NCERT Notes for Class 9 Maths Chapter 9 Quadrilaterals
![NCERT Notes for Class 9 Maths Chapter 9 Quadrilaterals NCERT Notes for Class 9 Maths Chapter 9 Quadrilaterals](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhQZZxdBSvfwLeMFtm2qJetAL1t3hl0oiYZxhjwQ-L4YW4wVopdBdWN5XwJ6AeISFT7Bqwu0PAR2VNxt2K8huoGA1T782fnvrj1LnisQ8o5bx5mtQzyiSTluN9t9eDpLTpYqhNCk-qm1uQ2NQcImSJH8GdLO52DvqIJ2TApyqe0BsiBlhw-88t_KpEES1ac/s16000/quadrilaterals-revision-notes-class9-maths.jpg)
Class 9 Maths Chapter 9 Quadrilaterals Notes
Chapter Name | Quadrilaterals Notes |
Class | CBSE Class 9 |
Textbook Name | Mathematics Class 9 |
Related Readings |
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Quadrilateral
A plane figure bounded by four line segments is called quadrilateral.
Properties:
- It has four sides.
- It has four vertices or comers.
- It has two diagonals.
- The sum of four interior angles is equal to 360°.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgVkVJW56q9qRXsYh2HEKosQlnxElzB_2Gnnlcok6Ehxe_SM83oKKAYmL5hgYaUxWlM7TqZuR3FWLl0wCqXe_BzmTU6zdkiRblnQv7i9CahFxYvIkEFBW3iARqGGHqOf3TNLI0cHJUh6kRVBgk07i00lmhDIzS58yjJ-xriLnd5RSivd3QAvUl7n3o8N1U/w141-h127/Ncert%20notes%20Class%209%20math%20Chapter%209%20Quadrilaterals%20img%201.jpg)
In quadrilateral ABCD, AB, BC, CD and DA are sides; AC and BD are diagonals and
∠ABC + ∠BCD + ∠CDA + ∠DAB = 360°.
Types of Quadrilaterals
1. Parallelogram
A quadrilateral whose each pair of opposite sides are parallel.
- AB || DC
- AD || BC
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiz7JdNRivjWGzps6Bdm-xDJQGmmuJRasNz4H1l2dMjGrA6BwdNJAXpt48MFyNJBrRXgeFWsW0bDkd21K2RC7KuxgsViZoiCJi6pdtyErgWY27e7o-N-bN7m2mXHtPEy6B4qnw24VrFMkVTll9MwsBb763yOPtEivo8sdAPSrfdOo8qAxmoMeSTbQc-ylM/w184-h108/Ncert%20notes%20Class%209%20math%20Chapter%209%20Quadrilaterals%20img%202.jpg)
2. Rectangle
A parallelogram whose one angle is 90°. Diagonals are equal.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi2wyrFySBAslN59WS47U0rM1BcBdR45hougfwrRtoopeC_8KUN-wXfCrVw92-45y0KLj9eStn-3XRmlAccFe4oQf6dyGCTJGQEVKSUESuIk0Fmzn2PGfbRKBuLmHzpxfjUDnID0qIohlUd1js-YuhQiuYRGKOFHqvTmPYDivbTgYfUQSGUofdCU1qolyQ/w204-h146/Ncert%20notes%20Class%209%20math%20Chapter%209%20Quadrilaterals%20img%203.jpg)
3. Rhombus
A parallelogram whose adjacent sides are equal.
Note: Diagonal bisect each other at 90°.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi9gEL2CzH-S1KeQIPPBt8bQyO6gKcIdwL0kCkFS5DSqhAhj4nGtHF0RI03hxrCgj3DmT-spUhMTFnxiC7NsRf9iib9KNNsXvXxpS7U0ARw_5zlZW2yWsHvQsxTZQtUz81OICKy09PXNwv5rOuv0NjdQX1X8V81vhB71JwT0APyvhBksNuSmoCP6dKkQe4/w169-h136/Ncert%20notes%20Class%209%20math%20Chapter%209%20Quadrilaterals%20img%204.jpg)
4. Square
A rectangle whose adjacent sides are equal (four sides are equal). Diagonal bisect each other at 90°.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEimShC8s4f1W6Ikjbr0TzGw6AIliV7hoUhkWqf0vtsq73vqtXB1uFLnZxA1dl2p8OjOTAx2IsYkTKhWpDeHVZ6qlpKSN85037P35ui100Z-VAlqlJXXge4MsauoIMxHaN8SybY0dfITDVNapWs7Yle2rs5epRKhxkUOD7xyGhgukSCCUqnuf0HQG5DcSdw/w162-h169/Ncert%20notes%20Class%209%20math%20Chapter%209%20Quadrilaterals%20img%205.jpg)
5. Trapezium
A quadrilateral whose one pair of opposite sides are parallel. AB || DC
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh5V9rBRPTDLSbux_JeLTi46qNoBBTR8IweZ8oEsonsgdNb00BHmSWjL4PVAcwLg7tOnl8YwGropBSf3WdCqq7bUc2lCULH6vJuRs7j620W7GAGOtAP70xlNanoOkeREhi2_UFwG8a7KPivLxkxpkBrI2Pfvg7ZyGEqHXkAe52MxpSO10f_3lQoD8m_8kA/w159-h150/Ncert%20notes%20Class%209%20math%20Chapter%209%20Quadrilaterals%20img%206.jpg)
6. Kite
It has two pair of adjacent sides that are equal in length but opposite sides are unequal.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgpc7gKyDrCnRVDyH0RNcmdLvv1PZ-HVgOyUhVlY8oCGujCKIsdInDNA1O3hO8sx7g8gRNdF9QEasi6Ov17KuyX8zFij9LuDpISjXF41H9-NYeDk5fGTVExEdnWWT4-_bm8RUrs1vO03QSEGJIkKBHSZ2gEEqacqxCI21lBqhLe1YU3RXLEKix87tUukiI/w173-h201/Ncert%20notes%20Class%209%20math%20Chapter%209%20Quadrilaterals%20img%207.jpg)
Note:
- One of the diagonal bisects the other at right angle.
