RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.1 Volume and Surface Areas of Solids

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.1: The Physics Wallah academic team has produced a comprehensive answer for Chapter 19 Volume and Surface Area of Solids in the RS Aggarwal class 10 textbook. Complete the NCERT exercise questions and utilise them as a guide. Solutions for Physics Wallah NCERT Class 10 Maths problems in the exercise require assistance to be completed. Class 10 Math NCERT solutions were uploaded by Physics Wallah.

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RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.1 Volume and Surface Areas of Solids Overview

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RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.1 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.1 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.1

Q. Two cubes each of volume 27 $c{m}^3$ are joined end to end to form a solid. Find the surface area of the resulting cuboid.

Solution

Given volume of one cube = 27 $c{m}^3$ side of a cube is = a $a^3$ = 27 a = $\sqrt[3]{327}$ a = 3 now we got side of cuboid l = 3+3= 6 b = 3 h = 3 surface area of cuboid = 2 (lb+bh+hl) = 2 (6 $\times$ 3+3 $\times$ 3+3 $\times$ 3) = 2 (36) = 2 $\times$ 36 = 72 $c{m}^2$

Q. The volume of a hemisphere is $2425\frac{1}{2}c{m}^3.$ Find its curved surface area.

Solution

Given : Volume of hemisphere = $2425\frac{1}{2}c{m}^3$ or $2425.5c{m}^3$ Volume of hemisphere = 2/3πr³ 2425.5 = $\frac{2}{3}\times \frac{22}{7}\times r^3$ r³ = $\frac{2425.5\times 21}{44}$ r³ = $\frac{50935.5}{44}$ r³ = 1157.625 r = 10.5 cm Now, Curved Surface Area of hemisphere = 2πr² = $2\times \frac{22}{7}\times 10.5\times 10.5$ = 693 $c{m}^2$ CSA of hemisphere is 693 $c{m}^2$

Q. (i) A 5-m-wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used at the rate of Rs 25 per metre.

Solution

TSA OF A HEMISPHERE = 3πr² 3πr²=462 3× $\frac{22}{7}$ ×r²=462 $\frac{66}{7}$ r²=462 r²=462× $\frac{7}{66}$ r²=49 r=√49=7 volume of hemisphere = $\frac{2}{3}$ πr³ $\frac{2}{3}$ × $\frac{22}{7}$ ×7×7×7=718.6cm³

(ii) The radius and height of a solid right-circular cone are in the ratio of 5 : 12. If its volume is 314 $c{\mathrm{m3,\; find}}^{}$ its total surface area. $[Take\pi =\mathrm{3.14.}]$ same as above Q. If the volumes of two cones are in the ratio of 1 : 4 and their diameters are in the ratio of 4 : 5, find the ratio of their heights.

Solution

Here, $\frac{d_1}{d_2}=\frac{4}{5}$ $\Rightarrow \frac{r_1}{r_2}=\frac{4}{5}$ Now, $\frac{V_1}{V_2}=\frac{\frac{1}{3}\pi r_1^2{h}_1}{\frac{1}{3}\pi r_2^2{h}_2}=\frac{r_1^2{h}_1}{r_2^2{h}_2}$ $\Rightarrow \frac{V_1}{V_2}=\frac{r_1^2}{r_2^2}\times \frac{h_1}{h_2}$ $\Rightarrow \frac{1}{4}={\left(\frac{4}{5}\right)}^2\times \frac{h_1}{h_2}$ $\Rightarrow \frac{h_1}{h_2}=\frac{25}{64}$

Q. The sum of the radius of the base and the height of a solid cylinder is 37 metres. If the total surface area of the cylinder be 1628 sq metres, find its volume.

Solution

Let r and h be the radius and height of the solid cylinder respectively. Given, r + h = 37 cm Total surface area of the cylinder = 1628 $c{m}^2$ (Given) ∴ 2 πr (r + h) = 1628 ⇒ 2 πr × 37 = 1628 ⇒ r = 7 cm Now, r + h = 37 cm ⇒7 + h = 37 ⇒ h = 37 – 7 = 30 cm Volume of the cylinder = $\pi r^{2}h=\frac{22}{7}\times 7^2\times 30=4620c{m}^3$

Q. The surface area of a sphere is 2464 $c{m}^2.$ If its radius be doubled, what will be the surface area of the new sphere?

