NCERT Solutions for Class 9 Maths Chapter 8 Triangles Exercise 8.1 Solutions

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NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1 Solutions are based on different types of quadrilaterals, and their properties.

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Check out NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.1 Solutions

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 8.2 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

Also check:

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CBSE X Related Questions

1.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals
- \(\frac{21}{4} \text{ cm}\)
- \(\frac{28}{3} \text{ cm}\)
- \(\frac{12}{7} \text{ cm}\)
- \(5.5 \text{ cm}\)
View Solution
2.
The HCF of 960 and 432 is :
- 48
- 54
- 72
- 36
View Solution
3.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).
View Solution
4.
The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :
- 0
- 1
- 3
- 2
View Solution
5.
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
6.
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

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