- One pair of opposite angles are equal.
Properties of a Parallelogram
- Opposite sides are equal.
g., AB = DC and AD = BC - Consecutive angles are supplementary.
g., ∠A + ∠D = 180° - Diagonals of parallelogram bisect each other.
- Diagonal divide it into two congruent triangles. A B
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg_-FvNpM_kSzu42ohrrAVq3I8OwG3G3Cs2aVjW_WAiV6NXQeCbt0zll8rxqcQbTOj-uey-UkXXMJRziDC2IHV5gcxWXsw6Go2MzI-7fx6t6rUenpMpKccBpQHsN1hX5dpE5WqtupnyMNq57DuVwBpByj76fnYVn1D5SPiYYn-BzDO23ISdqMGPEk7F_Dg/w199-h161/Ncert%20notes%20Class%209%20math%20Chapter%209%20Quadrilaterals%20img%208.jpg)
- Theorem 8.1: A diagonal of a parallelogram divides it into two congruent triangles.
- Theorem 8,2: In a parallelogram, opposite sides are equal.
- Theorem 8.3: If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
- Theorem 8.4: In a parallelogram, opposite angles are equal.
- Theorem 8.5: If in a quadrilateral, each pair of opposite angles of a quadrilateral is equal then it is a parallelogram.
- Theorem 8.6: The diagonals of a parallelogram bisect each other.
- Theorem 8.7: If the diagonals of quadrilateral bisect each other, then it is a parallelogram.
- Theorem 8.8: A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.
Mid-point Theorem
Theorem 8.9: The line segment joining the mid-points of two sides of a triangle is parallel to the third.
Given: A triangle ABC, E and F are mid-points of sides AB and AC respectively.
i.e., AE = EB and AF = FC
To Prove:
(i) EF || BC
(ii) EF = 12 BC
Construction: Draw a line through C parallel to AB and extend EF which intersect at D.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjBFkkDsbQ34lteFy6l4HtC71mcukr2R0jG3qLFvHPVUAf_8oyoMbgsN46ZcGW9b2uU4hZVdx8Bp3-dICNurwiRdNT-dhNHdpfBbE8Ygf7NWDKZ3sVjKKpqgSlbBfqmEqHkVqda18uyxUbXeCYZBqjZBueECCSWiPXcLebOicsShIZldme3cGJBJrooOG4/w236-h202/Ncert%20notes%20Class%209%20math%20Chapter%209%20Quadrilaterals%20img%209.jpg)
Proof: (i) In AAEF and ACDF,
AF = CF (F is the mid-point of AC)
∠AFE = ∠CFD (Vertically opposite angles)
∠EAF = ∠DCF (Alternate interior angles)
∴ ΔAEF = ΔCDF (by ASA congruency)
∴ AE = CD (by CPCT)
and BE = CD (AE = BE)
EF = FD (by CPCT);
Hence, BCDE is a parallelogram.
ED || BC )
∴ EF || BC
(ii) BCDE is a parallelogram.
DE = BC
EF + FD = BC
2EF = BC
EF = 1/2 BC
Converse of Mid-Point Theorem
Theorem 8.10: The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side. ‘
Given: ΔABC in which E is the mid point of AB.
EF || BC
To Prove: AF = FC
Construction: Draw CD || AB and extend EF which intersect at D.
Proof: EF || BC
∴ ED || BC
AB || CD
⇒ BE || CD
∴ BCDE is a parallelogram.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJMhsp56_Nc9GMs7cpYh9jERX6HdGJ7YC3tBjDFU4NLJdTd46kjzp7rCybu1Lq2zfZXXtBsFDQ0NCJMMy_6hVS8yhg7guXGSBbYbWjFrsxrckMUCYeBDD8JH2bYDOgsjm70egbC6TeYjZuWJULyrIV1rY1vCjRaXjEoyMDM140jSntOSBR2jHciHJw2GM/w236-h174/Ncert%20notes%20Class%209%20math%20Chapter%209%20Quadrilaterals%20img%2010.jpg)
Now in ΔAEF and ΔCDF, ∠AFE = ∠CFD (Vertically opposite angles)
∠EAF = ∠DCF (Alternate interior angles)
AE = CD (BE = AE opposite side of a parallelogram and BE = CD
∴ AAEF ≅ ACDF (by AAS congruency)
Hence AF = FC (by CPCT)