Two cubes each of side 4 cm are joined end to end find the volume of the resulting cuboid

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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150 Qs. 150 Marks 150 Mins

Given:

Side of cube = 4 cm

Formula used:

The surface area of the cuboid = 2(lb + bh + lh)

Here, l → Length, b → Breadth, h → height

Calculations:

After joining the two cubes length of side = 8 cm

Then, it can form a cuboid 

Total surface area of cuboid = 2[(8 × 4) + (4 × 4) + (8 × 4)]

⇒ 2[32 + 16 + 32] = 2 × 80

⇒ 160 cm2

∴ The resulting total surface area of a cuboid is 160 cm2

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Two cubes, each of side 4 cm are joined end to end. Find the surface area of the resulting cuboid.

Solution

If two cubes of sides $4cm$ are joined end to end, then the length $\left(l\right),$ breadth $\left(b\right)$ and height $\left(h\right)$ of the resulting cuboid are $8cm,4cm,$ and $4cm,$ respectively.

$\therefore$ Surface area of the resulting cuboid $=2(lb+bh+lh)$
$=2(8\times 4+4\times 4+8\times 4)c{m}^2$
$=2\times (32+16+32)c{m}^2$
$=160c{m}^2$
Thus, the surface area of the resulting cuboid is $160c{m}^2$.

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