NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13 1 Solutions

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.1 Solutions are based on the concept of Surface area of a cuboid and a cube.

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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	Surface Area of a Cylinder Formula	Area of Prism
Surface Area of a Cylinder	Area of Hollow Cylinder	Surface Area of a Cone Formula

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CBSE X Related Questions

1.
A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))
- \(314 \sqrt{2}\) \(\text{cm}^{2}\)
- \(314\) \(\text{cm}^{2}\)
- \(\frac{3140}{3}\) \(\text{cm}^{2}\)
- \(3140 \sqrt{2}\) \(\text{cm}^{2}\)
View Solution
2.
Prove that: \(\frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\csc^3 \theta}{\csc^2 \theta - 1} = \sec \theta \cdot \csc \theta (\sec \theta + \csc \theta)\)
View Solution
3.
PQ is tangent to a circle with centre O. If \(OQ = a\), \(OP = a + 2\) and \(PQ = 2b\), then relation between \(a\) and \(b\) is
- \(a^2 + (a + 2)^2 = (2b)^2\)
- \(b^2 = a + 4\)
- \(2a^2 + 1 = b^2\)
- \(b^2 = a + 1\)
View Solution
4.
Assertion (A) : If probability of happening of an event is \(0.2p\), \(p>0\), then \(p\) can't be more than 5.
Reason (R) : \(P(\bar{E}) = 1 - P(E)\) for an event \(E\).
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution
5.
Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is
- \(10^{\circ}\)
- \(60^{\circ}\)
- \(45^{\circ}\)
- \(100^{\circ}\)
View Solution
6.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is
- \(4 \text{ cm}\)
- \(4\sqrt{2} \text{ cm}\)
- \(8 \text{ cm}\)
- \(2\sqrt{2} \text{ cm}\)
View Solution

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13 1 Solutions

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

CBSE X Related Questions

1.
A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))

2.
Prove that: \(\frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\csc^3 \theta}{\csc^2 \theta - 1} = \sec \theta \cdot \csc \theta (\sec \theta + \csc \theta)\)

3.
PQ is tangent to a circle with centre O. If \(OQ = a\), \(OP = a + 2\) and \(PQ = 2b\), then relation between \(a\) and \(b\) is

4.
Assertion (A) : If probability of happening of an event is \(0.2p\), \(p>0\), then \(p\) can't be more than 5.
Reason (R) : \(P(\bar{E}) = 1 - P(E)\) for an event \(E\).

5.
Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is

6.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

Similar Mathematics Concepts

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NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13 1 Solutions

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

CBSE X Related Questions

1.A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))

2.Prove that: \(\frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\csc^3 \theta}{\csc^2 \theta - 1} = \sec \theta \cdot \csc \theta (\sec \theta + \csc \theta)\)

3.PQ is tangent to a circle with centre O. If \(OQ = a\), \(OP = a + 2\) and \(PQ = 2b\), then relation between \(a\) and \(b\) is

4.Assertion (A) : If probability of happening of an event is \(0.2p\), \(p>0\), then \(p\) can't be more than 5. Reason (R) : \(P(\bar{E}) = 1 - P(E)\) for an event \(E\).

5.Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is

6.If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))

2.
Prove that: \(\frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\csc^3 \theta}{\csc^2 \theta - 1} = \sec \theta \cdot \csc \theta (\sec \theta + \csc \theta)\)

3.
PQ is tangent to a circle with centre O. If \(OQ = a\), \(OP = a + 2\) and \(PQ = 2b\), then relation between \(a\) and \(b\) is

4.
Assertion (A) : If probability of happening of an event is \(0.2p\), \(p>0\), then \(p\) can't be more than 5.
Reason (R) : \(P(\bar{E}) = 1 - P(E)\) for an event \(E\).

5.
Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is

6.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is