NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.5 Solutions

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.5 Solutions are based on the Volume of a cuboid.

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Check out NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.5 Solutions

Read More: NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas and Volumes

Exercise Solutions of Class 9 Maths Chapter 13 Surface Areas and Volumes

Also check other Exercise Solutions of Class 9 Maths Chapter 13 Surface Areas and Volumes

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Important Chapter Related Topics
Surface Areas And Volumes	Cuboid	Area of Prism
Surface Area of a Cylinder	Area of Hollow Cylinder	Surface Area of a Cone Formula
Surface Area of a Cylinder Formula

Also Check:

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CBSE X Related Questions

1.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).
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2.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.
View Solution
3.
The natural number 1 is :
- a prime number.
- a composite number.
- prime as well as composite.
- neither prime nor composite.
View Solution
4.
A kite is flying at a height of \(60 \text{ m}\) above the ground level. Ravi, standing at the roof of the house is holding the string straight and observes the angle of elevation of kite as \(30^{\circ}\). From the bottom of the same building, the angle of elevation of kite is \(45^{\circ}\). Find the length of the string and height of roof from the ground. (Use \(\sqrt{3} = 1.73\))
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5.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals
- \(\frac{21}{4} \text{ cm}\)
- \(\frac{28}{3} \text{ cm}\)
- \(\frac{12}{7} \text{ cm}\)
- \(5.5 \text{ cm}\)
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6.
If \(\alpha, \beta\) are the zeroes of the polynomial \(p(x) = x^2 - 3x - 1\), then find the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\).
View Solution

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