = 18.84 \times 40 = 753.60 \, cm^2$.
Wait,re-evaluating: The article is open at the other end,so the base area is not included.
Surface Area = (Curved Surface Area of Cylinder) + (Curved Surface Area of Cone) = $2 \pi r h_{cyl} + \pi r l$.
Surface Area = $\pi r (2 h_{cyl} + l) = 3.14 \times 6 \times (2 \times 12 + 10) = 18.84 \times (24 + 10) = 18.84 \times 34 = 640.56 \, cm^2$.

(D) Let the radius of the hemispherical base be $r$ and the slant height of the cone be $l = 5 \, cm$.
The total surface area of the article is the sum of the curved surface area of the cone and the curved surface area of the hemisphere.
Total Surface Area $= \pi rl + 2\pi r^2 = 103.62 \, cm^2$.
Given $\pi = 3.14$,we have $3.14 \times r \times 5 + 2 \times 3.14 \times r^2 = 103.62$.
$15.7r + 6.28r^2 = 103.62$.
Dividing by $3.14$: $5r + 2r^2 = 33$.
$2r^2 + 5r - 33 = 0$.
Solving the quadratic equation: $2r^2 + 11r - 6r - 33 = 0 \implies r(2r + 11) - 3(2r + 11) = 0$.
$(r - 3)(2r + 11) = 0$. Since $r > 0$,$r = 3 \, cm$.
The height of the cone $h$ is given by $h = \sqrt{l^2 - r^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \, cm$.
The total height of the article is the sum of the height of the cone and the radius of the hemisphere: $H = h + r = 4 + 3 = 7 \, cm$.

(A) Given: Radius of the cone $(r)$ = $15 \, cm$,Height of the cone $(h)$ = $20 \, cm$.
First,calculate the slant height $(l)$ of the cone: $l = \sqrt{r^2 + h^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25 \, cm$.
The total surface area of the article consists of the curved surface area of the cone and the curved surface area of the hemisphere.
Curved surface area of the cone = $\pi rl = 3.14 \times 15 \times 25 = 1177.5 \, cm^2$.
Curved surface area of the hemisphere = $2\pi r^2 = 2 \times 3.14 \times 15^2 = 2 \times 3.14 \times 225 = 1413 \, cm^2$.
Total surface area = $1177.5 + 1413 = 2590.5 \, cm^2$.

(C) The total surface area of the tent is the sum of the curved surface area of the cylinder and the curved surface area of the cone.
Radius $(r)$ = $1.2\, m$.
Height of cylinder $(h_1)$ = $4\, m$.
Height of cone $(h_2)$ = $3.5\, m$.
Slant height of the cone $(l)$ = $\sqrt{r^2 + h_2^2} = \sqrt{(1.2)^2 + (3.5)^2} = \sqrt{1.44 + 12.25} = \sqrt{13.69} = 3.7\, m$.
Curved surface area of cylinder = $2\pi rh_1 = 2 \times \frac{22}{7} \times 1.2 \times 4 \approx 30.17\, m^2$.
Curved surface area of cone = $\pi rl = \frac{22}{7} \times 1.2 \times 3.7 \approx 13.95\, m^2$.
Total area = $30.17 + 13.95 = 44.12\, m^2$.
Rounding to the nearest whole number,the cloth used is $44\, m^2$.

(A) The article consists of a cylinder and two hemispheres attached at both ends.
Radius of the cylinder $(r)$ = $21 \, cm$.
Height of the cylinder $(h)$ = $18 \, cm$.
Radius of the hemispheres $(r)$ = $21 \, cm$.
The total surface area of the article = (Curved surface area of the cylinder) + $2 \times$ (Curved surface area of the hemisphere).
Curved surface area of the cylinder = $2 \pi r h = 2 \times \frac{22}{7} \times 21 \times 18 = 2 \times 22 \times 3 \times 18 = 2376 \, cm^2$.
Curved surface area of two hemispheres = $2 \times (2 \pi r^2) = 4 \pi r^2 = 4 \times \frac{22}{7} \times 21 \times 21 = 4 \times 22 \times 3 \times 21 = 5544 \, cm^2$.
Total surface area = $2376 + 5544 = 7920 \, cm^2$.

(C) Given: Radius of hemisphere $(r)$ = $3.5\, cm = \frac{7}{2}\, cm$.
Total height of the toy = $15.5\, cm$.
Height of the cone $(h)$ = Total height - Radius of hemisphere = $15.5 - 3.5 = 12\, cm$.
Slant height of the cone $(l)$ = $\sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\, cm$.
Total surface area of the toy = Curved surface area of cone + Curved surface area of hemisphere.
$= \pi rl + 2\pi r^2 = \pi r(l + 2r)$.
$= \frac{22}{7} \times 3.5 \times (12.5 + 2 \times 3.5) = 11 \times (12.5 + 7) = 11 \times 19.5 = 214.5\, cm^2$.

