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