Class 9 RD SHARMA Solutions Maths Chapter 13 - Quadrilaterals
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RD Sharma Solutions for CBSE Mathematics is a highly regarded resource. It provides comprehensive and detailed explanations for all the questions in the textbook. Breaking down larger concepts into manageable steps, these solutions not only provide better clarity but also improve the problem-solving skills of students. This comes in especially useful when solving complex or tricky questions.
One of the most important chapters covered in CBSE Class 9 Math is Quadrilaterals. In this resource, you can gain an in-depth understanding and step-by-step solutions to the problems related to this geometric shape.
Quadrilaterals are polygons that have four sides and four angles. The CBSE Class 9 Mathematics curriculum teaches students about the various types of quadrilaterals, such as parallelograms, rectangles, squares, rhombuses, trapeziums, and kites. Students learn the properties and characteristics of these quadrilaterals and solve related problems. With access to detailed explanations and examples in the RD Sharma solution, students gain a solid understanding of the unique features of each type of quadrilateral.
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Quadrilaterals Exercise Ex. 13.1
Solution 1
Solution 2
Solution 3
Since the sum of all interior angles of a quadrilateral is 360o.
3x + 5x + 9x + 13x = 360o30x = 360o
x = 12o
Hence, the angles are
3x = 3
12 = 36o5x = 5
12 = 60o9x = 9
12 = 108o13x = 13
12 = 156o Solution 4
Quadrilaterals Exercise Ex. 13.2
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
i. F
ii. T
iii. F
iv. F
v. T
vi. F
vii. F
viii. T
Solution 9
Solution 10
Quadrilaterals Exercise Ex. 13.3
Solution 1
C and D are cosecutive interior angles on the same side of the transversal CD. Therefore,
C + D = 180o
Solution 2
Solution 3
Since, diagonals of a square bisect each other at right angle. Therefore, 
AOB = 90o
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Quadrilaterals Exercise Ex. 13.4
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Join PQ, QR, RS, SP and BD.
In
ABD, S and P are mid points of AD and AB respectively. So, By using mid-point theorem, we can say that
SP || BD and SP =
BD ... (1) Similarly in
BCDQR || BD and QR =
BD ... (2) From equations (1) and (2), we have
SP || QR and SP = QR
As in quadrilateral SPQR one pair of opposite sides are equal and parallel to
each other.
So, SPQR is a parallelogram.
Hence, PR and QS bisect each other.
Solution 13
(i) isosceles
(ii) right triangle
(iii) parallelogram
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Quadrilaterals Exercise 13.70
Solution 1
ABCD is a Quadrilateral.
The opposite sides AB and DC, AD and BC have no common point.
Hence, correct option is (a).
Solution 2
Consecutive sides of a Quadrilateral ABCD are
AB and BC,
BC and CD,
CD and AD,
AD and AB,
which have only one point in common
i.e the joint point of their ends.
Hence, correct option is (b).
Quadrilaterals Exercise 13.71
Solution 3
Solution 4
For a rhombus, the angle between the diagonals is 90° and not 60°.
Hence, correct option is (d).
Solution 5
Diagonals necessarily bisect opposite angles in a square.
Hence, correct option is (d).
Solution 6
The two diagonals are equal in a rectangle (property).
Hence, correct option is (c).
Solution 7
Solution 8
Solution 9
AR, BR, CP, DP are the bisectors of angles of parallelogram.
Because two bisectors of adjacent angles make 90° between them So PQRS is a Rectangle
Because DP and BR are acute angle bisectors so the distance between them PQ < PS (The distance between other two bisectors)
So PQ ≠ PS (So PQRS is not a square, but only a rectangle)
Hence, correct option is (c).
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Quadrilaterals Exercise 13.73
Solution 31
Solution 32
Quadrilaterals Exercise 13.72
Solution 18
Solution 19
Solution 20
Sum of all angles of a Quadrilateral = 360°
4x + 7x + 9x + 10x = 360°
30x = 360°
x = 12°
So, sum of smallest and largest angle,
i.e. 4x + 10x = 14x = 14 × 12 = 168°
Hence, correct option is (c).
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 30
Solution 27
Solution 28
Solution 29
