Question

find the volume the lateral surface and the total surface area of the cubiod whose dimensions are : length = 26 CM, breadth =14 CM, and height = 6.5 ​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Given:

(i) length = 12 cm, breadth = 8 cm and height = 4.5 cm

(ii) length = 26 m, breadth = 14 m and height = 6.5 m

(iii) length = 15 m, breadth = 6 m and height = 5 dm

(iv) length = 24 m, breadth = 25 cm and height = 6 m

To find:

The volume, the lateral surface area and the total surface area of

the given dimensions of the cuboid

Formulas of cuboid required to solve the problem:

\begin{gathered}\boxed{\bold{Lateral \:Surface\:Area = 2h(l+b)}}\\\\\boxed{\bold{Total \:Surface\:Area = 2(bl+bh+hl)}}\\\\\boxed{\bold{Volume = l\times b\times h}}\\\end{gathered}

LateralSurfaceArea=2h(l+b)

TotalSurfaceArea=2(bl+bh+hl)

Volume=l×b×h

Solution:

(i). length = 12 cm, breadth = 8 cm and height = 4.5 cm

L.S.A. = 2 × 4.5 × [12 + 8] = 9 × 20 = 180 cm²

T.S.A. = 2[(12 × 8) + (8 × 4.5) + (4.5 × 12)] = 2[96 + 36 + 54)] = 2 × 186 = 372 cm²

Volume = 12 × 8 × 4.5 = 432 cm³

(ii). length = 26 m, breadth = 14 m and height = 6.5 m

L.S.A. = 2 × 6.5 × [26 + 14] = 13 × 40 = 520 m²

T.S.A. = 2[(26 × 14) + (14 × 6.5) + (6.5 × 26)] = 2[364 + 91 + 169)] = 2 × 624 = 1248 m²

Volume = 26 × 14 × 6.5 = 2366 m³

(iii). length = 15 m, breadth = 6 m and height = 5 dm

We know ⇒ 1 dm = 0.1 m

∴ 5 dm = 0.5 m

L.S.A. = 2 × 0.5 × [15 + 6] = 1 × 21 = 21 m²

T.S.A. = 2[(15 × 6) + (6 × 0.5) + (0.5 × 15)] = 2[90 + 3 + 7.5)] = 2 × 100.5 = 201 m²

Volume = 15 × 6 × 0.5 = 45 m³

(iv). length = 24 m, breadth = 25 cm and height = 6 m

We know ⇒ 1 cm = 0.01 m

∴ 25 cm = 0.25 m

L.S.A. = 2 × 6 × [24 + 0.25] = 12 × 24.25 = 291 m²

T.S.A. = 2[(24 × 0.25) + (0.25 × 6) + (6 × 24)] = 2[6 + 1.5 + 144)] = 2 × 151.5 = 303 m²

Volume = 24 × 0.25 × 6 = 36 m³