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Class 9 RD SHARMA Solutions Maths Chapter 13 - Quadrilaterals

Solutions are a valuable educational resource as they help students easily navigate their academic journey confidently. They provide a deeper understanding of the subject during self-study and enhance the learning outcomes. Mathematics is considered one of the most essential subjects as it plays an important role in science, business, and technology fields.

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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Solution 3

Let the common ratio between the angles is x. So, the angles will be 3x, 5x, 9x and 13x respectively.
Since the sum of all interior angles of a quadrilateral is 360o.
3x + 5x + 9x + 13x = 360o

30x = 360o
x = 12o
Hence, the angles are
3x = 3  12 = 36o
5x = 5  12 = 60o
9x = 9  12 = 108o
13x = 13  12 = 156o

i. F

ii. T

iii. F

iv. F

v. T

vi. F

vii. F

viii. T

Solution 1

C and D are cosecutive interior angles on the same side of the transversal CD. Therefore,

C + D = 180o

Solution 3

Since, diagonals of a square bisect each other at right angle. Therefore, AOB = 90o

Solution 12

Let ABCD is a quadrilateral in which P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively.
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 BCD
QR || 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.
Since, diagonals of a parallelogram bisect each other.
Hence, PR and QS bisect each other.

Solution 13

(i) isosceles

(ii) right triangle

(iii) parallelogram

Solution 1

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,

which have only one point in common

i.e the joint point of their ends.

Hence, correct option is (b).

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 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 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).