Solution

The surface area of a sphere is given by $4\pi r^2$ So, it is proportional to the square of the radius. If the radius is doubled, surface area will become 4 times. New surface area = $4\times 2464=9856c{m}^2$

Q. A military tent of height $8.25$ m is in the form of a right circular cylinder of base diameter $30m$ and height $5.5m$ surmounted by a right circular cone of same base radius. Find the length of canvas used in making the tent, if the breadth of the canvas is $1.5m.$

Solution

Given, Height of the military tent $=8.25m$ Height of the circular cylinder $4=5.5m$ As, Height of the military tent = Height of the circular cylinder + Height of the right circular cone So, Height of the right circular cone $=8.25-5.5$ Therefore, height of the right circular cone $=2.75m$ We know that, Slant height of the cone(l)= $\sqrt{h^2+r^2}$ ​where, $r=$ radius of base and $h=$ altitude height of cone radius of cylinder = radius of cone Therefore radius of cone $=\frac{30}{2}=15m$ $l=$ $\sqrt{h^2+r^2}$ $l=$ $\sqrt{{15}^2+{2.75}^2}$ $l=$ $\sqrt{225+7.5625}$ ​Slant height, $l=15.25m$ Surface area of cone= $\pi \times r\times l$ $=3.14\times 15\times 15.25=718.3$ Surface area of cylinder = 2 $\pi \times r\times h$ $=2\times 3.14\times 15\times 5.5=518.1$ Area of the canvas = Surface area of cone + Surface area of cylinder $=718.9+518.57$ $=1236.47m^2$ ∴ Length of canvas $=\frac{1236.47}{1.5}=824.3m$

Q. A tent is in the shape of a right circular cylinder up to a height of 3 m and conical above it. The total height of the tent is 13.5 m and the radius of its base is 14 m. Find the cost of cloth required to make the tent at the rate of Rs 80 per square metre. $[Take\pi =\frac{22}{7.}]$

Solution

CSA of cylinder $=2\pi rh$ $2\times \frac{22}{7}\times 14\times 3=264m^2$ radius = 14 m height = 13.5 - 3 = 10.5 $l^2=r^2+h^2$ $14{}^{}+10.5{}^{}196+110.25=306.25l\sqrt{306.25}l=17.5m$ CSA of cone $=\pi rl$ $\frac{22}{7}\times 14\times 17.5=770m^2$ total area $=264+770=1034m^2$ cost of cloth per square m = Rs 80 cost of cloth $=1034m^2=80\times 1034=82720$

Q. A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m and the slant height of the conical portion is 53 m, find the area of canvas needed to make the tent. $[Take\pi =\frac{22}{7}]$

Solution

Height = h = 3 metres Slant height = l = 53 metres Radius = r = 52.5 metres Area of canvas required = C.S.A of cone = $\pi rl$ = $\frac{22}{7}$ × 52.5 × 53 = 8745 m²

Q. A rocket is in the form of a circular cylinder closed at the lower end and a cone of the same radius is attached to the top. The radius of the cylinder is 2.5 m, its height is 21 m and the slant height of the cone is 8 m. Calculate the total surface area of the rocket.

Solution

Cylindrical portion: Radius, r = 2.5 m Height, h = 21 m Surface area $=2\pi rh=2\times \frac{22}{7}\times 2.5\times 21=330m^2$ Conical portion: Radius, r = 2.5 m Slant height, l = 8 m Curved surface area $=\pi rl=\frac{22}{7}\times 2.5\times 8=62.86m^2$ Circular top: Radius = 2.5 m Area $=\pi r^2=\frac{22}{7}\times {2.5}^2=19.64m^2$ Total surface area of the rocket $=330+62.86+19.64=412.5m^2$

Q. A solid is in the shape of a cone surmounted on a hemisphere, the radius of each of them being 3.5 cm and the total height of the solid is 9.5 cm. Find the volume of the solid.

