NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.9 Solutions

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.9 Solutions are based on the description of the construction of various objects, like cube, cuboid, sphere, cylinders and more. It covers the summary of the topic.

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Check out NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.9 Solutions

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

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

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CBSE X Related Questions

1.
In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \).
Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).
View Solution
2.
In \(\triangle ABC, \angle B = 90^\circ\). If \(\frac{AB}{AC} = \frac{1}{2}\), then \(\cos C\) is equal to
- \(\frac{3}{2}\)
- \(\frac{1}{2}\)
- \(\frac{\sqrt{3}}{2}\)
- \(\frac{1}{\sqrt{3}}\)
View Solution
3.
If \(\alpha, \beta\) are zeroes of the polynomial \(8x^2 - 5x - 1\), then form a quadratic polynomial in x whose zeroes are \(\frac{2}{\alpha}\) and \(\frac{2}{\beta}\).
View Solution
4.
Nidhi received simple interest of ₹1,200 when she invested ₹x at 6% per annum and ₹y at 5% per annum for 1 year. Had she invested ₹x at 3% per annum and ₹y at 8% per annum for that year, she would have received simple interest of ₹1,260.Find the values of x and y.
View Solution
5.
Prove that: \[ \frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0 \]
View Solution
6.
Find the zeroes of the polynomial: \[ q(x) = 8x^2 - 2x - 3 \] Hence, find a polynomial whose zeroes are 2 less than the zeroes of \(q(x)\)
View Solution

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.9 Solutions

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

CBSE X Related Questions

1.
In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \).
Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).

2.
In \(\triangle ABC, \angle B = 90^\circ\). If \(\frac{AB}{AC} = \frac{1}{2}\), then \(\cos C\) is equal to

3.
If \(\alpha, \beta\) are zeroes of the polynomial \(8x^2 - 5x - 1\), then form a quadratic polynomial in x whose zeroes are \(\frac{2}{\alpha}\) and \(\frac{2}{\beta}\).

4.
Nidhi received simple interest of ₹1,200 when she invested ₹x at 6% per annum and ₹y at 5% per annum for 1 year. Had she invested ₹x at 3% per annum and ₹y at 8% per annum for that year, she would have received simple interest of ₹1,260.Find the values of x and y.

5.
Prove that: \[ \frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0 \]

6.
Find the zeroes of the polynomial: \[ q(x) = 8x^2 - 2x - 3 \] Hence, find a polynomial whose zeroes are 2 less than the zeroes of \(q(x)\)

Similar Mathematics Concepts

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.9 Solutions

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

CBSE X Related Questions

1.In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \). Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).

2.In \(\triangle ABC, \angle B = 90^\circ\). If \(\frac{AB}{AC} = \frac{1}{2}\), then \(\cos C\) is equal to

3.If \(\alpha, \beta\) are zeroes of the polynomial \(8x^2 - 5x - 1\), then form a quadratic polynomial in x whose zeroes are \(\frac{2}{\alpha}\) and \(\frac{2}{\beta}\).

4. Nidhi received simple interest of ₹1,200 when she invested ₹x at 6% per annum and ₹y at 5% per annum for 1 year. Had she invested ₹x at 3% per annum and ₹y at 8% per annum for that year, she would have received simple interest of ₹1,260.Find the values of x and y.

5.Prove that: \[ \frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0 \]

6.Find the zeroes of the polynomial: \[ q(x) = 8x^2 - 2x - 3 \] Hence, find a polynomial whose zeroes are 2 less than the zeroes of \(q(x)\)

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \).
Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).

2.
In \(\triangle ABC, \angle B = 90^\circ\). If \(\frac{AB}{AC} = \frac{1}{2}\), then \(\cos C\) is equal to

3.
If \(\alpha, \beta\) are zeroes of the polynomial \(8x^2 - 5x - 1\), then form a quadratic polynomial in x whose zeroes are \(\frac{2}{\alpha}\) and \(\frac{2}{\beta}\).

4.
Nidhi received simple interest of ₹1,200 when she invested ₹x at 6% per annum and ₹y at 5% per annum for 1 year. Had she invested ₹x at 3% per annum and ₹y at 8% per annum for that year, she would have received simple interest of ₹1,260.Find the values of x and y.

5.
Prove that: \[ \frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0 \]

6.
Find the zeroes of the polynomial: \[ q(x) = 8x^2 - 2x - 3 \] Hence, find a polynomial whose zeroes are 2 less than the zeroes of \(q(x)\)