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ML Aggarwal Solutions Class 9 Mathematics Solutions for Mensuration Exercise 16.4 in Chapter 16 - Mensuration

Question 3 Mensuration Exercise 16.4

The volume of a cuboid is 3600 \mathrm{cm}^{3} and its height is 12 cm. The cross-section is a rectangle whose length

and breadth is in the ratio 4: 3. Find the perimeter of the cross-section.

Answer:

It is given that

Volume of a cuboid = 3600 \mathrm{cm}^{3}

Height of cuboid = 12 cm

We know that

The volume of cuboid = Area of rectangle × height

Substituting the values

3600 = area of rectangle × 12

By further calculation

Area of rectangle = 3600/ 12

Area of rectangle = 300 \mathrm{cm}^{2} \ldots \ldots

Here

Ratio of length and breadth of rectangle = 4: 3

Consider

Length of rectangle = 4x

Breadth of rectangle = 3x

Area of rectangle = length × breadth

Substituting the values

Area of rectangle = 4x × 3x

So we get

M L Aggarwal - Understanding ICSE Mathematics - Class 9 chapter Mensuration Question 3 Solution image

Here

Length of rectangle = 4 × 5 = 20 cm

Breadth of rectangle = 3 × 5 = 15 cm

Perimeter of the cross section = 2 (l + b)

Substituting the values

= 2 (20 + 15)

= 2 × 35

= 70 cm

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