NCERT Solutions for Class 9 Maths Chapter 8 : Quadrilaterals

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The NCERT Solutions for Class 9 Mathematics have been provided in this article. Quadrilaterals include more than one type of shape, but all quadrilaterals have four sides. The most common type of quadrilateral is square, but rectangles and rhombuses are considered quadrilaterals too. Quadrilaterals are also called quadrangles and tetragons.

Class 9 Maths Chapter 8 Quadrilaterals belong to Unit 4 Geometry which has a weightage of 27 marks in the Class 9 Maths Examination. NCERT Solutions for Class 9 Maths for Chapter 8 cover the following important concepts:

Download: NCERT Solutions for Class 9 Mathematics Chapter 8 pdf

NCERT Solutions for Class 9 Maths Chapter 8

Important Topics in Class 9 Maths Chapter 8 Quadrilaterals

Important Topics in Class 9 Maths Chapter 8 Quadrilaterals are elaborated below:

Cyclic Quadrilateral

Cyclic quadrilateral is a type of quadrilateral which has all of its four vertices lying on the circumference of a circle. Cyclic quadrilteral is a unique four-sided quadrilateral circumscribed in a circle.

Important Points of a Cyclic Quadrilateral:

The sides of a cyclic quadrilateral that are found to touch the circumference of a circle are the same four chords of the circle.
Area of a cyclic quadrilateral = √(s−a)(s−b)(s−c)(s−d) where a, b, c, and d are the four sides and S is the semi perimeter.
Perimeter of a cyclic quadrilateral = 2S, where S = (1/2) × (a + b + c + d)

Quadrilateral Angle Sum Property

Quadrilateral Angle sum property states that the sum of all four interior angles is equal to 360^o.

Example: Assume that the sum of three interior angles of a quadrilateral is 240^o. With the given data, determine its fourth angle.

Solution: First, consider that the fourth angle is x
Now, since the sum of four interior angles of a quadrilateral is 360^o, we can say,
⇒ x + 240^o = 360^o
⇒ x = 120^o

Perimeter of a Paralellogram

The perimeter of a parallelogram can be defined as the sum of the total distance outside its shape. A paralellogram is a four-sided polygon with parallel sides that never intersect.

Perimeter of Paralellogram Formula:
Perimeter of parallelogram = a + b + c + d
⇒ 2a + 2b= 2(a+b)
⇒ P = 2(a+b).

NCERT Solutions for Class 9 Maths Chapter 8 Exercises:

The detailed solutions for all the NCERT Solutions for Quadrilaterals under different exercises are:

Also Read:

Unit Related Topics
Quadrilateral formulas	Types of Quadrilaterals	Properties of a Parallelogram
Difference between Rhombus and Parallelogram	Area of a Trapezoid Formula	Tangential Quadrilateral Formula
Class 9 Maths Chapter 8 Quadrilaterals MCQs	Angles of a Parallelogram	Area of a Square

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Math Study Guides
Maths Important Formulas	Maths MCQs	Comparison Articles in Maths
Difference Between Topics in Maths	Math Study Material	Algebra
Trigonometry	Number Systems	Mensuration
Math Formula	NCERT Solutions for Class 9 Maths	Geometry

CBSE X Related Questions

1.
Find the sums given below :
\(7 + 10\frac 12+ 14 + ....... + 84\)
\(34 + 32 + 30 + ....... + 10\)
\(–5 + (–8) + (–11) + ....... + (–230)\)

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2.
Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, . . . .
(ii) \(2, \frac{5}{2},3,\frac{7}{2}\), . . . .
(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . .
(iv) – 10, – 6, – 2, 2, . . .
(v) 3, \(3 + \sqrt{2} , 3 + 3\sqrt{2} , 3 + 3 \sqrt{2}\) . . . .
(vi) 0.2, 0.22, 0.222, 0.2222, . . . .
(vii) 0, – 4, – 8, –12, . . . .
(viii) \(\frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}\), . . . .
(ix) 1, 3, 9, 27, . . . .
(x) a, 2a, 3a, 4a, . . . .
(xi) a, \(a^2, a^3, a^4,\) . . . .
(xii) \(\sqrt{2}, \sqrt{8} , \sqrt{18} , \sqrt {32}\) . . . .
(xiii) \(\sqrt {3}, \sqrt {6}, \sqrt {9} , \sqrt {12}\) . . . . .
(xiv) \(1^2 , 3^2 , 5^2 , 7^2\), . . . .
(xv) \(1^2 , 5^2, 7^2, 7^3\), . . . .

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3.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

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4.
Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km.
(v) A fraction becomes\(\frac{ 9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

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5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

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6.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :
Length (in mm)
Number of leaves
118 - 126
3
127 - 135
5
136 - 144
9
145 - 153
12
154 - 162
5
163 - 171
4
172 - 180
2
Find the median length of the leaves.
(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

Length (in mm)	Number of leaves
118 - 126	3
127 - 135	5
136 - 144	9
145 - 153	12
154 - 162	5
163 - 171	4
172 - 180	2

View Solution

NCERT Solutions for Class 9 Maths Chapter 8 : Quadrilaterals

NCERT Solutions for Class 9 Maths Chapter 8

Important Topics in Class 9 Maths Chapter 8 Quadrilaterals

Cyclic Quadrilateral

Quadrilateral Angle Sum Property

Perimeter of a Paralellogram

NCERT Solutions for Class 9 Maths Chapter 8 Exercises:

CBSE X Related Questions

1.
Find the sums given below :
\(7 + 10\frac 12+ 14 + ....... + 84\)
\(34 + 32 + 30 + ....... + 10\)
\(–5 + (–8) + (–11) + ....... + (–230)\)

3.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

Similar Mathematics Concepts

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NCERT Solutions for Class 9 Maths Chapter 8 : Quadrilaterals

NCERT Solutions for Class 9 Maths Chapter 8

Important Topics in Class 9 Maths Chapter 8 Quadrilaterals

Cyclic Quadrilateral

Quadrilateral Angle Sum Property

Perimeter of a Paralellogram

NCERT Solutions for Class 9 Maths Chapter 8 Exercises:

CBSE X Related Questions

1. Find the sums given below :\(7 + 10\frac 12+ 14 + ....... + 84\)\(34 + 32 + 30 + ....... + 10\)\(–5 + (–8) + (–11) + ....... + (–230)\)

3. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Find the sums given below :
\(7 + 10\frac 12+ 14 + ....... + 84\)
\(34 + 32 + 30 + ....... + 10\)
\(–5 + (–8) + (–11) + ....... + (–230)\)

3.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)