Class 9 RD SHARMA Solutions Maths Chapter 13 - Quadrilaterals
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 = 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
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 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.
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