Question

1. Find the volume, the lateral surface area and the total surface areas and f a cuboid whose dimensions are given below.

(i) length = 15 m, breadth = 6 m and height = 5 dm
(ii) length = 24 m, breadth = 25 cm and height = 6 m

2. A matchbox measures 4 cm x 2.5 cm x 1.5 cm. What is the
packet containing 12 such matchboxes?
​

Answer 1

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1)

$\bold{ =2(1+b)×h}$

$\bold{= 2(15 + 6) \times 0.6}$

$\bold{ = (30 +12 )\times 0.6 }$

$\bold{ = 32 \times 0.6}$

❥ 19.2

༆ⓗⓞⓟⓔ ⓘⓣⓢ ⓗⓔⓛⓟⓕⓤⓛ࿐

Answer 2

Answer:

1. ( i )

The volume of the cuboid is 45 m³.

The lateral surface area of the cuboid is 21 m².

The total surface area of the cuboid is 201 m².

( ii )

The volume of the cuboid is 36 m³.

The lateral surface area of the cuboid is 291 m².

The total surface area of the cuboid is 303 m².

2. The volume of the packet containing 12 matchboxes is 180 m³.

Step-by-step-explanation:

1.

We have given the dimensions of a cuboid.

We have to find volume, lateral surface area and total surface area of the cuboid.

( i )

Length ( l ) = 15 m

Breadth ( b ) = 6 m

Height ( h ) = 5 dm = 5 * 0.1 m = 0.5 m

Now, we know that,

Volume of cuboid = Length * Breadth * Height

⇒ Volume of cuboid = 15 * 6 * 0.5

⇒ Volume of cuboid = 90 * 0.5

⇒ Volume of cuboid = 45 m³

∴ The volume of the cuboid is 45 m³.

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Now, we know that,

Lateral surface area of cuboid = 2 h ( l + b )

⇒ Lateral surface area of cuboid = 2 * 0.5 ( 15 + 6 )

⇒ Lateral surface area of cuboid = 1 * 21

⇒ Lateral surface area of cuboid = 21 m²

∴ The lateral surface area of the cuboid is 21 m².

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Now, we know that,

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

⇒ Total surface area of cuboid = 2 [ ( 15 * 6 ) + ( 6 * 0.5 ) + ( 15 * 0.5 ) ]

⇒ Total surface area of cuboid = 2 ( 90 + 3 + 7.5 )

⇒ Total surface area of cuboid = 2 * ( 93 + 7.5 )

⇒ Total surface area of cuboid = 2 * 100.5

⇒ Total surface area of cuboid = 201 m²

∴ The total surface area of the cuboid is 201 m².

─────────────────────────

( ii )

Length ( l ) = 24 m

Breadth ( b ) = 25 cm = 25 ÷ 100 m = 0.25 m

Height ( h ) = 6 m

Now, we know that,

Volume of cuboid = Length * Breadth * Height

⇒ Volume of cuboid = 24 * 0.25 * 6

⇒ Volume of cuboid = 24 * 1.5

⇒ Volume of cuboid = 4 * 6 * 1.5

⇒ Volume of cuboid = 4 * 9

⇒ Volume of cuboid = 36 m³

∴ The volume of the cuboid is 36 m³.

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Now, we know that,

Lateral surface area of cuboid = 2 h ( l + b )

⇒ Lateral surface area of cuboid = 2 * 6 ( 24 + 0.25 )

⇒ Lateral surface area of cuboid = 12 * 24.25

⇒ Lateral surface area of cuboid = 291 m²

∴ The lateral surface area of the cuboid is 291 m².

─────────────────────────

Now, we know that,

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

⇒ Total surface area of cuboid = 2 [ ( 24 * 0.25 ) + ( 0.25 * 6 ) + ( 24 * 6 ) ]

⇒ Total surface area of cuboid = 2 ( 6 + 1.5 + 144 )

⇒ Total surface area of cuboid = 2 ( 7.5 + 144 )

⇒ Total surface area of cuboid = 2 * 151.5

⇒ Total surface area of cuboid = 303 m²

∴ The total surface area of the cuboid is 303 m².

─────────────────────────

2.

We have given that,

The dimensions of a matchbox are 4 cm, 2.5 cm and 1.5 cm.

We have to find the volume of a packet containing such 12 matchboxes.

Now, we know that,

Volume of cuboidal matchbox = Length * Breadth * Height

⇒ Volume of cuboidal matchbox = 4 * 2.5 * 1.5

⇒ Volume of cuboidal matchbox = 10 * 1.5

⇒ Volume of cuboidal matchbox = 15 m³

Now,

Volume of packet containing 12 matchboxes = 12 * Volume of one matchbox

⇒ Volume of packet containing 12 matchboxes = 12 * 15

⇒ Volume of packet containing 12 matchboxes = 180 m³

∴ The volume of the packet containing 12 matchboxes is 180 m³.