It is given that length = 15m, breadth = 6m and height = 5dm = 0.5m

We know that

Volume of cuboid = l × b × h

By substituting the values we get

Volume of cuboid = 15 × 6 × 0.5

By multiplication

Volume of cuboid = 45 m³

We know that

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

By substituting the values

Lateral surface area of a cuboid = 2 (15 + 6) × 0.5

On further calculation

Lateral surface area of a cuboid = 2 × 21 × 0.5

By multiplication

Lateral surface area of a cuboid = 21 m²

We know that

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

By substituting the values

Total surface area of cuboid = 2 (15 × 6 + 6 × 0.5 + 15 × 0.5)

On further calculation

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

So we get

Total surface area of cuboid = 2 × 100.5

By multiplication

Total surface area of cuboid = 201 m²

It is given that length = 26m, breadth = 14m and height = 6.5m

We know that

Volume of cuboid = l × b × h

By substituting the values we get

Volume of cuboid = 26 × 14 × 6.5

By multiplication

Volume of cuboid = 2366 m³

We know that

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

By substituting the values

Lateral surface area of a cuboid = 2 (26 + 14) × 6.5

On further calculation

Lateral surface area of a cuboid = 2 × 40 × 6.5

By multiplication

Lateral surface area of a cuboid = 520cm²

We know that

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

By substituting the values

Total surface area of cuboid = 2 (26 × 14 + 14 × 6.5 + 26 × 6.5)

On further calculation

Total surface area of cuboid = 2 (364 + 91 + 169)

So we get

Total surface area of cuboid = 2 × 624

By multiplication

Total surface area of cuboid = 1248 m²

 It is given that length = 12cm, breadth = 8cm and height = 4.5cm

We know that

Volume of cuboid = l × b × h

By substituting the values we get

Volume of cuboid = 12 × 8 × 4.5

By multiplication

Volume of cuboid = 432 cm³

We know that

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

By substituting the values

Lateral surface area of a cuboid = 2 (12 + 8) × 4.5

On further calculation

Lateral surface area of a cuboid = 2 × 20 × 4.5

By multiplication

Lateral surface area of a cuboid = 180cm²

We know that

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

By substituting the values

Total surface area of cuboid = 2 (12 × 8 + 8 × 4.5 + 12 × 4.5)

On further calculation

Total surface area of cuboid = 2 (96 + 36 + 54)

So we get

Total surface area of cuboid = 2 × 186

By multiplication

Total surface area of cuboid = 372 cm²

It is given that (2x – 5) ^o and (x – 10) ^o are the complementary angles.

So we can write it as

(2x – 5) ^o + (x – 10) ^o = 90^o

2x – 5^o + x – 10^o = 90^o

On further calculation

3x – 15^o = 90^o

So we get

3x = 105^o

By division

x = 105/3

x = 35^o

Therefore, the value of x for which the angles (2x – 5) ^o and (x – 10) ^o are the complementary angles is 35^o.

Consider the required angle as x^o and 90^o – x^o

According to the question it can be written as

x^o/ 90^o – x^o = 4/5

By cross multiplication we get

5x = 4 (90 – x)

5x = 360 – 4x

On further calculation we get

5x + 4x = 360

9x = 360

By division

x = 360/9

So we get

x = 40

Therefore, the angles are 40^o and 90^o – x^o = 90^o – 40^o = 50^o

Consider the required angle as x^o

We know that the complement can be written as 90^o – x^o and the supplement can be written as 180^o – x^o

90^o – x^o = 1/3(180^o – x^o)

We can also write it as

90^o – x^o = 60^o – (1/3) x^o

So we get

x^o – (1/3)x^o = 90^o – 60^o

(2/3) x^o = 30^o

By division we get

x^o = ((30×3)/2)

x^o = 45^o

Therefore, the angle whose complement is one third of its supplement is 45^o.

Consider the required angle as x^o

We know that the complement can be written as 90^o – x^o and the supplement can be written as 180^o – x^o

180^o – x^o = 4(90^o – x^o)

We can also write it as

180^o – x^o = 360^o – 4x^o

So we get

4x^o – x^o = 360^o – 180^o

3x^o = 180^o

By division we get

x^o = 180/3

x^o = 60^o

Therefore, the angle whose supplement is four times its complement is 60^o.

Consider the required angle as x^o

We know that the supplement can be written as 180^o – x^o

x^o = 5(180^o – x^o)

We can also write it as

x = 900 – 5x

So we get

6x = 900

By division we get

x = 900/6

x^o = 150^o

Therefore, the angle which is five times its supplement is 150^o.

Consider the required angle as x^o

We know that the complement can be written as 90^o – x^o

x^o = 4(90^o – x^o)

We can also write it as

x = 360 – 4x

So we get

5x = 360

By division we get

x = 360/5

x^o = 72^o

Therefore, the angle which is four times its complement is 72^o.

Consider the required angle as x^o

We know that the supplement can be written as 180^o – x^o

x^o = (180^o – x^o) – 30^o

We can also write it as

x + x = 180 – 30

So we get

2x = 150

By division we get

x = 150/2

x^o = 75^o

Therefore, the measure of an angle which is 30^o more than its supplement is 75^o