Surface Areas and Volumes Class 10 Formulas PDF

Surface Areas and Volumes Class 10 Formulas PDF Download

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Surface Areas and Volumes Class 10 Formulas PDF Details
Surface Areas and Volumes Class 10 Formulas
PDF Name Surface Areas and Volumes Class 10 Formulas PDF
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Surface Areas and Volumes Class 10 Formulas

Dear users, today we are going to present Surface Areas and Volumes Class 10 Formulas PDF for all of you. If you are also looking for it then you have arrived on the right page. Through this post, you can easily download surface area and volume for Class 10 from a direct download link which is given below.

In this article, we are going to discuss the surface area and volume of different solid shapes such as the cube, cuboid, cone, cylinder, etc. The surface area can be generally classified into Lateral Surface Area (LSA), Total Surface Area (TSA), and Curved Surface Area (CSA). Here you can also know about the procedure to find the volume and its surface area in detail.

Surface Areas and Volumes Class 10 Formulas PDF

Below given is the table for the calculating Surface area and Volume for the basic geometrical figures:

Name Perimeter Total Surface Area Curved Surface Area/Lateral Surface Area Volume Figure
Square 4b b2 —- —- Square
Rectangle 2(w+h) w.h —- —- Rectangle
Parallelogram 2(a+b) b.h —- —- Parallelogram
Trapezoid a+b+c+d 1/2(a+b).h —- —- Trapezoid
Circle 2 π r π r2 —- —- Circle
Ellipse 2π√(a2 + b2)/2       π a.b —- —- Ellipse
Triangle a+b+c 1/2 * b * h —- —- Triangle
Cuboid 4(l+b+h) 2(lb+bh+hl) 2h(l+b) l * b * h Cuboid
Cube 6a  6a2 4a2 a3 Cube
Cylinder —- 2 π r(r+h) 2πrh π r2 h Cylinder
Cone —- π r(r+l) π r l 1/3π r2 h Cone
Sphere —- 4 π r2 4π r2 4/3π r3 Sphere
Hemisphere —- 3 π r2 2 π r2 2/3π r3 Hemisphere

Surface Areas and Volumes Class 10 Formulas – Solved Examples

Q.1. What is the curved surface area of a hemisphere if the diameter is 12cm. 12cm.

Ans: The diameter is 12cm.12cm.

The radius is 122cm=6cm122cm=6cm

The curved surface area of the hemisphere =2πr2=2×227×6×6cm2=2πr2=2×227×6×6cm2

=226.28cm2=226.28cm2

Q.2. Find the length of the sides of the cube if its volume is 216cm3.216cm3.

Ans: Given, the volume of the cube =216cm3=216cm3

Let the length of the sides is a.a.

We know,

The volume of a cube =(side)3=(side)3

Substituting the value we get,

a3=216a3=216

⇒a=216−−−√3⇒a=6cm⇒a=2163⇒a=6cm

Hence, the side of the cube is 6cm.6cm.

Q.3. If the diagonal is 43–√cm.43cm. Find the volume of the cube.

Ans: We know, V=d333√⇒V=(43√)333√=64×33√33√⇒V=64cm3V=d333⇒V=(43)333=64×3333⇒V=64cm3

Hence, the volume of the cube is 64cm3.64cm3.

Q.4. The slant height of a frustum of a cone is 4cm,4cm, and the perimeters (circumference) of its circular ends are 18cm18cm and 6cm.6cm. Find the curved surface area of the frustum.

Ans: Given, l=4cml=4cm

Circumference of a circular end =18cm=18cm

⇒2πr1=18cm⇒2πr1=18cm

⇒πr1=9cm⇒πr1=9cm

Circumference of another circular end =6cm=6cm

⇒2πr2=6cm⇒2πr2=6cm

⇒πr2=3cm⇒πr2=3cm

We know that curved surface area =πl(r1+r2)=l(πr1+πr2)=πl(r1+r2)=l(πr1+πr2)

=4×(9+3)cm2=4×(9+3)cm2

=48cm2=48cm2

Hence, the curved surface area of the frustum is 48cm2.48cm2.

Q.5. Madhu prepares a birthday cap with a piece of paper in the form of a right circular cone of radius 33 inches and a height of 44 inches. Find the slant height of the birthday cap made by her.

Ans: Given that, the birthday cap prepared by Madhu is in the shape of a right circular cone.

The radius of the cone (r)=3(r)=3 inches and the height of the cone (h)=4(h)=4 inches.

It is known that the slant height of the cone is given by l=r2+h2−−−−−−√.l=r2+h2.

⇒l=32+42−−−−−−√⇒l=32+42

⇒l=9+16−−−−−√⇒l=9+16

⇒l=25−−√=5⇒l=25=5 inches

Hence, the slant height of the birthday cap is 55 inches.

You can download Surface Areas and Volumes Class 10 Formulas PDF by clicking on the following download link.


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