(D) The article consists of a cone placed on a hemisphere. The total surface area is the sum of the curved surface area of the cone and the curved surface area of the hemisphere.
$1$. Slant height of the cone $(l)$: $l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm$.
$2$. Curved surface area of the cone $(CSA_{cone})$: $\pi rl = 3.14 \times 5 \times 13 = 204.1 \, cm^2$.
$3$. Curved surface area of the hemisphere $(CSA_{hemi})$: $2\pi r^2 = 2 \times 3.14 \times 5^2 = 2 \times 3.14 \times 25 = 157 \, cm^2$.
$4$. Total surface area: $CSA_{cone} + CSA_{hemi} = 204.1 + 157 = 361.1 \, cm^2$.

(D) The $CSA$ (Curved Surface Area) of a cylinder is the area of the curved surface excluding the two circular bases.
For a cylinder with radius $r$ and height $h$,the $CSA$ is calculated by multiplying the circumference of the base $(2 \pi r)$ by the height $(h)$.
Therefore,$CSA = 2 \pi r h$.

(A) hemisphere is half of a sphere.
Since the total surface area of a sphere is $4 \pi r^{2}$,the curved surface area $(CSA)$ of a hemisphere is half of that value.
$CSA \text{ of a hemisphere} = \frac{1}{2} \times (4 \pi r^{2}) = 2 \pi r^{2}$.

(C) The surface area of a sphere with radius $r$ is given by the formula $A = 4 \pi r^{2}$.
Therefore,the correct option is $C$.

(D) The curved surface area $(CSA)$ of a cone with radius $r$ and slant height $l$ is given by the formula $\pi r l$.
Here,$r$ represents the radius of the circular base and $l$ represents the slant height of the cone.

(B) hemisphere is half of a sphere.
The curved surface area $(CSA)$ of a hemisphere is $2 \pi r^{2}$.
The base of a hemisphere is a circle with an area of $\pi r^{2}$.
The total surface area $(TSA)$ is the sum of the curved surface area and the base area.
$TSA = CSA + \text{Base Area} = 2 \pi r^{2} + \pi r^{2} = 3 \pi r^{2}$.

(D) The Total Surface Area $(TSA)$ of a cylinder consists of the curved surface area plus the area of the two circular bases.
Curved Surface Area $(CSA)$ $= 2 \pi r h$
Area of two circular bases $= 2 \times (\pi r^{2}) = 2 \pi r^{2}$
Total Surface Area $(TSA)$ $= 2 \pi r h + 2 \pi r^{2}$
Factoring out $2 \pi r$,we get $TSA = 2 \pi r(r+h)$.

(B) The total surface area of a cone is the sum of its curved surface area and the area of its circular base.
Curved Surface Area $= \pi rl$
Base Area $= \pi r^{2}$
Total Surface Area $= \pi rl + \pi r^{2} = \pi r(l + r)$
Where $r$ is the radius of the base and $l$ is the slant height of the cone.

(B) $5$-rupee coin is shaped like a cylinder with a very small height $h$.
The total surface area of a cylinder is given by the sum of the areas of its two circular bases and its curved surface area.
Total Surface Area $= 2 \times (\text{Area of base}) + \text{Curved Surface Area}$.
Total Surface Area $= 2(\pi r^2) + 2 \pi r h$.
Factoring out $2 \pi r$,we get $2 \pi r(r + h)$.
Thus,the correct formula is $2 \pi r(r + h)$.

(B) Let the radii of the two spheres be $r_1$ and $r_2$. Given that the ratio of their radii is $r_1 : r_2 = 3 : 5$.
The surface area of a sphere is given by the formula $S = 4\pi r^2$.
The ratio of the surface areas of the two spheres is $\frac{S_1}{S_2} = \frac{4\pi r_1^2}{4\pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2$.
Substituting the given ratio: $\frac{S_1}{S_2} = \left(\frac{3}{5}\right)^2 = \frac{9}{25}$.
Thus,the ratio of their surface areas is $9: 25$.

(C) Given: Radii of the frustum are $R = 10 \, cm$ and $r = 7 \, cm$. The height $h = 4 \, cm$.
First,we calculate the slant height $l$ of the frustum using the formula $l = \sqrt{h^2 + (R - r)^2}$.
Substituting the values: $l = \sqrt{4^2 + (10 - 7)^2} = \sqrt{16 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, cm$.
The curved surface area of a frustum of a cone is given by the formula $CSA = \pi (R + r) l$.
Substituting the values: $CSA = \pi (10 + 7) \times 5 = \pi (17) \times 5 = 85\pi \, cm^2$.
Thus,the curved surface area is $85\pi \, cm^2$.

