NCERT Solutions for Class 9 Maths Chapter 8 Triangles Exercise 8.2 Solutions

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NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2 Solutions are based on the Mid-point Theorem.

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Read More: NCERT Solutions For Class 9 Maths Chapter 8 Quadrilaterals

Exercise Solutions of Class 9 Maths Chapter 8 Quadrilaterals

Also check other Exercise Solutions of Class 9 Maths Chapter 8 Quadrilaterals

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1 Solutions.

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Important Chapter Related Topics
Quadrilaterals	Cyclic quadrilateral	Quadrilateral angle sum property
Perimeter of a parallelogram	Area of a Rectangle	Area of a Square

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CBSE X Related Questions

1.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.
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2.
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
- \(4 \text{ cm}\)
- \(4\sqrt{2} \text{ cm}\)
- \(8 \text{ cm}\)
- \(2\sqrt{2} \text{ cm}\)
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3.
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.
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4.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :
- \( 130^\circ \)
- \( 50^\circ \)
- \( 120^\circ \)
- \( 90^\circ \)
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5.
The natural number 1 is :
- a prime number.
- a composite number.
- prime as well as composite.
- neither prime nor composite.
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6.
The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :
- 0
- 1
- 3
- 2
View Solution

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