Two cubes each of edge 5 cm are joined end to end.the surface area of the resulting cuboid is

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Solution:

We will find the length of the edge of each cube by using the formula for the volume of a cube = a3, where the length of the edge is 'a'.

As the cubes are joined end to end, they will appear as follows:

Using the formula for the surface area of a cuboid = 2(lb + bh + lh), where l, b, and h are length, breadth, and height respectively.

Let the length of the edge of each cube be 'a'

Therefore, volume of the cube = a3

Volume of the cube, a3 = 64 cm3

a3 = 64 cm3

a = ∛(64 cm3)

a = 4 cm

Therefore,

Length of the resulting cuboid, l = a = 4 cm

Breadth of the resulting cuboid, b = a = 4 cm

Height of the resulting cuboid, h = 2a = 2 × 4 cm = 8 cm

Surface area of the resulting cuboid = 2 (lb + bh + lh)

= 2 (4 cm × 4 cm + 4 cm × 8 cm + 4 cm × 8 cm)

= 2 (16 cm2 + 32 cm2 + 32 cm2)

= 2 × 80 cm2

= 160 cm2

Thus, the surface area of the resulting cuboid is 160 cm2.

☛ Check: NCERT Solutions for Class 10 Maths Chapter 13

Video Solution:

2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

NCERT Solutions Class 10 Maths Chapter 13 Exercise 13.1 Question 1

Summary:

The surface area of the resulting cuboid if 2 cubes each of volume 64 cm3 are joined end to end is 160 cm2.

☛ Related Questions:

Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Let h be the height, l the slant height and r1 and r2 the radii of the circular bases of the frustum ABB’ A’ shown in Fig. such that r1 > r2.
Let the height of the cone VAB be h1 and its slant height be i.e., VO = h1 and VA = VB = l1
∴ VA’ = VA – AA’ = l1– l
and VO’ = VO – OO’ = h1– hHere, ΔVOA ~ ΔVO‘A’

Now,Height of the cone VA‘B’

Let S denote the curved surface area of the frustum of cone. Then,S = Lateral (curved) surface area of cone VAB

- Curved surface area of cone VA‘B’

Curved surface area of the frustum
= π(r1 + r2)lTotal surface area of the frustum= Lateral (curved) surface area+ Surface area of circular bases

= π (r1 + r2) I + πr12 + πr22
= π {(r1 + r2) l + r12 + r22}.

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Three cubes each of side 5 centimetre are joined end to end. Find the surface area of the resulting cuboid

Solution

When three cubes are joined side by side, the resulting cuboid has the same width (breadth), $b=5cm$ and height, $h=5cm.$ but length, $l=5+5+5=15cm.$

Surface Area of the resulting cuboid $=2(lb+bh+lh)$

$=2(15\times 5+5\times 5+5\times 15)$

$=2(75+25+75)=2\times 175=350c{m}^2.$