(A) The formula for the curved surface area $(CSA)$ of a cone is given by $CSA = \pi r l$,where $r$ is the radius and $l$ is the slant height.
Given: $r = 12 \, cm$ and $l = 20 \, cm$.
Substituting the values into the formula:
$CSA = \pi \times 12 \times 20$
$CSA = 240 \pi \, cm^2$.
Therefore,the curved surface area is $240 \pi \, cm^2$.

(C) cube has $6$ faces in total. Each face of a cube with side length $l$ has an area of $l^2$.
If the cube is open at one end,it means one face is missing.
Therefore,the total surface area of the cube open at one end is the sum of the areas of the remaining $5$ faces.
Total surface area $= 5 \times l^2 = 5l^2$.

(A) cube is a three-dimensional shape having $6$ identical square faces.
If the side length of the cube is $l$,then the area of one square face is $l \times l = l^{2}$.
Since there are $6$ such faces in a closed cube,the total surface area is the sum of the areas of all $6$ faces.
Therefore,Total Surface Area $= 6 \times l^{2} = 6 l^{2}$.

(A) cylinder has two circular bases.
Since the base of a cylinder is a circle with radius $r$,the area of the base is given by the formula for the area of a circle.
Area of a circle $= \pi r^{2}$.
Therefore,the area of the base of a cylinder is $\pi r^{2}$.

(A) The curved surface area of a cylinder is the area of the side surface excluding the top and bottom circular bases.
For a cylinder with radius $r$ and height $h$,the curved surface area is calculated by multiplying the circumference of the base $(2 \pi r)$ by the height $(h)$.
Therefore,the formula is $2 \pi r h$.

(B) The total surface area of a cylinder consists of the area of the two circular bases and the curved surface area.
Area of two circular bases $= 2 \times (\pi r^2) = 2 \pi r^2$.
Curved surface area of the cylinder $= 2 \pi r h$.
Total surface area $= 2 \pi r^2 + 2 \pi r h$.
Factoring out $2 \pi r$,we get $2 \pi r(r + h)$.

(D) The curved surface area $(CSA)$ of a cone is given by the formula $\pi rl$,where $r$ is the radius of the base and $l$ is the slant height of the cone.

(C) The total surface area of a cone consists of the sum of the curved surface area and the area of the circular base.
$1$. The curved surface area of a cone is given by $\pi r l$,where $r$ is the radius and $l$ is the slant height.
$2$. The area of the circular base is $\pi r^{2}$.
$3$. Therefore,the total surface area $= \text{Curved surface area} + \text{Base area} = \pi r l + \pi r^{2}$.
$4$. Factoring out $\pi r$,we get $\pi r(l + r)$.
Thus,the correct option is $C$.

(B) The surface area of a sphere with radius $r$ is given by the formula $A = 4 \pi r^{2}$.
This represents the total area covered by the outer boundary of the sphere.

(C) sphere has a total surface area of $4 \pi r^{2}$.
Since a hemisphere is exactly half of a sphere,its curved surface area is half of the total surface area of the sphere.
Curved surface area of a hemisphere $= \frac{1}{2} \times (4 \pi r^{2}) = 2 \pi r^{2}$.

(C) In a right circular cone,the slant height $(l)$,the vertical height $(h)$,and the radius of the base $(r)$ form a right-angled triangle.
According to the Pythagorean theorem,the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Therefore,$l^{2} = h^{2} + r^{2}$.
Taking the square root on both sides,we get $l = \sqrt{h^{2} + r^{2}}$.

(C) hemisphere is half of a sphere. The curved surface area of a sphere is $4\pi r^{2}$,so the curved surface area of a hemisphere is $2\pi r^{2}$.
Additionally,the base of a hemisphere is a circle with an area of $\pi r^{2}$.
Therefore,the total surface area of a hemisphere is the sum of the curved surface area and the base area: $2\pi r^{2} + \pi r^{2} = 3\pi r^{2}$.

(D) cone is a three-dimensional geometric shape that tapers smoothly from a flat base to a point called the apex or vertex.
By definition,the base of a right circular cone is a circle.
Therefore,the correct option is $D$.

(B) The Total Surface Area $(TSA)$ of a cone is given by the formula: $TSA = \pi r(r + l) = \pi r^2 + \pi rl$.
The Curved Surface Area $(CSA)$ of a cone is given by the formula: $CSA = \pi rl$.
To find how much the $TSA$ exceeds the $CSA$,we subtract the $CSA$ from the $TSA$:
Difference $= TSA - CSA$
Difference $= (\pi r^2 + \pi rl) - (\pi rl)$
Difference $= \pi r^2 + \pi rl - \pi rl$
Difference $= \pi r^2$.
Thus,the total surface area of a cone exceeds its curved surface area by $\pi r^2$.