Solution

Total height of the solid $=9.5cm$ Radius of the cone = Radius of the hemisphere $=r=3.5cm$ Radius of the hemisphere = height of hemisphere $=3.5$ cm Height of cone,( h) = total height of the solid - height of the hemisphere $h=9.5-3.5=6cm$ The volume of the solid = volume of cone + volume of the hemisphere $=\frac{1}{3}\pi r^{2}h+\frac{2}{3}\pi r^3$ $=\frac{1}{3}\pi r^2(h+2r)$ $=\frac{1}{3}\times \frac{22}{7}\times 3.5\times 3.5\times (6+2\times 3.5)$ $=\frac{1}{3}\times \frac{22}{7}\times 3.5\times 3.5\times (6+7)$ $=\frac{1}{3}\times \frac{22}{7}\times 3.5\times 3.5\times \left(13\right)$ $=\frac{1}{3}\times 22\times .5\times 3.5\times \left(13\right)$ $=\frac{500.5}{3}$ $=166.83c{m}^3$ Hence, the volume of the solid is $166.83c{m}^3$ Q. A toy is in the form of a cone of radius 3. 5 cm mounted on a hemisphere of same radius on its circular face. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution

Radius of cone = radius of hemisphere, r = 3.5 cm Height of the cone, h = 15.5 -3.5 = 12 cm slant height, $l=\sqrt{r^2+h^2}=\sqrt{{3.5}^2+{12}^2}=\sqrt{12.25+144}=\sqrt{156.25}=12.5cm$ Total surface area of the toy = CSA of hemisphere + CSA of cone $=2\pi r^2+\pi rl=\pi r[2r+l]=\frac{22}{7}\times 3.5[2\times 3.5+12.5]=11[7+12.5]=11\times 19.5=214.5c{m}^2$

Q. A wooden article was made by scooping out a hemisphere from each end of a cylinder, as shown in the figure. If the height of the cylinder is $20cm$ and its base is of diameter $7cm$ , find the total surface area of the article when it is ready.

Solution

Height of the cylinder $=20cm$ Base radius of the cylinder $=3.5cm$ The total surface area would be the sum of the curved surface area of cylinder and the surface areas of 2 hemispheres which is given by $2\pi rh+2\times 2\pi r^2=2\pi rh+4\pi r^2$ $=2\pi r(h+2r)$ $=2\times \frac{22}{7}\times 3.5\times (20+2\times 3.5)=22\times 27=594c{m}^2$

Q. From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out. Find the volume of the remaining solid. Also, find the total surface area of the remaining solid. $[Take\pi =\mathrm{3.14.}]$

Solution

Volume of the remaining solid = Volume of cylinder - volume of the cone $\text{Volume of the remaining solid}=\pi r^{2}h-\frac{1}{3}\pi r^{2}h=\frac{2}{3}\pi r^{2}h=\frac{2}{3}\times 3.14\times (6{)}^2\times 8=602.88c{m}^3$ Slant height of cone, $AB=\sqrt{B{C}^2+A{C}^2}=\sqrt{8^2+6^2}=\sqrt{64+36}=10cm$ Total surface area of the remaining solid = curved surface area of cylinder + area of base of cylinder + curved surface area of cone $=2\pi rh+\pi r^2+\pi rl=2\pi \times 6\times 8+\pi \times 6^2+\pi \times 6\times 10=96\pi +36\pi +60\pi =192\pi =192\times 3.14=602.88c{m}^2$

Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.1

Structured Learning : The solutions provide a structured approach to learning the concepts of volume and surface areas of different solids. Each problem is systematically solved, helping students understand step-by-step how to apply formulas and methods to solve problems.

Clarity and Explanation : The solutions offer clear explanations for each step, making it easier for students to grasp the underlying concepts. This clarity aids in building a strong foundation in geometry, which is essential for higher-level math and real-world applications.

Practice and Application : The exercise provides ample practice problems that cover various types of solids such as cubes, cuboids, cylinders, cones, and spheres. This extensive practice helps reinforce learning and enhances problem-solving skills.

Variety of Problems : RS Aggarwal Solutions include a variety of problems ranging from straightforward calculations to more complex problems involving multiple steps. This variety ensures that students are exposed to different types of scenarios, preparing them to tackle diverse challenges.

Exam Preparation : The solutions are designed to align with the exam pattern and requirements, making them an excellent resource for exam preparation. Students can practice solving problems similar to those that may appear in their exams, thereby boosting their confidence.