## Envision Math Grade 8 Volume 1 Chapter 8 Solve Problems Involving Surface Area And Volume

**Page 412 Exercise 1 Answer**

A three dimensional object is in three dimensions having length, width and height.

Example: Cubes, prisms, pyramids, spheres, cones, and cylinders are all examples of three-dimensional objects.

Volume is the product of length, width and height.

Surface areas is the sum of products of length, width and height taken two at a time.

Volume of a prism = cross-sectional area × length.

The surface area of a 3D shape is the total area of all its faces.

**Page 415 Exercise 1 Answer**

The figure of circle is shown as

Here r is know as radius from center to circle which is constant throughout.

The radius is the distance from the center to the edge of a circle.

**Page 415 Exercise 2 Answer**

The objects around you are three-dimensional.

A three dimensional object is in three dimensions having length, width and height.

Example: Cubes, prisms, pyramids, spheres, cones, and cylinders are all examples of three-dimensional objects.

Day to day life examples: cup, mobile phone, ball and bottle.

A shape that has length, width, and height is three dimensional.

**Page 415 Exercise 3 Answer**

A three dimensional object is in three dimensions having length, width and height.

Example: Cubes, prisms, pyramids, spheres, cones, and cylinders are all examples of three-dimensional objects.

Any side of a cube can be considered a base.

Any side of a cube can be considered a base.

**Page 415 Exercise 5 Answer**

Given:

The ________________ of a circle is a line segment that passes through its center and has endpoints on the circle.

The distance from one point on a circle through the center to another point on the circle.

It is also the longest distance across the circle.

The diameter of a circle is a line segment that passes through its center and has endpoints on the circle.

Hence, the diameter of a circle is a line segment that passes through its centre and has endpoints on the circle.

**Page 415 Exercise 7 Answer**

Given:

9⋅3.14

To find the product:

Multiply 9 with 3.14.

We have,

9⋅3.14

Multiply the numbers:

= 28.26

It follows that the product is 28.26.

Hence, the product is 28.26.

**Page 415 Exercise 8 Answer**

Given:

4.2⋅10.5

To find the product:

Multiply 4.2 with 10.5.

We have,

4.2⋅10.5

Multiply the numbers:

= 44.10

It follows that the product is 44.10.

Hence, the product is 44.10.

**Page 415 Exercise 9 Answer**

Given:

Hence, the area of the given circle is 100.48cm^{2}.

**Page 415 Exercise 10 Answer**

Given:

To find the area of the circle:

First, find r using the diameter and then plug the value of r,π in the area of the circle formula.

So, the area of the circle is 56.52 cm^{2}

Hence, the area of the given circle is 56.52cm^{2}.

**Page 415 Exercise 11 Answer**

Given:

To find the missing value of x:

Use the Pythagorean Theorem and apply it to the given right triangle.

Hence, the missing value is x = 5in.

**Page 415 Exercise 12 Answer**

Given:

To find the missing value of x:

Use the Pythagorean Theorem and apply it to the given right triangle.

Hence, the missing value is x = 18m.