(B) $1-$ rupee coin is shaped like a thin cylinder.
For a cylinder with radius $r$ and height $h$,the curved surface area $(CSA)$ is given by the formula $A = 2 \pi r h$.
Here,$h$ represents the thickness of the coin.

(A) frustum of a cone is formed when a cone is cut by a plane parallel to its base.
Let $r_{1}$ and $r_{2}$ be the radii of the two circular ends of the frustum,and $l$ be the slant height of the frustum.
The formula for the Curved Surface Area $(CSA)$ of a frustum of a cone is derived from the difference between the lateral surface area of the original cone and the smaller cone removed from the top.
The formula is given by: $CSA = \pi l(r_{1} + r_{2})$.

(A) The total surface area $(TSA)$ of a frustum of a cone consists of the lateral surface area plus the areas of the two circular bases.
$1$. The lateral surface area of a frustum is given by $\pi l(r_{1} + r_{2})$,where $l$ is the slant height and $r_{1}, r_{2}$ are the radii of the two circular bases.
$2$. The area of the top circular base is $\pi r_{1}^{2}$.
$3$. The area of the bottom circular base is $\pi r_{2}^{2}$.
$4$. Therefore,the total surface area $(TSA)$ is the sum of these three parts: $TSA = \pi l(r_{1} + r_{2}) + \pi r_{1}^{2} + \pi r_{2}^{2}$.

(B) $5$-rupee coin is shaped like a thin cylinder.
The total surface area $(TSA)$ of a cylinder with radius $r$ and height $h$ is the sum of its curved surface area and the areas of its two circular bases.
$TSA = \text{Curved Surface Area} + 2 \times \text{Area of circular base}$
$TSA = 2 \pi r h + 2 \pi r^2$
By taking $2 \pi r$ as a common factor,we get:
$TSA = 2 \pi r(r + h)$
Therefore,the correct formula is $A = 2 \pi r(r + h)$.

(B) The base of a cylinder is a circle.
Given,the radius of the base $r = 7 \, cm$.
The area of the base of a cylinder is given by the formula $\text{Area} = \pi r^2$.
Substituting the value of $r$ and $\pi = \frac{22}{7}$:
$\text{Area} = \frac{22}{7} \times 7 \times 7$
$\text{Area} = 22 \times 7 = 154 \, cm^2$.
Thus,the area of the base is $154 \, cm^2$.

(C) The formula for the curved surface area $(CSA)$ of a cylinder is $2 \pi r h$.
Given:
Radius $(r)$ = $1.4 \, cm = \frac{14}{10} \, cm$
Height $(h)$ = $10 \, cm$
Substituting the values into the formula:
$CSA = 2 \times \frac{22}{7} \times \frac{14}{10} \times 10$
$CSA = 2 \times 22 \times 2$
$CSA = 88 \, cm^2$
Therefore,the curved surface area is $88 \, cm^2$.

(B) The base of a cylinder is a circle. The area of a circle is given by the formula $A = \pi r^{2}$.
Given that the area of the base is $616\,cm^{2}$.
So,$616 = \frac{22}{7} \times r^{2}$.
Rearranging the equation to solve for $r^{2}$:
$r^{2} = 616 \times \frac{7}{22}$.
$r^{2} = 28 \times 7$.
$r^{2} = 196$.
Taking the square root of both sides:
$r = \sqrt{196} = 14\,cm$.
Therefore,the radius of the base of the cylinder is $14\,cm$.

(C) For a cone,the slant height $l$ is given by the formula $l = \sqrt{r^2 + h^2}$.
Given,radius $r = 6 \, cm$ and height $h = 8 \, cm$.
Substituting the values into the formula:
$l = \sqrt{6^2 + 8^2}$
$l = \sqrt{36 + 64}$
$l = \sqrt{100}$
$l = 10 \, cm$.
Therefore,the slant height of the cone is $10 \, cm$.

(C) The formula for the $CSA$ (Curved Surface Area) of a cone is given by $CSA = \pi r l$.
Given radius $r = 4 \, cm$ and slant height $l = 6 \, cm$.
Substituting these values into the formula:
$CSA = \pi \times 4 \times 6 = 24 \pi \, cm^{2}$.

(B) The formula for the $CSA$ (Curved Surface Area) of a cone is given by $CSA = \pi r l$.
Given radius $r = 7\, cm$ and slant height $l = 15\, cm$.
Substituting the values into the formula:
$CSA = \frac{22}{7} \times 7 \times 15$
$CSA = 22 \times 15 = 330\, cm^2$.