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’
Slant 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’
[Using (A) and (C)]
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 4 cm are joined end to end, then the length (l), breadth (b) and height (h) of the resulting cuboid are 8 cm,4 cm, and 4 cm, respectively.
∴ Surface area of the resulting cuboid =2(lb+bh+lh)
=2(8×4+4×4+8×4) cm2
=2×(32+16+32) cm2
=160 cm2
Thus, the surface area of the resulting cuboid is 160 cm2.
NCERT Previous Years Papers
14